The pH of the solution prepared by dissolving 1.30g of sodium acetate in 60.5 mL of 0.20 M acetic acid is approximately 3.34.
To calculate the pH of the solution, we need to consider the equilibrium between the acetic acid (CH3COOH) and its conjugate base, acetate ion (CH3COO-), using the Ka value provided. We will also take into account the presence of sodium acetate (CH3COONa), which will provide additional acetate ions.
First, let's determine the moles of acetic acid present in the solution:
Moles of acetic acid = concentration x volume
Moles of acetic acid = 0.20 M x 60.5 mL x (1 L / 1000 mL)
Moles of acetic acid = 0.0121 mol
Next, let's determine the moles of sodium acetate:
Moles of sodium acetate = mass / molar mass
Moles of sodium acetate = 1.30 g / 82.03 g/mol
Moles of sodium acetate = 0.0158 mol
Since sodium acetate dissociates completely, the concentration of acetate ions (CH3COO-) will be equal to the moles of sodium acetate divided by the total volume of the solution:
Concentration of acetate ions = moles of sodium acetate / total volume
Concentration of acetate ions = 0.0158 mol / 60.5 mL x (1 L / 1000 mL)
Concentration of acetate ions = 0.260 M
Now, we can set up an ICE (Initial, Change, Equilibrium) table to calculate the concentrations of acetic acid and acetate ions at equilibrium:
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
Initial: 0.0121 M 0 M 0 M
Change: -x +x +x
Equilibrium: 0.0121 - x x x
The Ka expression for acetic acid is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Using the given Ka value, we can set up the equation:
1.75 x 10^-5 = (x)(x) / (0.0121 - x)
Since x is expected to be small compared to 0.0121, we can approximate the denominator as 0.0121:
1.75 x 10^-5 = x^2 / 0.0121
Simplifying the equation:
x^2 = 1.75 x 10^-5 x 0.0121
x^2 = 2.1175 x 10^-7
x ≈ 4.6 x 10^-4
Now, we can calculate the concentration of H3O+ ions:
[H3O+] = x = 4.6 x 10^-4 M
Finally, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(4.6 x 10^-4)
pH ≈ 3.34
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how the Sun is responsible for most of the energy on Earth by explaining it's connection to photosynthesis and fossil fuels.
Answer:
The sun provides most of the energy on earth because it is the earth's (and the solar system's) greatest source of heat and light.
Explanation:
The process of photosynthesis in plants uses the energy from the sun and converts that energy to be used for plants.
The burning of fossil fuels produces greenhouse gases which trap the heat in our earth's atmosphere (causing global warming).
how would you draw a bohr rutherford diagram for Fluorine-19 ? Explain your thought process and say how many electrons it has and where each electron would go?
To draw the Bohr-Rutherford diagram for fluorine-19, we begin by identifying the atomic number of fluorine, which is 9. This means that a neutral fluorine atom has 9 electrons.
The Bohr-Rutherford diagram represents the electron shells or energy levels of an atom, with each shell holding a certain number of electrons. The first shell holds a maximum of 2 electrons, while the second shell holds a maximum of 8 electrons.
Based on this information, we divide the 9 electrons of fluorine into the appropriate shells:
1. The first two electrons occupy the first shell (closest to the nucleus). This will completely fill the first shell.
2. The remaining 7 electrons occupy the second shell. However, since the second shell can hold a maximum of 8 electrons, one electron will be placed in each orbital (subshell) before the electrons are paired.
The electron configuration of fluorine-19, arranged in the Bohr-Rutherford diagram, can be represented as follows:
1st shell: 2nd shell:
(2) (7)
The numbers in parentheses indicate the number of electrons in each shell. In this case, the first shell has 2 electrons and the second shell has 7 electrons.
It is important to note that the Bohr-Rutherford diagram provides a simplified representation of the electron distribution and does not account for the more complex orbital arrangements. However, it provides a basic visual understanding of the electron configuration in an atom.
Briefly, fluorine-19 has a Bohr-Rutherford diagram with 2 electrons in the first shell and 7 electrons in the second shell, for a total of 9 electrons.
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Which statement about balanced chemical equations is true?
OA. The mass of the new atoms that are formed equals the mass of
the atoms that made up the reactants.
OB. The total mass of the reactants equals the total mass of the
products.
OC. The total number of moles of products equals the total number of
moles of reactants,
OD. The mass of the products is greater than the mass of the
reactants when the number of moles increases.
SUBMIT
The total mass of the reactants equals the total mass of the products the statement about balanced chemical equations is true. Hence, option B is correct.
This is known as the Law of Conservation of Mass, which states that matter can neither be created nor destroyed in a chemical reaction. In other words, the mass of the reactants must equal the mass of the products in a balanced chemical equation.
While the identities of the atoms may change during a reaction, the total number of atoms of each element on both sides of the equation must be the same, thus leading to the conservation of mass.
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A 20.0 L helium balloon initially at 25°C and 1.00 atm is heated in the sunlight to a final temperature of 41.6°C. During this process the balloon expands in such a way that the pressure remains constant. What is the change in entropy of the gas during this process? Answer in J/K to three decimal places. R = 8.314 J/mol K = 0.08206 L-atm/mol K
We can apply the following equation to determine the gas's change in entropy:
S = nR T f /Ti ln
Given: 20.0 L is the initial volume (Vi).
Initial temperature (Ti) = 25 °C, which equals 25 + 273.15 K, or 298.15 K.
Final temperature = 41.6°C, which is equal to 41.6 + 273.15 K, or 314.75 K.
Pressure does not change (P)
(R) = 8.314 J/mol K for the gas constant.
The ideal gas law must first be used to calculate the quantity of helium gas (n):
PV = nRT
When we rewrite the equation, we get:
n = PV / RT
replacing the specified values:
n is equal to (1.00 atm) x (20.0 L) / (0.08206 L-atm/mol K x 298.15 K).
n ≈ 0.813 mol
Now we can determine the entropy change:
S = ln(314.75 K / 298.15 K) * (0.813 mol) * (8.314 J/mol K)
ΔS ≈ 1.071 J/K
As a result, the gas's entropy changed by about 1.071 J/K during this operation.
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Consider the balance redox reaction below
K2Cr2O7 + 7H2SO4 + 6KBr ----> 3Br2O + 4K2SO4 + Cr2(SO4)3 + 7H2O
WHich element is oxidized in the reaction?
A.) Br in KBr B.) S in H2S04
C.) H in H2SO4 D.) Cr in K2CR2O7
In the redox reaction, [tex]K_{2} Cr_{2} O_{7}[/tex] + 7[tex]H_{2} SO_{4}[/tex] + 6KBr → [tex]3Br_{2}O[/tex] + [tex]4K_{2} SO_{4}[/tex] + [tex]Cr_{2} (SO_{4})_{3}[/tex] + [tex]7H_{2}O[/tex], the element that is oxidized is Option A. Br in KBr.
In the given reaction, [tex]K_{2} Cr_{2} O_{7}[/tex], also known as potassium dichromate, is a strong oxidizing agent that causes oxidation of the other reactants. Oxidation refers to the loss of electrons by an atom or molecule. Similarly, reduction refers to the gain of electrons by an atom or molecule.
In [tex]K_{2} Cr_{2} O_{7}[/tex], the oxidation state of chromium (Cr) is +6. In the product, [tex]Cr_{2} (SO_{4})_{3}[/tex], the oxidation state of Cr is +3. Therefore, Cr has been reduced, meaning it has gained electrons and is not oxidized.
On the other hand, [tex]H_{2} SO_{4}[/tex], the oxidation state of sulfur (S) is +6. In the product, [tex]K_{2} SO_{4}[/tex], the oxidation state of S is +6. Therefore, S has not been oxidized or reduced.
Similarly, [tex]H_{2} SO_{4}[/tex], the oxidation state of hydrogen (H) is +1. In the product, [tex]H_{2}O[/tex], the oxidation state of H is 0. Therefore, H has been reduced, meaning it has gained electrons and is not oxidized.
Finally, in KBr, the oxidation state of Br is -1. In the product, [tex]Br_{2}O[/tex], the oxidation state of Br is +1. Therefore, Br has been oxidized, meaning it has lost electrons.
In conclusion, the element that oxidized in the given reaction is Br in KBr (Option A).
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During a volcanic eruption, lava flowed at a rate of 37 m/min. At this rate how far in kilometers
can lava travel in 45 minutes?
Answer:
The lava can travel approximately 1.665 kilometers in 45 minutes.
Explanation:
6 of 28
Attempt 2
If 7.66 g of CuNO, is dissolved in water to make a 0.140 M solution, what is the volume of the solution in milliliters?
The volume of a 0.140 M solution of Cu(NO3)2 that contains 7.66 g of the compound, volume of the solution is 292.9 mL.
To determine the volume of a 0.140 M solution of Cu(NO3)2 that contains 7.66 g of the compound, we can use the following formula:
Molarity = moles of solute / volume of solution in liters
First, we need to calculate the number of moles of Cu(NO3)2 in the given mass of the compound:
moles of Cu(NO3)2 = mass / molar mass
The molar mass of Cu(NO3)2 can be calculated by adding the atomic masses of copper, nitrogen, and six oxygen atoms:
1 x Cu = 63.55 g/mol
2 x N = 14.01 g/mol x 2 = 28.02 g/mol
6 x O = 15.99 g/mol x 6 = 95.94 g/mol
Molar mass of Cu(NO3)2 = 63.55 g/mol + 28.02 g/mol + 95.94 g/mol = 187.51 g/mol
Now, we can calculate the number of moles of Cu(NO3)2:
moles of Cu(NO3)2 = 7.66 g / 187.51 g/mol = 0.0409 moles
Finally, we can use the formula above to calculate the volume of the solution:
0.140 M = 0.0409 moles / volume of solution in liters
Volume of solution in liters = 0.0409 moles / 0.140 M = 0.2929 L
Converting to milliliters, we get:
Volume of solution in milliliters = 0.2929 L x 1000 mL/L = 292.9 mL
Therefore, the volume of the solution is 292.9 mL.
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express views on how the law , the citizen and community can proteand support victims of human rights violations?
Answer:
On the support of victims of human rights violation, the law should clearly state and express the repercussions of violating human rights. The citizen on the other hand should be well versed with human rights and should report any violator of human rights to the relevant authorities for immediate action to be taken.
References
Use the References to access important values if needed for this question.
When 22.0 ml, of a 5.25 x 10-4 M cobalt(II) fluoride solution is combined with 12.0 mL of a 6.18 x 10-4 M sodium sulfide solution does a precipitate form?
Kp (COS) 5.9 x 10-21)
=
O Yes, the precipitate forms.
O No, the precipitate doesn't form.
For these conditions the Reaction Quotient, Q, is equal to
When 22.0 ml, of a [tex]5.25 x 10^{-4[/tex] M cobalt(II) fluoride solution is combined with 12.0 mL of a [tex]6.18 * 10^{-4[/tex] M sodium sulfide solution does a precipitate form, the answer is: No, the precipitate doesn't form.
The balanced chemical equation for the reaction between cobalt(II) fluoride and sodium sulfide is:
[tex]CoF_2[/tex](aq) + Na2S(aq) → CoS(s) + 2NaF(aq)
To determine if a precipitate will form, we need to calculate the reaction quotient, Q, using the initial concentrations of the reactants. The expression for Q is:
Q = [tex][CoS][NaF]^2[/tex] / [CoF2]
where the square brackets denote molar concentrations.
Using the given volumes and concentrations, we can calculate the initial number of moles of each species:
moles CoF2 = [tex](22.0 mL / 1000 mL/L) * (5.25 * 10^-^4 M) = 1.155 * 10^-^5[/tex] mol
moles Na2S = [tex](12.0 mL / 1000 mL/L) * (6.18 * 10^-^4 M) = 7.416 * 10^-^6[/tex] mol
Because the reaction stoichiometry shows that 1 mol of CoF2 reacts with 1 mol of Na2S to form 1 mol of CoS, the amount of CoS formed will be equal to the lesser of these two values, which is 7.416 x 10^-6 mol.
The molar concentration of CoS in the resulting solution will be:
[CoS] = [tex](7.416 * 10^-^6 mol) / (34.0 mL / 1000 mL/L) = 2.181 * 10^-^4 M[/tex]
The molar concentrations of NaF and CoF2 in the resulting solution will be:
[NaF] = [tex](2 x 7.416 x 10^-6 mol) / (34.0 mL / 1000 mL/L) = 4.363 x 10^-4 M[/tex]
[CoF2] = [tex](1.155 x 10^-5 mol) / (34.0 mL / 1000 mL/L) = 3.398 x 10^-4 M[/tex]
Substituting these values into the expression for Q, we get:
Q = [tex](2.181 * 10^-^4)(4.363 * 10^-^4)^2 / (3.398 * 10^-^4) = 1.251[/tex]
Comparing the value of Q to the equilibrium constant Kp, we can see that Q is much greater than Kp (Q > Kp), indicating that the reaction will proceed in the reverse direction to reach equilibrium, and no precipitate will form.
Therefore, the answer is: No, the precipitate doesn't form.
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When a 20.2 mL sample of a 0.382 M aqueous hydrocyanic acid solution is titrated with a 0.421 M aqueous barium hydroxide solution, what is the pH after 13.7 mL of barium hydroxide have been added?
The concept molarity is used here to determine the pH after adding 12.6 mL of the base. The term molarity is an important method which is used to calculate the concentration of a solution. Here the pH is 1.23.
The term molarity is defined as the number of moles of the solute dissolved per litre of the solution. It is also called the molar concentration. It is represented as 'M' and its unit is mol / L.
Molarity is given as:
M = Number of moles / Volume of solution in liters
'n' of HCN = 20.2 × 1 L / 1000 mL × 0.382 = 0.0077 mol
'n' of Ba(OH)₂ = 13.7 × 1L / 1000 mL × 0.421 = 0.0057 mol
Excess H⁺ = 0.002
Total volume = 20.2 + 13.7 = 33.9 mL = 0.0339 L
Concentration of H⁺ = 0.002 / 0.0339 = 0.058
So pH is:
pH = - log[H⁺]
pH = - log[ 0.058] = 1.23
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Initially, a 400.3 m³ spring-loaded piston-cylinder assembly contains R-134a at 600 kPa and 150°C. The refrigerant temperature was cooled to -30°C and the volume was 0.1 m³. Calculate the transfer and the work produced by the refrigerant during this process.
An Earth scientist is testing how various soil mixtures affect plant growth. She begins with 10 soil types. For each mixture, she combines equal
amounts of 2 soil types.
How many soil mixtures must the scientist create in order to test all possible combinations of the 10 soil types?
You may use the calculator.
OA. 20
OB. 45
OC. 55
OD. 90
The scientist needs to create 45 soil mixtures in order to test all possible combinations of the 10 soil types. Option B.
Combination problemTo calculate the number of unique soil mixtures, we can use the formula for the number of combinations of n objects taken k at a time, which is given by:
C(n,k) = n! / (k! * (n-k)!)
In this case, we have 10 soil types and we want to take 2 at a time to create unique mixtures. Thus, we substitute n=10 and k=2:
C(10,2) = 10! / (2! * (10-2)!)
= 10! / (2! * 8!)
= (10 x 9) / 2
= 45
In other words, the scientist needs to create 45 soil mixtures in order to test all possible combinations of the 10 soil types.
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Calculate the frequency of the =4
line in the Lyman series of hydrogen.
The frequency of the =4 line in the Lyman series of hydrogen is 3.09 x 10¹⁵ Hz.
What is the frequency of the n = 4 line in the Lyman series of hydrogen?The energy levels in the Lyman series of hydrogen are given by the formula:
E = -13.6/n²where
E is the energy of the level and n is an integer representing the level number.
The transition from level n to level 1 produces a photon with a frequency given by:
[tex]v = (E_n - E_1)/h[/tex]
where
v is the frequency of the photon,h is Planck's constant, and [tex]E_n[/tex] and [tex]E_1[/tex] are the energies of levels n and 1, respectively.For the n = 4 line in the Lyman series, the initial level is n = 4 and the final level is n = 1.
The energy of the initial level is:
[tex]E_4[/tex] = -13.6/4²
[tex]E_4[/tex] = -0.85 eV
The energy of the final level is:
[tex]E_1[/tex]= -13.6/1²
[tex]E_1[/tex] = -13.6 eV
The energy difference between the levels is:
[tex]E_4 - E_1[/tex] = -0.85 - (-13.6)
[tex]E_4 - E_1[/tex] = 12.75 eV
Converting to joules:
v = (12.75 x 1.6 x 10⁻¹⁹ J)/6.626 x 10⁻³⁴ J s
v = 3.09 x 10¹⁵ Hz
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Consider the following reaction:
2N2(g) + O2(g) ⇌ 2NO2(g)
A reaction mixture initially contains 3.21 atm N2 and 6.21 atm O2. Determine the equilibrium pressure of NO2 if Kp for the reaction at this temperature is 3.2 × 10-28.
The equilibrium pressure of the nitrogen oxide is given as 4 atm.
What is the Kp?Kp is the equilibrium constant for a chemical reaction in terms of partial pressures. It is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their stoichiometric coefficients
We know that;
Kp = [tex]pNO_{2} ^2/pN_{2} ^2 . pO_{2}[/tex]
[tex]3.2 * 10^-28 = (2x)^2/(3.21 - 2x) (6.21 - x)\\3.2 * 10^-28 = 4x^2/19.9 - 3.21x - 12.42x + 2x^2\\3.2 * 10^-28(19.9 - 15.63x + 2x^2) = 4x^2\\6.4 * 10^-27 - 5 * 10^-27 x + 6.4 * 10^-28x^2 = 4x^2\\x = 4 atm[/tex]
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please help with this! ty
The mole concept is an important method which is used to calculate the amount of the substance. 1 mole is defined as a number which is equal to 6.022 × 10²³ particles also called the Avogadro's constant.
One mole of a substance is that amount of it which contains as many particles or entities as there are atoms in exactly 12 g of Carbon-12.
The equation used to calculate the number of moles is:
Number of moles = Given mass / Molar mass
Molar mass of NaCl = 58.44 g/mol
1. n = 8 / 58.44 = 0.13
2. n = 2.3 / 58.44 = 0.039
3. n = 9.59 / 58.44 = 0.16
4. n = 38.44 / 58.44 = 0.65
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Please Help!!50 points and I’ll mark as brainliest!
Tasks are in the picture.
1) The pH is 2.5
2) The pH is 11.5
3) The initial concentration is[tex]2.1 * 10^-14[/tex]M
What is the pH?pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm of the concentration of hydrogen ions in the solution.
1) The pH of the solution can be gotten from;
K = [tex]x^2[/tex]/0.65 - x
[tex]1.754 * 10^-5[/tex](0.65 - x) = [tex]x^2[/tex]
[tex]1.14 * 10^-5 - 1.754 * 10^-5x = x^2\\x^2 + 1.754 * 10^-5x - 1.14 * 10^-5 = 0[/tex]
x = 0.003 M
pH = -log(0.003)
= 2.5
2) Kb = [tex]x^2[/tex]/0.35 - x
[tex]1.8 * 10^-5 (0.35 - x) = x^2\\6.3 * 10^-6 - 1.8 * 10^-5x = x^2\\x^2 + 1.8 * 10^-5x - 6.3 * 10^-6 = 0[/tex]
x = 0.003 M
pOH = -log (0.003)
= 2.5
pH = 14 - 2.5 = 11.5
3) Hydrogen ion concentration = Antilog (-11.5)
= 3.2 * 10^-12 M
[tex]4.9 * 10^-10 = ( 3.2 * 10^-12)^2/x \\4.9 * 10^-10x = ( 3.2 * 10^-12)^2\\x = ( 3.2 * 10^-12)^2/4.9 * 10^-10\\x = 2.1 * 10^-14 M[/tex]
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In which of the following compounds does sulfur have the highest (i.e most positive) oxidation number? a) CuS b) SO2 c) K2SO3 d) NA2SO4
Answer:
c) K2SO3
Explanation:
The oxidation number of S in K2SO4 K 2 S O 4 is +6. So, this is the highest oxidation number of S amongst the oxidation number of S in all the given compounds.
I NEED HELP ASAP PLS
A 638.3 g sample of nickel releases 12337 joules of heat. The specific heat capacity of nickel is
0.5024 J/(g.°C). By how much did the temperature of this sample change, in degrees Celsius?
A 0.6949 mol sample of aluminium absorbs 5281.1 joules of heat. The specific heat capacity of aluminium is 24.2 J/(mol K). By how much did the temperature of this sample change, in kelvins?
A 0.4363 mol sample of indium experiences a temperature change of +109.71 °C while absorbing 1279.9 joules of heat. What is the specific heat capacity of indium?
A 824.5 g sample of silver undergoes a temperature change of +1078.87 K while absorbing
207260 joules of heat. What is the specific heat capacity of silver?
A 616.7 g sample of iridium is subjected to a temperature change of -2020.47 °C while releasing 1617300 joules of heat. What is the specific heat capacity of iridium?
A 875.8 g sample of uranium absorbs 15291 joules of heat. The specific heat capacity of uranium is 0.116 J/(g K). By how much did the temperature of this sample change, in kelvins?
A 990.1 g sample of molybdenum absorbs 179910 joules of heat. The specific heat capacity of molybdenum is 0.2772 J/(g.°C). By how much did the temperature of this sample change, in degrees Celsius?
A 3.596 mol sample of barium is subjected to a temperature change of -568.85 K. The specific heat capacity of barium is 28.07 J/(mol K). How many joules of heat were transferred by the sample?
A sample of rhodium experiences a temperature change of -1311.77 °C while releasing 203580 joules of heat. The specific heat capacity of rhodium is 0.2428 J/(g.°C). What is the mass of this sample?
A sample of methane goes through a temperature change of +330.22 °C while absorbing 672220 joules of heat. The specific heat capacity of methane is 2.191 J/(g.°C). What is the mass of this sample?
Could someone help me solve this
The mass of CO₂ needed to fill the zip-top bag by converting the moles is 4.4817 g CO₂
How to determine mass?To determine the mass of CO₂, know the number of moles of CO₂ and its molar mass. The formula to calculate the mass of CO₂ is:
mass of CO₂ = number of moles of CO₂ x molar mass of CO₂
The number of moles of CO₂ can be calculated using the Ideal Gas Law, which relates the number of moles of a gas to its pressure, volume, temperature, and gas constant.
Once calculated the number of moles of CO₂, multiply it by the molar mass of CO₂ to obtain the mass of CO₂. The molar mass of CO₂ is approximately 44 g/mol.
For this case the mass of CO₂ after conversion is 0.102 mol CO₂ x 44.0099 g/mol CO₂ = 4.4817 g CO₂
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The gas phase reaction between nitric oxide and hydrogen occurs in three elementary steps:
NO + NO <--> N2O2 (fast equilibrium)
N2O2 + H2 --> H2O + N2O (slow)
N2O + H2 --> N2 + H2O (fast)
Based on this mechanism we would expect the rate law to be:
a) Second order in NO, first order in H2, and zero order in H2O
b) Second order in both NO and H2, and zero order in H2O
c) Second order in both NO and H2, and first order in H2O
d) First order in NO, first order in H2, and -1 order in H2O
The correct option is (b) Second order in both NO and H2, and zero order in H2O.
The rate law for a chemical reaction can be determined from its elementary steps and the rate-determining step (RDS). In this mechanism, the RDS is the second step, which is the slowest step, so the rate law will be based on this step.
According to the second step of the mechanism, the stoichiometry of the reaction is 1:1 for N2O2 and H2. Therefore, the rate law should have a second order dependence on both N2O2 and H2. The rate law should also reflect the stoichiometry of the reaction, so it will be:
Rate = k[N2O2][H2]
The final product is H2O, which is not involved in the RDS, so it will not appear in the rate law. The correct answer is (b) Second order in both NO and H2, and zero order in H2O. The fast equilibrium step does not affect the overall rate of the reaction, as it simply establishes an equilibrium concentration of N2O2. The rate law only reflects the RDS, which involves the consumption of N2O2 and H2 to form N2 and H2O.
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define tetragonal unit cell
Answer:
The tetragonal unit cell is distinguished by an axis of fourfold symmetry, about which a rotation of the cell through an angle of 90° brings the atoms into coincidence with their initial positions. The elements boron and tin can crystallize in tetragonal form, as can some minerals such as zircon.
Explanation:
whats an example of a low amount of force being applied to an object explain in 1-2 sentences.
An example of a low amount of force being applied to an object is as follows; a child pushing a barrel that is not moving.
What is force?Applied force is the force which is applied to an object by another object.
Ideally, a force is a physical quantity that denotes ability to push, pull, twist or accelerate a body. Work is kind of force that causes a body to move.
For example, a person pushing a barrel is an example of applied force. When the person pushes the barrel then there is an applied force acting upon the barrel.
However, when a child pushes against the barrel and it doesn't move, it can be said that a low amount of force is applied.
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PbI2 is sparingly soluble in water (Ksp = 8.7 x 10-9). Which two of the following statements are true?
- More solid will dissolve when 10 g of PbI2(s) is added to 1.0 L of water than when 5 g of PbI2(s) is added.
- Adding HNO3(aq) to a saturated solution of PbI2(s) will cause more solid to precipitate
- PbI2(s) will be more soluble in pure water than in a solution containing KI(aq)
- PbI2(s) powder will be more soluble in water than large crystals of PbI2(s)
- The solubility product of PbI2(s) will increase with increasing temperature
Two statements that are true are:
More solid will dissolve when 10 g of PbI2(s) is added to 1.0 L of water than when 5 g of PbI2(s) is added.
The solubility product of PbI2(s) will increase with increasing temperature.
1- More solid will dissolve when 10 g of PbI2(s) is added to 1.0 L of water than when 5 g of PbI2(s) is added.
This statement is true because the solubility of PbI2 is limited by its Ksp value. Therefore, the more solid PbI2 added, the closer the solution gets to being saturated and the more PbI2 will dissolve.
5-The solubility product of PbI2(s) will increase with increasing temperature.
This statement is true because the solubility of most ionic solids increases with increasing temperature. A higher temperature increases the kinetic energy of the solvent molecules, allowing them to better solvate and break apart the solid PbI2. As a result, more ions are available to react and the Ksp value increases.
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Initially, a 0.3 m³ spring-loaded piston-cylinder assembly contains R-134a at 600 kPa and 150°C. The refrigerant temperature was cooled to -30°C and the volume was 0.1 m³. Calculate the transfer of 151 and the work produced by the refrigerant during this process.
The work produced by the refrigerant during this process is 163.27 kJ, and the transfer of heat is -825.63 kJ. The negative sign indicates that heat is being removed from the refrigerant.
To calculate the transfer of heat and work produced during this process, we can use the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat added to the system minus the work done by the system. First, we need to determine the initial and final states of the refrigerant. The initial state is 600 kPa and 150°C, and the final state is -30°C and a volume of 0.1 m³. We can use the refrigerant tables to determine the specific volume and internal energy of the refrigerant at each state. From the tables, we find that the specific volume of the refrigerant at the initial state is 0.0551 m³/kg and the internal energy is 770.68 kJ/kg. At the final state, the specific volume is 0.001344 m³/kg and the internal energy is 108.32 kJ/kg. Using the first law of thermodynamics, we can calculate the transfer of heat and work produced during this process as follows:
ΔU = Q - W
where ΔU is the change in internal energy, Q is the transfer of heat, and W is the work produced by the refrigerant.
ΔU = U2 - U1 = 108.32 kJ/kg - 770.68 kJ/kg = -662.36 kJ/kg
Q = ΔU + W
W = -Q + ΔU = -mCp(T2 - T1) + ΔU
where m is the mass of the refrigerant, Cp is the specific heat capacity of the refrigerant, T1 is the initial temperature, and T2 is the final temperature.
Assuming a mass of 1 kg for the refrigerant, the specific heat capacity of R-134a at constant pressure (Cp) is 1.51 kJ/kgK. Plugging in the values, we get:
W = -mCp(T2 - T1) + ΔU
W = -1 kg x 1.51 kJ/kgK x (-30°C - 150°C) + (-662.36 kJ/kg)
W = 163.27 kJ
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Relationship between uv light and activity of object
write and balance an equation for a reaction in which iron (Fe) and Hydrochloric Acid (HCL) react to form Iron cloride (FeCl2) and Hydrogen (H2)
The balanced equation for the reaction between iron (Fe) and hydrochloric acid (HCl) to form iron chloride (FeCl2) and hydrogen (H2) is: Fe + 2HCl → FeCl2 + H2
This equation represents a single-displacement reaction in which iron displaces hydrogen from hydrochloric acid, resulting in the formation of iron chloride and hydrogen gas. The coefficients in the equation show that one molecule of iron reacts with two molecules of hydrochloric acid to produce one molecule of iron chloride and one molecule of hydrogen gas. To balance this equation, we need to ensure that the number of atoms of each element on both sides of the equation is the same. In this case, we have one iron atom on both sides, two hydrogen atoms on the reactant side, and two hydrogen atoms on the product side. We also have two chlorine atoms on the product side and none on the reactant side. To balance the equation, we add a coefficient of 2 in front of hydrochloric acid and in front of hydrogen gas: Fe + 2HCl → FeCl2 + 2H2
This balanced equation shows that one molecule of iron reacts with two molecules of hydrochloric acid to produce one molecule of iron chloride and two molecules of hydrogen gas.
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Shown above is the phase diagram for water as it is heated. Which section represents the phase of water with the highest kinetic energy?
The section that represents the phase of water with the highest kinetic energy is the gas phase or vapor phase.
Gas phase or vapor phase section is above the boiling point curve, which separates the liquid and gas phases. At this point, the temperature is at or above 100°C (at standard atmospheric pressure), and the kinetic energy of the water molecules is sufficient to overcome the intermolecular forces holding them in the liquid phase and escape into the gas phase. The gas phase has the highest kinetic energy because the water molecules in this phase are more widely separated and move more rapidly than in the liquid or solid phases. The gas phase is also characterized by the highest entropy or disorder, as the molecules are free to move in any direction and occupy a large volume. The section that represents the phase of water with the highest kinetic energy is gas phase or vapor phase.
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Write a balanced chemical equation for each of the following.
Gaseous carbon monoxide reacts with hydrogen gas to form gaseous methane (CH4) and liquid water.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
CO+3H2———> CH4+H2O
Explanation:
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what is the hyphen notation and nuclear symbol for oxygen
The element name or symbol is followed with a hyphen and the mass number. On the periodic table of elements, oxygen is represented by the symbol O. It has an atomic number of 8 because it has 8 protons in its nucleus
Examples:Carbon-14 or C-14 (meaning the isotope of carbon that has a mass number of 14)
Circle O is shown. Line segments E O and E O are radii with length 24. Angle D O E is 45 degrees. The area within sector D O E is shaded.
What is the area of the shaded sector?
24π
45π
72π
576π
In the given circle, the area of the shaded sector is 72π, hence option C is correct.
The region inside the portion of the circle formed by two radii and an arc is known as the area of a sector. It only covers a small portion of the entire circle's surface.
To find the area of the shaded sector, use the formula:
According to the diagram in the question
OE is the radius
The central angel of the sector is 45⁰
A = θ/360 πr²
Put the values in the formula, we get
45 ÷ 60 × π × 24²
1/8 × π × 24 × 24
3 × 24 × π
= 72π
Thus, the area of the shaded sector is 72π.
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