The pressure of SO₂Cl₂ after 60 seconds is given by [SO₂Cl₂]ₜ ≈ 300 Torr * [tex]e^{-12.6}[/tex], and the time at which the pressure declines to 1/2 its initial value is approximately 3.3 seconds.
(a) To determine the pressure of SO₂Cl₂ after 60 seconds, we need to calculate the concentration of SO₂Cl₂ at that time and then convert it to pressure using the ideal gas law.
Given:
Rate constant (k) =[tex]2.1 * 10^{-1} s^{-1}[/tex]
Initial pressure ([SO₂Cl₂]₀) = 300 Torr
Time (t) = 60 seconds
Using the rate equation, we can rearrange it to solve for [SO₂Cl₂]ₜ:
[SO₂Cl₂]ₜ = [SO₂Cl₂]₀ * [tex]e^{-kt}[/tex]
Plugging in the values:
[SO₂Cl₂]ₜ = 300 Torr * [tex]e^{-2.1 * 10^{-1} s^{-1} * 60 s}[/tex]
Calculating the exponential term:
[SO₂Cl₂]ₜ ≈ 300 Torr *[tex]e^{-12.6}[/tex]
(b) To find the time at which the pressure of SO₂Cl₂ declines to 1/2 its initial value, we need to solve for t in the rate equation when [SO₂Cl₂]ₜ = [SO₂Cl₂]₀ / 2.
Using the rate equation:
[SO₂Cl₂]ₜ = [SO₂Cl₂]₀ *[tex]e^{-kt}[/tex]
[SO₂Cl₂]₀ / 2 = [SO₂Cl₂]₀ * [tex]e^{-kt}[/tex]
1/2 = [tex]e^{-kt}[/tex]
Taking the natural logarithm of both sides:
ln(1/2) = -kt
Solving for t:
t = -ln(1/2) / k
t ≈ 0.693 / k
Plugging in the value of k:
t ≈ 0.693 / (2.1 × 10^(-1) s^(-1))
Simplifying:
t ≈ 3.3 seconds
Therefore, the pressure of SO₂Cl₂ after 60 seconds is given by [SO₂Cl₂]ₜ ≈ 300 Torr * [tex]e^{-12.6}[/tex], and the time at which the pressure declines to 1/2 its initial value is approximately 3.3 seconds.
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5. A compound that contains \( \mathrm{C}, \mathrm{H} \), and \( \mathrm{N} \) and has a molecular ion with an \( \mathrm{m} / \mathrm{z} \) value of 45 . Which of the following is the molecular formu
The molecular formula of a compound that contains C,H, and N and has a molecular ion with an m/z value of 45 is CH₅N₂. The correct answer is C.
The molecular ion with an m/z value of 45 suggests that the compound contains a molecular weight of 45 amu. We need to find a molecular formula that satisfies this molecular weight.
Let's calculate the molecular weights of the given options:
a) C₂H₇O: (2 * 12.01) + (7 * 1.01) + 16.00 = 45.08 amu
b) C₂H₇F: (2 * 12.01) + (7 * 1.01) + 19.00 = 47.08 amu
c) CH₅N₂: 12.01 + (5 * 1.01) + (2 * 14.01) = 45.07 amu
d) C₂H₇N: (2 * 12.01) + (7 * 1.01) + 14.01 = 43.08 amu
Among the given options, the molecular formula that corresponds to a molecular weight of 45 amu is option c) CH₅N₂.
Therefore, the correct answer is c) CH₅N₂.
Complete Question:
A compound that contains C,H, and N and has a molecular ion with an m/z value of 45 . Which of the following is the molecular formula? a) C2H7O b) C2H7F c) CH5N2 d) C2H7N
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answer each part please
One possible mechanism for the gas phase reaction of hydrogen with nitrogen monoxide is: step 1 slow: \( \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\ma
The gas phase reaction between hydrogen (H₂) and nitrogen monoxide (NO) proceeds through a slow step where one molecule of H₂ reacts with two molecules of NO to form nitrogen oxide (N₂O).
The reaction between hydrogen and nitrogen monoxide in the gas phase follows the following mechanism:
Step 1 (Slow):
The slow step involves the reaction between one molecule of hydrogen gas (H₂) and two molecules of nitrogen monoxide (NO). This reaction results in the formation of nitrogen oxide (N₂O). The balanced equation for this step can be represented as:
H₂(g) + 2NO(g) → N₂O(g)
It's important to note that this is just one possible mechanism for the reaction between hydrogen and nitrogen monoxide. The actual reaction mechanism may involve multiple steps, and additional reactions or intermediates may be present. The given equation represents the rate-determining step, which is the slowest step in the overall reaction.
Overall, the gas phase reaction of hydrogen with nitrogen monoxide proceeds through a slow step in which one molecule of H₂ reacts with two molecules of NO to produce nitrogen oxide (N₂O).
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Which of the following aqueous solutions are good buffer systems? (Select all that apply.)
1.
0.13 M hypochlorous acid + 0.19 M potassium hypochlorite
0.30 M hydrobromic acid + 0.19 M sodium bromide
0.28 M ammonia + 0.32 M ammonium bromide
0.16 M potassium hydroxide + 0.22 M potassium chloride
0.35 M sodium iodide + 0.24 M sodium nitrate
2.
0.21 M hypochlorous acid + 0.14 M sodium hypochlorite
0.29 M perchloric acid + 0.20 M potassium perchlorate
0.28 M ammonium nitrate + 0.33 M ammonia
0.18 M potassium hydroxide + 0.24 M potassium bromide
0.40 M hydrocyanic acid + 0.30 M potassium cyanide
3.
0.21 M hydrofluoric acid + 0.17 M sodium fluoride
0.21 M hydrochloric acid + 0.18 M sodium chloride
0.28 M ammonia + 0.39 M barium hydroxide
0.12 M potassium hydroxide + 0.29 M potassium bromide
0.35 M sodium iodide + 0.23 M barium iodide
Only the solutions in the first and third sets are good buffer systems. A buffer system is a solution that resists changes in pH when small amounts of acid or base are added. Buffer systems are made up of a weak acid and its conjugate base, or a weak base and its conjugate acid.
In the first set, the solutions contain a weak acid and its conjugate base. For example, the solution containing 0.13 M hypochlorous acid and 0.19 M potassium hypochlorite is a buffer system because hypochlorous acid is a weak acid and potassium hypochlorite is the conjugate base of hypochlorous acid.
In the third set, the solutions contain a weak base and its conjugate acid. For example, the solution containing 0.21 M hydrofluoric acid and 0.17 M sodium fluoride is a buffer system because hydrofluoric acid is a weak acid and sodium fluoride is the conjugate base of hydrofluoric acid.
The solutions in the second set are not good buffer systems because they contain a strong acid or base. For example, the solution containing 0.29 M perchloric acid and 0.20 M potassium perchlorate is not a buffer system because perchloric acid is a strong acid.
The solutions in the fourth and fifth sets are not good buffer systems because they do not contain a weak acid or base. For example, the solution containing 0.12 M potassium hydroxide and 0.29 M potassium bromide is not a buffer system because potassium hydroxide is a strong base.
Therefore, 1 and 3 are correct answers.
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An ionic compound can only dissolve in water if its heat of
solution in water is exothermic. true or false
An ionic compound can only dissolve in water if its heat ofsolution in water is exothermic and the statement is False.
An ionic compound can dissolve in water regardless of whether its heat of solution is exothermic or endothermic. The solubility of an ionic compound in water is determined by the balance between the energy required to break the ionic bonds in the solid and the energy released when the ions interact with water molecules.
If the overall process of dissolving is energetically favorable, the compound will dissolve, regardless of whether it is exothermic or endothermic.
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Calculate the standard free energy change (AG°) for each of the following reactions (after first calculating the cell potential). (a) 3 Pb(s) + 2 Au³+ (aq). → 3 Pb²+ (aq) + 2 Au(s) (b) 2 Cr(s) + 3 Cd2+ (aq) → 2 Cr³+ (aq) + 3 Cd(s) Which reaction has the larger standard free energy change? reaction (a) O reaction (b)
The standard free energy change (ΔG°) for reaction (a) is larger than that for reaction (b).
To calculate the standard free energy change (ΔG°) for each reaction, we need to first calculate the cell potential (E°cell) using standard reduction potentials. The formula for ΔG° in terms of E°cell is ΔG° = -nF E°cell, where n is the number of moles of electrons transferred and F is the Faraday constant.
(a) 3 Pb(s) + 2 Au³+(aq) → 3 Pb²+(aq) + 2 Au(s)
The half-reaction for the reduction of Au³+ is:
Au³+(aq) + 3e⁻ → Au(s) E° = +1.498 V
The half-reaction for the reduction of Pb²+ is:
Pb²+(aq) + 2e⁻ → Pb(s) E° = -0.126 V
The overall cell potential is the difference between the reduction potentials of the two half-reactions:
E°cell = E°cathode - E°anode
E°cell = 1.498 V - (-0.126 V)
E°cell = 1.624 V
The number of moles of electrons transferred is 3 in both half-reactions, so n = 3.
Using the equation ΔG° = -nF E°cell, we can calculate the standard free energy change for reaction (a).
(b) 2 Cr(s) + 3 Cd²+(aq) → 2 Cr³+(aq) + 3 Cd(s)
The half-reaction for the reduction of Cd²+ is:
Cd²+(aq) + 2e⁻ → Cd(s) E° = -0.403 V
The half-reaction for the oxidation of Cr(s) is:
Cr(s) → Cr³+(aq) + 3e⁻ E° = -0.744 V
The overall cell potential is the difference between the reduction potentials of the two half-reactions:
E°cell = E°cathode - E°anode
E°cell = -0.403 V - (-0.744 V)
E°cell = 0.341 V
The number of moles of electrons transferred is 6 in both half-reactions, so n = 6.
Using the equation ΔG° = -nF E°cell, we can calculate the standard free energy change for reaction (b).
Comparing the calculated ΔG° values, we find that the standard free energy change for reaction (a) is larger than that for reaction (b).
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Pure liquid water consists of H 2
O molecules________________: 1) held in a rigid three-dimentional network. 2) with local preference for linear geometry. 3) with large numbers of strained or broken hydrogen bonds. 4) which do not switch H-bonds readily. 5) all are true.
Pure liquid water consists of H₂O molecules option 5 all are true
Pure liquid water consists of H₂O molecules, which are held in a rigid three-dimensional network. This network is formed through hydrogen bonding between water molecules. Each water molecule can form up to four hydrogen bonds, resulting in the formation of a network with a three-dimensional structure.
The local preference for linear geometry refers to the tendency of water molecules to align in a linear fashion when forming hydrogen bonds. This arrangement allows for efficient hydrogen bonding between neighboring water molecules.
Water molecules in the liquid state have a significant number of strained or broken hydrogen bonds due to the constant motion and interactions among the molecules. However, new hydrogen bonds can form and broken bonds can be repaired due to the dynamic nature of water.
Therefore, all of the given statements are true for pure liquid water which is option 5.
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A patient was prescribed 20.0mg of lithium every 12 hours to treat their medical condition. If the lithium comes in the form of lithium carbonate, which is 17.3% lithium by mass, what mass in grams of lithium carbonate must the patient consume per day? (hint: write the percentage as "parts per hundred" before starting problem.)
To obtain a daily dose of 20.0 mg of lithium, the patient needs to consume approximately 231.21 mg of lithium carbonate per day, considering that lithium carbonate is 17.3% lithium by mass.
To solve this problem, we can use the concept of the percentage composition of lithium carbonate.
Prescribed dose of lithium = 20.0 mg every 12 hours
Percentage of lithium in lithium carbonate = 17.3%
Let's calculate the mass of lithium carbonate the patient must consume per day:
First, let's convert the prescribed dose to grams per day:
Prescribed dose = 20.0 mg * 2 doses per day = 40.0 mg/day
Now, we need to determine the mass of lithium carbonate that contains this amount of lithium.
Let's assume the mass of lithium carbonate to be x grams.
The mass of lithium in x grams of lithium carbonate is given by:
Mass of lithium = x grams * (17.3 parts lithium/100 parts lithium carbonate)
Since the percentage is expressed as "parts per hundred," we can set up the following equation:
Mass of lithium = x * (17.3/100)
We want the mass of lithium to be equal to the prescribed dose, which is 40.0 mg.
Therefore, we can set up the equation:
40.0 mg = x * (17.3/100)
To solve for x, we can rearrange the equation:
x = (40.0 mg) / (17.3/100)
Now, let's calculate the value of x:
x = (40.0 mg) / (17.3/100) = 40.0 * (100/17.3) mg
x ≈ 231.21 mg
Therefore, the patient must consume approximately 231.21 mg of lithium carbonate per day to obtain a dose of 20.0 mg of lithium.
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Write structural formulas for:
1.) Ester with the formula C3H6O2 whose alkoxyl group (-OR) is a methyl.
2.) Ester with the formula C3H6O2 whose alkyl group (-OR) is an ethyl.
The RCOO is formed from the carboxylic acid of 3 carbon atom i.e., prop- and R' is formed from the alcohol of ethyl alcohol - C2H5O.
The structural formulas for the following esters can be given as follows:1. Ester with the formula C3H6O2 whose alkoxyl group (-OR) is a methyl.The structural formula of ester is represented as RCOOR'. Here R represents the alkyl group of carboxylic acid and R' represents the alkyl group of alcohol.So, for the given ester C3H6O2:Here RCOO is formed from the carboxylic acid of 3 carbon atom i.e., prop- and R' is formed from the alcohol of methanol - CH3OH
Hence, the structural formula of the ester C3H6O2 with alkoxyl group (-OR) methyl is given as follows:2. Ester with the formula C3H6O2 whose alkyl group (-OR) is an ethyl.The structural formula of ester is represented as RCOOR'. Here R represents the alkyl group of carboxylic acid and R' represents the alkyl group of alcohol.So, for the given ester C3H6O2:Here RCOO is formed from the carboxylic acid of 3 carbon atom i.e., prop- and R' is formed from the alcohol of ethyl alcohol - C2H5O.
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Threonine has pka1 = 2.09 and pka2 = 9.1. Use Henderson-hasselbalch eq to calc the ratio of protonated and neutral forms at ph = 1.5. calculate the ratio of neutral and deprotonated forms at ph = 10.00.
At pH 1.5, the ratio of protonated to neutral forms of threonine is approximately 0.448 and at pH 10.00, the ratio of neutral to deprotonated forms of threonine is approximately 7.943.
The Henderson-Hasselbalch equation is used to calculate the ratio of protonated and neutral forms of a compound at a given pH. For threonine, with [tex]pKa_1[/tex] = 2.09 and [tex]pKa_2[/tex] = 9.1, we can calculate the ratios at pH 1.5 and pH 10.00.
At pH 1.5 (acidic conditions), the pH is lower than both pKa values. In this case, the amino acid will predominantly exist in the protonated form. The ratio of protonated (H₃N⁺-CH(CH₃)OH-COO⁻) to neutral (H₃N-CH(CH₃)OH-COOH) forms can be calculated using the Henderson-Hasselbalch equation:
ratio = [tex]10^{(pH - pKa)}[/tex]
ratio = [tex]10^{(1.5 - 2.09)}[/tex]= 0.448
Therefore, at pH 1.5, the ratio of protonated to neutral forms of threonine is approximately 0.448.
At pH 10.00 (alkaline conditions), the pH is higher than both pKa values. In this case, the amino acid will predominantly exist in the deprotonated form. To calculate the ratio of neutral (H₃N-CH(CH₃)OH-COOH) to deprotonated (H₃N-CH(CH₃)OH-COO⁻) forms, we use the Henderson-Hasselbalch equation again:
ratio = [tex]10^{(pH - pKa)}[/tex]
ratio = [tex]10^{(10.00 - 9.1)}[/tex] = 7.943
Therefore, at pH 10.00, the ratio of neutral to deprotonated forms of threonine is approximately 7.943.
In conclusion, at pH 1.5, the ratio of protonated to neutral forms of threonine is approximately 0.448 and at pH 10.00, the ratio of neutral to deprotonated forms of threonine is approximately 7.943.
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Based on how bonds were defined in Ch. 4, which combination(s) of elements would result in the formation of a polar covalent bond? You may select more than one option. - Cu and C - Si and F - C and H - C and Br - S and Br
The combination(s) of elements that would result in the formation of a polar covalent bond are: Si and F, C and H, and S and Br.
A polar covalent bond is formed when two atoms with different electronegativities share electrons unequally, resulting in a partial positive and partial negative charge on the bonded atoms.
In the given combinations:
- Si and F: Fluorine (F) is highly electronegative, while silicon (Si) has a lower electronegativity. Their difference in electronegativity leads to the formation of a polar covalent bond.
- C and H: Carbon (C) and hydrogen (H) have different electronegativities. Although the difference is not as significant as in other combinations, a polar covalent bond can still be formed.
- S and Br: Bromine (Br) is more electronegative than sulfur (S), resulting in the formation of a polar covalent bond between them.
On the other hand:
- Cu and C: Copper (Cu) and carbon (C) have similar electronegativities, so they would form a nonpolar covalent bond.
- C and Br: Carbon (C) and bromine (Br) also have similar electronegativities, leading to the formation of a nonpolar covalent bond.
Therefore, Si and F, C and H, and S and Br combinations would result in the formation of polar covalent bonds due to the electronegativity differences between the atoms involved.
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Which of the following is a steroid based hormone? a) testosterone b) epinepherine c) vasopressin d) glucagon e) thyroxine
The steroid-based hormone among the options provided is testosterone (a).
Testosterone is a steroid hormone produced primarily in the testes in males and in smaller amounts in the ovaries and adrenal glands in females. It plays a crucial role in the development of male reproductive tissues and secondary sexual characteristics.
Epinephrine (b), also known as adrenaline, is a catecholamine hormone and a neurotransmitter produced in the adrenal glands. It is involved in the "fight or flight" response, increasing heart rate and blood flow to muscles.
Vasopressin (c), also called antidiuretic hormone (ADH), is a peptide hormone produced in the hypothalamus and released from the posterior pituitary gland. It regulates water reabsorption in the kidneys and helps maintain fluid balance.
Glucagon (d) is a peptide hormone produced by the alpha cells of the pancreas. It raises blood glucose levels by promoting the breakdown of glycogen in the liver.
Thyroxine (e), also known as T4, is a thyroid hormone produced by the thyroid gland. It plays a vital role in regulating metabolism and growth.
Among the given options, only testosterone (a) is a steroid hormone, while the others belong to different hormone classes.
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A student is given a sample of red cobalt sulfate hydrate. She weighed the sample in a dry covered crucible and obtained a mass of 26.148 g for the crucible, cover, and sample. Before adding the sample, the crucible and cover weighed 24.322 g. She then heated the crucible to drive off the water of hydration, keeping the crucible at red heat for about 10 minutes with the cover slightly ajar. She then let the crucible cool, and found it had a lower mass; the crucible, cover and contents then weighed 25.329 g. In the process the sample was converted to blue anhydrous CoSO 4
. Show all calculations necessary to answer the following questions. 1. What was the mass of the hydrate sample? g hydrate 26.148−24.322 2. What is the mass of the anhydrous CoSO 4
? gCoSO 4
3. How much water was driven off? gH 2
O 4. What is the percentage of water in the hydrate? % water = mass of hydrate sample mass of water in sample
×100 %H 2
O 5. How many grams of water would there be in 100.0 g of hydrate? How many moles? gH 2
O moles H 2
O 6. How many grams of CoSO 4
are there in 100.0 g of hydrate? How many moles? What percentage of the hydrate is CoSO 4
? Convert the mass of CoSO 4
to moles. The molar mass of CoSO 4
is 154.996 g. gCoSO 4
moles CoSO 4
%CoSO 4
in hydrate 7. How many moles of water are present per mole of CoSO 4
? moles H 2
O/ moles CoSO 4
8. What is the formula of the hydrate?
[tex]\frac{Mass of CoSO4 in 100.0 g of hydrate}{Molar mass of CoSO4}[/tex]The given question is based on gravimetric estimation which is done to estimate the amount of water in the given sample.
For the gravimetric estimation of cobalt sulphate hydrate, the calculations can be done as:
1. Mass of the hydrate sample = (Mass of crucible, cover, and sample) - (Mass of crucible and cover)
= 26.148 g - 24.322 g = 1.826 g
2. Mass of the anhydrous CoSO₄ = (Mass of the crucible, cover, and anhydrous CoSO₄) - (Mass of the crucible and cover)
= 25.329 g - 24.322 g = 1.007 g
3. Mass of water driven off = Mass of the hydrate sample - Mass of the anhydrous CoSO₄
= 1.826 g - 1.007 g = 0.819 g
4. Percentage of water in the hydrate = [tex]\frac{Mass of water driven off}{Mass of hydrate sample}[/tex] × 100
= [tex]\frac{0.819}{1.826}[/tex] × 100 = 44.90%
5. Mass of water in 100.0 g of hydrate = [tex]\frac{Mass of water driven off}{Mass of hydrate sample}[/tex] × 100
= [tex]\frac{0.819}{1.826}[/tex] × 100 = 44.90 g
Moles of water in 100.0 g of hydrate = [tex]\frac{Mass of water in 100.0 g of hydrate}{Molar mass of water}[/tex]
= [tex]\frac{44.9}{18}[/tex] = 2.49 mol
6. Mass of CoSO₄ in 100.0 g of hydrate = [tex]\frac{Mass of CoSO4}{Mass of hydrate sample}[/tex] × 100
= [tex]\frac{1.007}{1.826}[/tex] × 100 = 55.15 g
Moles of CoSO₄ in 100.0 g of hydrate = [tex]\frac{Mass of CoSO4 in 100.0 g of hydrate}{Molar mass of CoSO4}[/tex]
= [tex]\frac{55.15}{154.996}[/tex] = 0.356 mol
Percentage of CoSO₄ in 100.0 g of hydrate = [tex]\frac{Mass of CoSO4 in hydrate}{Mass of hydrate sample}[/tex]
= [tex]\frac{1.007}{1.826}[/tex] × 100 = 55.15%
7. Moles of water per mole of CoSO₄ = [tex]\frac{Moles of water in 100g sample}{Moles of CoSO4 in 100g sample}[/tex]
= [tex]\frac{2.49}{0.356}[/tex] = 6.99 ≈ 7 moles H₂O/ mole CoSO₄
8. From the above answer, it is derived that for every mole of CoSO₄, there are 7 moles of water.
Thus, the empirical formula for the hydrate will be CoSO₄.7H₂O
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Solutions of sodium sulfate, zinc sulfate, and iron(II) sulfate are mixed. Write a balanced net ionic equation for the predicted redox reaction. Identify if the reaction is spontancous or non-spontaneous
Equation ZnSO4 + FeSO4 → Zn(s) + Fe2(SO4)3 The reaction is spontaneous because the oxidizing ability of Fe3+ is more significant than the oxidizing ability of Zn2+. The redox reaction is spontaneous and moves forward without any external intervention.
In the given question, the solutions of sodium sulfate, zinc sulfate, and iron(II) sulfate are mixed. The balanced net ionic equation for the predicted redox reaction and identification if the reaction is spontaneous or non-spontaneous are discussed below. Balanced Net Ionic Equation The ions that take part in the reaction are Na+, Zn2+, Fe2+ , SO42-.As Na+ and SO42- ions are present as reactants and products on both sides of the equation, they do not take part in the reaction. Hence, they are known as spectator ions.Zn2+ ions get oxidized to Zn(s), and Fe2+ ions get reduced to Fe(s).Zn2+(aq) + Fe2+(aq) → Zn(s) + Fe3+(aq)The oxidation state of Zn changes from +2 to 0, while that of Fe changes from +2 to +3.
Therefore, Zn is being oxidized, and Fe is being reduced. This redox reaction's balanced net ionic equation can be represented as follows:Zn2+(aq) + Fe2+(aq) → Zn(s) + Fe3+(aq) Overall Equation ZnSO4 + FeSO4 → Zn(s) + Fe2(SO4)3The reaction is spontaneous because the oxidizing ability of Fe3+ is more significant than the oxidizing ability of Zn2+. Hence, the redox reaction is spontaneous and moves forward without any external intervention. Furthermore, the Gibbs free energy change of the reaction can be negative.
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1.) A technician needs 1.55 moles of nitrogen to fill a 58.5 L container in an imaging apparatus. How many moles of nitrogen will she need to fill a 700.0 L container?
2.) The volume of a balloon is 2.85 L at 1.00 atm. What pressure is required to compress the balloon to a volume of 1.70 L?
3.)3.) A balloon has a volume of 43.0 L at 25oC. What is its volume at -8oC?
In these calculations, we used fundamental gas laws to determine various quantities. Therefore,
1) The technician will need approximately 18.64 moles of nitrogen to fill a 700.0 L container.
2) A pressure of approximately 1.68 atm is required to compress the balloon to a volume of 1.70 L.
3) The volume of the balloon at -8°C is approximately 38.19 L.
1) The volume of the container is directly proportional to the number of moles of nitrogen. Therefore, we can use the ratio of volumes to determine the number of moles of nitrogen needed.
Using the volume ratio, we can set up the following proportion:
V₁ / M₁ = V₂ / M₂
Solving for M₂:
M₂ = (M₁ * V₂) / V₁
M₂ = (1.55 mol * 700.0 L) / 58.5 L
M₂ ≈ 18.64 mol
Therefore, the technician will need approximately 18.64 moles of nitrogen to fill a 700.0 L container.
2) According to Boyle's law, the pressure and volume of a gas are inversely proportional, assuming constant temperature. We can use this relationship to solve the problem.
Using Boyle's law equation:
P₁ * V₁ = P₂ * V₂
Solving for P₂:
P₂ = (P1 * V₁) / V₂
P₂ = (1.00 atm * 2.85 L) / 1.70 L
P₂ ≈ 1.68 atm
Therefore, a pressure of approximately 1.68 atm is required to compress the balloon to a volume of 1.70 L.
3) Charles's law states that the volume of a gas is directly proportional to its temperature, assuming constant pressure. We can use this relationship to solve the problem.
Using the volume ratio, we can set up the following proportion:
V₁ / T₁ = V₂ / T₂
Solving for Volume₂:
V₂ = (V₁ * T₂) / T₁
V₂ = (43.0 L * 265 K) / 298 K
V₂ ≈ 38.19 L
Therefore, the volume of the balloon at -8°C is approximately 38.19 L.
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a buffer solution is prepared by combing some concentration of a ----------- and its ---------. give three examples.
A buffer solution is formed by combining a weak acid or base with its conjugate salt. This combination helps the solution resist changes in pH when small amounts of acid or base are added. Three examples of buffer solutions include acetic acid/sodium acetate, ammonia/ammonium chloride, and carbonic acid/sodium bicarbonate.
A buffer solution consists of a weak acid or base and its conjugate salt. When a weak acid is combined with its conjugate base or a weak base is combined with its conjugate acid, it creates a buffer system. The weak acid or base can donate or accept protons, while the conjugate salt helps maintain the pH by neutralizing any added acid or base.
Three examples of buffer solutions are:
1. Acetic acid/sodium acetate: Acetic acid (CH3COOH) is a weak acid, and sodium acetate (CH3COONa) is its conjugate salt.
2. Ammonia/ammonium chloride: Ammonia (NH3) is a weak base, and ammonium chloride (NH4Cl) is its conjugate salt.
3. Carbonic acid/sodium bicarbonate: Carbonic acid (H2CO3) is a weak acid, and sodium bicarbonate (NaHCO3) is its conjugate salt.
These buffer solutions help maintain a relatively constant pH when small amounts of acid or base are added to the solution.
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SARS-CoV-2 is the pathogen that cuases COVID-19. SARS-CoV-2 is a membraned single-stranded RNA virus. The viral RNA is linear, about 30,000 bases long, wrapped around by the viral nucleocapsid (N) protein. Like other coronaviruses, the composition of the SARS-CoV-2 viral RNA is biased against guanine (19.6\%) and cytosine (18.3\%). The serine residues in the N protein are heavily phosphorylated at multiple sites. The N protein is essential for viral RNA replication and packaging into virion particles. Which of the following statments about the virus are true based on the information provided? A. The viral RNA will contain four different nucleobases: adenine, guanine, cytosine, and thymine. B. The viral RNA is stable because it does not contain 2'-hydroxyl group in the ribose moiety. C. The viral RNA, when purified and removed of the N protein, will have a strong UV absorbance at 280 nm. D. The complex between the N protein and the viral RNA is a tertiary structure. E. The viral N protein must be rich in lysine and arginine. F. The single stranded viral RNA will contain local secondary structure regions with nucleobases paired through hydrogen bonds.
Based on the information provided, the correct statements are statement D (The complex between the N protein and the viral RNA is a tertiary structure.) and statement F (The single-stranded viral RNA will contain local secondary structure regions with nucleases paired through hydrogen bonds.)
A. The statement is false. The viral RNA of SARS-CoV-2 is a single-stranded RNA, not a double-stranded DNA. Therefore, it does not contain thymine. Instead, it contains uracil, which is the RNA equivalent of thymine.
B. The statement is false. The stability of the viral RNA is not solely determined by the presence or absence of the 2'-hydroxyl group. Other factors such as secondary and tertiary structures, as well as interactions with proteins, contribute to its stability.
C. The statement is false. The presence of the N protein does not necessarily result in a strong UV absorbance at 280 nm. UV absorbance at 280 nm is primarily associated with the presence of aromatic amino acids such as tryptophan and tyrosine.
D. The statement is true. The complex between the N protein and the viral RNA can be considered a tertiary structure. The N protein binds to the viral RNA, facilitating its replication and packaging into virion particles.
E. The statement is false. While the N protein may have phosphorylated serine residues, it does not necessarily mean that it must be rich in lysine and arginine. The amino acid composition of the N protein can vary and is not solely determined by phosphorylation.
F. The statement is true. Single-stranded RNA molecules can form local secondary structure regions through the pairing of nucleobases via hydrogen bonds. These secondary structures can play important roles in the function and regulation of the viral RNA.
Hence, the true statements based on the provided information are D and F.
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which of the following statements is true? excited electrons do not return to ground state until they move away from the heat of the flame. only one electron can be excited at a time. an electron may fall back to ground state in a single step or in multiple steps. each element emits a single, characteristic wavelength of light during the flame test.
The statement that is true among the options provided is: "An electron may fall back to the ground state in a single step or in multiple steps."
Higher energy levels are attained by excited electrons in atoms. The extra energy is finally released as light when they reach their ground state, though. The electron may return to its ground state in a single step, emitting a photon with a certain wavelength, or it may do so in several phases, releasing photons with various wavelengths.
The following additional options' statements are untrue:
Without having to leave the heat of the flame, excited electrons can return to the ground state. The flame's heat is what initially excite the electrons, yet the atom can still return to its ground state while it is still in the flame.
The claim that only one electron can be excited at a time is untrue since several electrons can be excited simultaneously.
In a flame test, each element emits a variety of recognizable light wavelengths. The energy differences between the excited states and the ground state of various electrons in the atom are reflected in the specific wavelengths that are emitted.
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How can you show using Pauli's exclusion principle that p sub shell can have only 6 electrons?
Calculate the percent ionization of formic acid (HCO2H) in a solution that is 0.190 M in formic acid and 0.310 M in sodium formate (NaHCO2). The Ka of formic acid is 1.77 x 10-4. Calculate the percent ionization of formic acid (HCO2H) in a solution that is 0.190 M in formic acid and 0.310 M in sodium formate (NaHCO2). The Ka of formic acid is 1.77x10-4.
20.8
0.0580
3.958
1.06 x 10-3
62.0
The percent ionization of formic acid in the given solution is 81.58%. This means that 81.58% of the formic acid molecules have dissociated into H+ ions and HCO2- ions.
The percent ionization of formic acid can be calculated using the formula:
Percent ionization = (concentration of H+ ions / initial concentration of formic acid) x 100%
Given that the initial concentration of formic acid is 0.190 M, we can calculate the concentration of H+ ions using the equilibrium expression for formic acid:
Ka = [H+][HCO2-] / [HCO2H]
Since sodium formate is a salt of formic acid, it will dissociate in water to release HCO2- ions. Therefore, the concentration of H+ ions can be calculated by subtracting the concentration of HCO2- ions from the initial concentration of sodium formate:
[H+] = [NaHCO2] - [HCO2-]
Given that the concentration of sodium formate is 0.310 M, we need to find the concentration of HCO2- ions. HCO2- ions can be considered as the conjugate base of formic acid, and their concentration can be determined by using the formula:
[HCO2-] = [H+] = [NaHCO2] - [HCO2-]
Substituting the given values, we have:
[HCO2-] = [NaHCO2] - [HCO2-]
[HCO2-] = 0.310 M - [HCO2-]
Now we can solve for [HCO2-]:
2[HCO2-] = 0.310 M
[HCO2-] = 0.155 M
Substituting the value of [HCO2-] into the equation for [H+], we can find the concentration of H+ ions:
[H+] = [NaHCO2] - [HCO2-]
[H+] = 0.310 M - 0.155 M
[H+] = 0.155 M
Finally, we can calculate the percent ionization:
Percent ionization = ([H+] / [HCO2H]) x 100%
Percent ionization = (0.155 M / 0.190 M) x 100%
Percent ionization = 81.58%
Therefore, the main answer is 81.58%.
To calculate the percent ionization of formic acid, we first need to find the concentration of H+ ions in the solution. This can be done by subtracting the concentration of HCO2- ions (from the dissociation of sodium formate) from the initial concentration of sodium formate. We then use the equation for percent ionization to calculate the percentage. In this case, the percent ionization is found to be 81.58%.
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Draw the Lewis structure for each of the following organic compounds. State the molecular geometry around each carbon atom. a. C 2 H 2
The Lewis structure for each of the following organic compounds is given in the explanation part.
The Lewis structure for C₂H₂ (ethyne) can be represented as:
H-C≡C-HEach carbon atom is bonded to one hydrogen atom and to each other by a triple bond.
The molecular geometry around each carbon atom in C₂H₂ is linear. The triple bond between the carbon atoms forms a straight line, resulting in a linear shape.
Thus, the structure is H-C≡C-H.
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Below is listed a set of conditions which might exist in an imagined mass spectrometry experiment. i) The electron beam inside the ionisation chamber must be of low enough energy for an electron to be captured by M to give M ii) The electron beam inside the ionisation chamber must be of high enough energy M to be ionised to give M +
iii) The magnetic field in the mass analyser must be high enough to allow ions M +
and higher through to the detector iv) The sample must be reasonably volatile Which ONE of the following sets of conditions is desirable for a successful experiment? A. i), ii) and iii) B. ii) and iv) C. ii) and iii) D. i) and iii) E. i), iii) and iv)
The set of conditions that is desirable for a successful mass spectrometry experiment is option B: ii) and iv).
In mass spectrometry, the ionization process is crucial to generate ions from the sample being analyzed. Condition ii) states that the electron beam inside the ionization chamber must be of high enough energy to ionize the sample molecules, resulting in the formation of M+ ions. This condition ensures that the sample is successfully ionized, which is necessary for further analysis.
Condition iv) states that the sample must be reasonably volatile. Volatility refers to the ability of a substance to vaporize and form a gas. In mass spectrometry, the sample is typically introduced into the instrument in the gas phase. Having a reasonably volatile sample ensures that it can be easily vaporized and introduced into the mass spectrometer for analysis.
The other conditions (i) and (iii) are not necessary for a successful experiment. Condition i) refers to the low energy of the electron beam to capture an electron by M, which is not essential for ionization. Condition iii) refers to the magnetic field strength in the mass analyzer, which is not directly related to the ionization or volatility of the sample.
Therefore, the combination of conditions ii) and iv) (Option B) is desirable for a successful mass spectrometry experiment, as they ensure effective ionization of the sample and the presence of a volatile analyte for analysis.
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A sample of gas has a volume of 2.31 L at a temperature of 57.60 ∘
C. The 935 sample is heated to a temperature of 129.00∘C (assume pressure and amount of gas are held constant). Predict whether the new volume is greater or less than the original volume, and calculate the new volume
The new volume of the gas is greater than the original volume. The new volume is calculated to be 2.80 L.
The volume of a gas sample at constant pressure is directly related to its absolute temperature (Charles’s law) and the volume of a gas is inversely related to its pressure at constant temperature (Boyle’s law).
We need to determine the new volume.
Let the beginning volume(V1) be = 2.31L
and the beginning temperature(T1) be = 57.60°C
= (273 + 57.60)K = 330.6 K, last temperature(t2) = 129°C
= (273 +129)K = 402k
Supposing the last volume to be V2, as per Charles's regulation,
V1/T1 = V2/T2
=> V2 = (V1 × T2)/T1
= 2.31× 402/330.6
V2 = 2.80 L
So, New volume = 2.80 L
Thus, the new volume is bigger than the original volume.
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Give the electron configuration using the crystal field theory of the following complexes and show your configuration in the the crystal field diagram (N.B calculate the oxidation number of metals 1 st
). 3.1. [Co(NH 3
) 5
Cl]Br 2
3.2. Na 3
[Fe(CN) 6
] (6) 4. Give a detailed mechanism for a reaction of ethylamine and acetyl chloride under basic conditions. (6)
The electron configuration using the crystal field theory of the following complexes and show your configuration in the the crystal field diagram (N.B calculate the oxidation number of metals
3.1. [Co(NH3)5Cl]Br2:
To determine the oxidation state of cobalt (Co), we need more information. Once we know the oxidation state, we can determine the electron configuration and draw the crystal field diagram.
3.2. Na3[Fe(CN)6]:
The oxidation state of iron (Fe) in this complex is +3.
Electron configuration: [Ar] 3d^5 4s^0
Crystal field diagram:
| dxz |
| dz^2 |
| dyz |
- - - - - - - - - - - - -
| dxy |
| dx^2-y^2 |
| dzx |
The actual electron configuration and crystal field diagram may vary depending on the coordination geometry and ligand field strength.
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Need help.
Question: How does HNMR help in determining which stereochemistry is preferred?
HNMR, or proton nuclear magnetic resonance spectroscopy, can provide valuable information about the stereochemistry of a molecule. It can help determine which stereochemistry is preferred by examining the chemical shifts and coupling patterns of protons in the molecule.
1. Chemical shifts: In HNMR, protons in different chemical environments exhibit different chemical shifts. The chemical shifts can be influenced by the neighboring atoms and the stereochemistry of the molecule. By comparing the chemical shifts of protons in different stereoisomers, it is possible to identify patterns and trends that can indicate the preferred stereochemistry.
2. Coupling patterns: HNMR also provides information about the coupling of protons with neighboring protons. The number and arrangement of neighboring protons can affect the splitting patterns observed in the NMR spectrum. By analyzing the coupling patterns, one can infer the relative arrangement of protons in the molecule, which can help determine the preferred stereochemistry.
By combining the information obtained from chemical shifts and coupling patterns in HNMR spectra, researchers can gain insights into the preferred stereochemistry of a molecule. This can be particularly useful when studying chiral compounds or investigating stereoisomerism.
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Describe two ways that the effects of acid rain can be
mitigated. In other words, what two things can be done to lessen
the acidity of rain?
Two ways to mitigate the effects of acid rain are reducing emissions of sulfur dioxide and nitrogen oxides and implementing limestone or lime neutralization methods.
To lessen the acidity of rain and mitigate the effects of acid rain, two effective approaches can be taken.
1. Reducing emissions of sulfur dioxide (SO2) and nitrogen oxides (NOx): Acid rain is primarily caused by the release of these pollutants into the atmosphere from sources like power plants, industrial processes, and vehicles.
By implementing stringent air pollution control measures, such as using cleaner fuels, installing scrubbers in industries, and employing catalytic converters in vehicles, the emissions of SO2 and NOx can be significantly reduced. This, in turn, helps decrease the acidity of rain.
2. Implementing limestone or lime neutralization methods: Adding limestone (calcium carbonate) or lime (calcium oxide or hydroxide) to bodies of water or affected soil can neutralize the acidity caused by acid rain.
Limestone or lime reacts with the acidic components in the rain, such as sulfuric acid or nitric acid, and forms less harmful compounds, like calcium sulfate or calcium nitrate.
This neutralization process helps restore the pH balance and reduces the detrimental effects of acid rain on aquatic ecosystems and soil quality.
By adopting these strategies, it is possible to mitigate the harmful effects of acid rain by reducing the emissions of acidifying pollutants and neutralizing the acidity in affected areas.
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A chemist would like to conduct a dihydroxylation reaction. Which of the following reagents would produce this diol? A. sodium hydoxide B. Hot alkaline KMnO4
C. Cold alkaline KMnO4
D. Cl2 and UV
The reagent which would produce a diol in a dihydroxylation reaction is hot alkaline [tex]KMnO_4[/tex]. Therefore, the correct option is B.
Dihydroxylation is the process of a molecule gaining two hydroxyl (-OH) groups. Potassium permanganate, often referred to as hot alkaline [tex]KMnO_4[/tex], is often employed as a reagent in dihydroxylation processes. It forms a diol (a molecule containing two alcohol functional groups) when it combines with unsaturated compounds, such as alkenes or alkynes, by linking two hydroxyl groups in a double or triple bond.
Therefore, the correct option is B.
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a. How many ATOMS of xenon are present in \( 6.80 \) grams of xenon trioxide? atoms of xenon. b. How many GRAMS of oxygen are present in molecules of xenon trioxide? grams of oxygen.
A. The number of atoms of Xenon in 0.0347 mol of XeO3 is 6.28 × 10^21 atoms of xenon.
B. The oxygen present in molecules of xenon trioxide is 1.00 × 10^24 grams.
a. The mass of Xenon Trioxide can be calculated by summing the atomic masses of the constituent elements. It is given that the mass of Xenon Trioxide is 6.80 g.
Using the periodic table, we have:
Xenon (Xe) = 131.293 g/mol
Oxygen (O) = 15.999 g/mol
The molecular formula of Xenon trioxide is XeO3. Therefore, the mass of one mole of XeO3 can be calculated as follows:
Mass of XeO3 = (1 × 131.293) + (3 × 15.999) g/mol
= 196.29 g/mol
The number of moles of XeO3 in 6.80 g of XeO3 can be calculated using the formula:
n = m/M
where n is the number of moles, m is the mass, and M is the molar mass of the substance.
n = 6.80/196.29 = 0.0347 mol
There are three atoms of Xenon in one molecule of XeO3. Therefore, the number of atoms of Xenon in 0.0347 mol of XeO3 is:
No of Xe atoms = 0.0347 × 3 × (6.02 × 10^23)
= 6.28 × 10^21 atoms of xenon.
b. The mass of oxygen present in one molecule of Xenon Trioxide (XeO3) can be calculated as follows:
Molecular mass of XeO3 = (1 × 131.293) + (3 × 15.999) g/mol
= 196.29 g/mol
The mass of oxygen in one molecule of XeO3 = (3 × 15.999) g/mol
= 47.997 g/mol
The number of molecules of XeO3 in 6.80 g of XeO3 can be calculated using the formula:
n = m/M
where n is the number of moles, m is the mass, and M is the molar mass of the substance.
n = 6.80/196.29 = 0.0347 mol
The number of molecules of XeO3 in 0.0347 mol can be calculated as follows:
Number of molecules = 0.0347 × (6.02 × 10^23) = 2.09 × 10^21 molecules
The mass of oxygen present in 2.09 × 10^21 molecules of XeO3 can be calculated as follows:
Mass of oxygen = 2.09 × 10^21 × 47.997 g/mol
= 1.00 × 10^24 g
Therefore, 1.00 × 10^24 grams of oxygen are present in molecules of xenon trioxide.
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"Examine the following structure. Which of the following is the correct classification of this substance? a. complex lipid
a. 4uinigedinger b. phosphorlpid c) gfrycerophothtwiliad d) all of theses are correct classification
The option d. all of these are correct classification is right.
Lipids are the non-polar substances that are hydrophobic in nature. The lipids are the structures comprising of glycerol that serves as backbone. Attached to it are two fatty acids and phosphate group. This makes the structure a glycerphospholipid.
The complex lipids are of another type as well, which is sphingophospholipids. The fatty acids and glycerol bind together through ester bonds. The same bond is also seen between glycerol and phosphoric acid, which further forms phosphoester bond with alcohol.
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the
pH of a diet Coca Cola solution is 3.12. What is the concentration
of [OH^-] present in the solution?
The concentration present is [tex]1.467 * 10^(-11) M[/tex].
For calculating the concentration of hydroxide ions ([OH-]) in a solution, we need to use the relationship between pH and pOH, which is given by:
pH + pOH = 14
Since we have the pH value of the solution (pH = 3.12), we can find the pOH:
pOH = 14 - pH
= 14 - 3.12
= 10.88
The pOH is related to the hydroxide ion concentration ([OH-]) by the equation:
pOH = -log10[OH-]
Rearranging the equation, we find:
[OH-] = [tex]10^{-pOH}[/tex]
Substituting the value of pOH:
[OH-] =[tex]10^{-10.88}[/tex]
Using a calculator, we find:
[OH-] ≈ [tex]1.467 * 10^(-11) M[/tex].
Therefore, the concentration of hydroxide ions ([OH-]) present in the diet Coca Cola solution is approximately
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How many parts per million did the level of CO 2
increase per century over the 18,000 -year period leading up to the Industrial Revolution? To do this, take 18,000 years and put it in centwies (100 years per century). Then divide your answer from Q1 by the number of centuries. 3ow much has the level of CO 2
increased from 1850 until present? (Top circle-middlecircle) 4. How many parts per million ( ppm ) did the level increase per century over the last 172 years
0.75 ppm increase per century over the 18,000-year period leading up to the Industrial Revolution. The current level of [tex]CO_2[/tex] is around 415 ppm. The level of [tex]CO_2[/tex] has increased by approximately 78.5 ppm per century over the last 172 years.
The answer to the question is as follows:
1. To calculate how many parts per million (ppm) the level of [tex]CO_2[/tex] increased per century over the 18,000-year period leading up to the Industrial Revolution, we first divide 18,000 by 100 to get the number of centuries. The answer is 180.
Then we divide the increase in [tex]CO_2[/tex] from pre-industrial levels of 280 ppm to current levels of 415 ppm by the number of centuries.
415 ppm - 280 ppm = 135 ppm
increase over 180 centuries (or 18,000 years)
135 ppm ÷ 180 centuries = 0.75 ppm
0.75 ppm increase per century over the 18,000-year period leading up to the Industrial Revolution.
2. The level of [tex]CO_2[/tex] has increased by approximately 135 ppm since preindustrial levels of 280 ppm. Therefore, the current level of [tex]CO_2[/tex]is around 415 ppm.
3. To calculate how many parts per million (ppm) the level of [tex]CO_2[/tex] increased per century over the last 172 years, we first divide 172 by 100 to get the number of centuries.
The answer is 1.72. Then we divide the increase in [tex]CO_2[/tex] from 280 ppm (preindustrial levels) to 415 ppm (current levels) by the number of centuries.
415 ppm - 280 ppm = 135 ppm
increase over 1.72 centuries (or 172 years)
135 ppm ÷ 1.72 centuries = 78.5 ppm
increase per century over the last 172 years.
Therefore, the level of [tex]CO_2[/tex] has increased by approximately 78.5 ppm per century over the last 172 years.
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