Answers
Answer:
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:
Mg: 1 moleHCl: 2 molesMgCl₂: 1 moleH₂: 1 moleBeing:
Mg: 24. 31 g/moleH: 1 g/moleCl: 35.45 g/molethe molar mass of the compounds participating in the reaction is:
Mg: 24.31 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleMgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/moleH₂: 2*1 g/mole= 2 g/moleThen, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:
Mg: 1 mole* 24.31 g/mole= 24.31 gHCl: 2 moles* 36.45 g/mole= 72.9 gMgCl₂: 1 mole* 95.21 g/mole= 95.21 gH₂: 1 mole* 2 g/mole= 2 gThen you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?
[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]
moles of H₂= 0.134
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
Let's consider the balanced equation between Mg and HCl.
Mg + 2 HCl ⇒ MgCl₂ + H₂
The molar mass of Mg is 24.3 g/mol. The moles corresponding to 3.25 g of Mg are:
[tex]3.25 g \times \frac{1mol}{24.3g} = 0.134 mol[/tex]
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ formed by 0.134 moles of Mg are:
[tex]0.134 mol Mg \times \frac{1molH_2}{1molMg} = 0.134molH_2[/tex]
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
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Related Questions
how many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
please help! due in a bit
Answers
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
Smooth muscle myosin is a motor protein that plays a crucial role in the contraction of smooth muscle. If this protein has a molar mass of 480,000 grams/mol, what is the mass, in grams, of 27 moles of smooth muscle myosin
Answers
Answer: Thus the mass, in grams, of 27 moles of smooth muscle myosin is 12960000 grams
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass, occupies 22.4 L at STP contains avogadro's number [tex](6.023\times 10^{23})[/tex] of particles.
Molecular mass of protein = 480,000 g/mol
Thus 1 mole of protein weighs = 480,000 g
So 27 moles of protein weighs = [tex]\frac{480,000}{1}\times 27=12960000g[/tex]
Thus the mass, in grams, of 27 moles of smooth muscle myosin is 12960000 grams
A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound
Answers
Answer:
Na₂SO₄
Explanation:
Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.
Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:
Na₂SO₄
Which of the following is a conjugate acid-base pair in the reaction represented by the
equation below?
H2PO4 + H20 H3PO, + OH
H2PO, and H2O
b) H,PO, and OH
c) H2PO, and H3PO,
None of the above
Answers
Answer: [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For the given reaction:
[tex]H_2PO_4^-+H_2O\rightleftharpoons H_3PO_4+OH^-[/tex]
Here, [tex]H_2O[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]OH^-[/tex] which is a conjugate base.
Similarly , [tex]H_2PO_4^-[/tex] is gaining a proton, thus it is considered as an base and after gaining a proton, it forms [tex]H_3PO_4[/tex] which is a conjugate acid.
Thus [tex]H_2PO_4[/tex] and [tex]H_3PO_4[/tex] is a conjugate acid-base pair in the reaction represented by the equation below
The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
Thus, a proton is being lost, making it an acid, and once a proton is lost, a conjugate base is formed. Similar to that, is gaining a proton, making it a base, and then it produces a conjugate acid after gaining a proton.
The Brnsted-Lowry hypothesis, often known as the proton theory of acids and bases, is an independent theory of acid-base reactions that was put forth in 1923 by Johannes Nicolaus Brnsted and Thomas Martin Lowry.
This theory's central idea is that when an acid and a base interact, the acid creates its conjugate base and the base creates its conjugate acid by exchanging a proton (the hydrogen cation, or H+).
Thus, The Bronsted-Lowry conjugate acid-base hypothesis defines an acid as a substance that loses protons and donates them to another chemical to produce conjugate base, and a base as a substance that takes protons to generate conjugate acid.
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Water was poured over a large oil fire to extinguish it. What would happen and why?
Answers
Answer:
I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water
Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.
What happens when you pour water on an oil fire?A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.
Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.
The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.
Thus, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.
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If complications arise after cataract surgery, the ophthalmologist will use a Nd:YAG laser to perform a posterior capsulotomy. If the wavelength of the laser used is 1064 nm (infrared), and the pulse duration is 2.00 x 10–6 s whose energy is 0.245 J per pulse, how many photons are produced in each pulse?
Answers
Answer: 1.311 × 10^18 photons are produced in each pulse
Explanation: Please see the attachments below
Compounds A and BB are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 gg of sulfur with 5.99 gg of oxygen, and compound BB results from combining 8.60 gg of sulfur with 12.88 gg of oxygen. Show that the mass ratios in the two compounds are simple multiples of each other.
Answers
Answer:
Mass ratio of sulfur and oxygen in compounds A and B is 3:2 which confirms that the mass ratios in the two compounds are simple multiples of each other
Explanation:
This question seeks to establish/confirm the law of multiple proportions which posits that elements combine to form different substances which are whole number multiples of each other. Best example of this plays out in the formation of several oxides of the same element. Looking at the ratio in which the elements combine in each of the oxides, we can assume that these ratios are simple whole number multiples of each other.
Now back to the question.
In substance A, we have 6 g of sulfur combining with 5.99 g of oxygen
Now, lest us calculate the ratio of the mass of sulfur to that of oxygen = 6g/5.99g = 1
Now let us calculate the mass ratio of sulfur to oxygen in the second compound = 8.6/12.88 = 0.668
Now the ratios in both compounds are 1 to 0.668. 0.668 to fraction is approximately 1/1.5.
So therefore, the ratio we are having would be 1:1/1.5 or 1:0.668
This is same as 1/(1÷1.5) which is 1.5/1 or simply 3/2
This gives a ratio of approximately 1.5 to 1 or 3 to 2
The ratio 3 to 2 indicates that the mass ratios in both com pounds are simple multiples of each other
Indicate whether the following represents a Chemical or Physical change: Milk sours
Answers
Answer:
Chemical Change
Explanation:
Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.
The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by:
Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)
ΔH rxn = 1676
If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?
Answers
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
What is a thermochemical equation?A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.
Step 1: Write the thermochemical equation.Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g) ΔH rxn = 1676 kJ
Step 2: Calculate the moles of Al formed when 1.000 × 10³ kJ of heat is transferred.According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.
1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al
Step 3: Calculate the mass corresponding to 1.193 moles of AlThe molar mass of Al is 26.98 g/mol.
1.193 mol × 26.98 g/mol = 32.19 g
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
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5. Rubbing alcohol is a commonly used disinfectant and has a cooling effect when applied to the skin. The active ingredient in rubbing alcohol is isopropanol. In drugstores, the most common concentration of rubbing alcohol sold contains 70% (vol/vol) isopropanol in water. Assuming the rubbing alcohol manufacturer uses a 100% isopropanol solution, what volume of pure isopropanol is required to produce a 200-mL bottle of rubbing alcohol
Answers
Answer:
Explanation:
70% (vol/vol) means
cotnaimns 70 %(vol/vol) 70 ml of isoprapnol is there in 100 ml of Rubbing sold alcohol.
if it is 200 ml then obvouly it has the 70*2 =140 ml of isoproanol required.
Alcohol is an organic compound that when rubbed on the skin it evaporates quickly leaving a cool effect on the skin. The reason why it evaporates is because it has loosely bound molecules and a low boiling temperature.
The volume of pure isopropanol required to produce a 200-ml bottle of rubbing alcohol is 140 ml
From the question:
Alcohol sold contains 70%(vol/vol). This means 70 ml of the solute of isopropanol can be found in 100 ml of solution.
Hence:
100ml of solution = 70ml of isopropanol
200ml of solution = ?
Cross Multiply
200 ml x 70 ml / 100 ml
= 140 ml
Therefore, the volume of pure isopropanol required to produce a 200-ml bottle of rubbing alcohol is 140 ml
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An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experiment, 13.19 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction
Answers
Answer:
93.15 %
Explanation:
We have to start with the chemical reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]
Now, we can balance the reaction:
[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]
Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:
1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).
2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).
3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).
[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]
Finally, we can calculate the yield percent:
[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]
I hope it helps!
The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%
We'll begin by writing the balanced equation for the reaction. This is given below: [tex]Na_{2}CO_{3} + CaCl_{2} - > CaCO_{3} + 2NaCl[/tex]Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol
Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
SUMMARY
From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃
Next, we shall determine the theoretical yield of of CaCO₃. This can be obtained as follow:From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃.
Therefore,
15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.
Thus, the theoretical yield of of CaCO₃ is 14.15 g
Finally, we shall determine the percentage yield. This can be obtained as follow:Actual yield of CaCO₃ = 13.19 g
Theoretical yield of CaCO₃ = 14.15 g
Percentage yield =?[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]
= 93.2%Therefore, the percentage yield of the reaction is 93.2%
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a binary ionic compound is made of two components name one of them
Answers
Answer:
CATION
Explanation:
It's one is the action and the mother is a cation.
Propane (C3H8) burns in a combustion reaction. How many grams of C3H8 are needed to produce 80.3 mols CO2 ?
Answers
Answer:
1177.88g of C3H8
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Next we shall determine the number of mole of C3H8 required to produce 80.3 moles of CO2. This is illustrated below:
From the balanced equation above,
1 mole of C3H8 reacted to produce 3 moles of CO2.
Therefore, Xmol of C3H8 will react to produce 80.3 moles of CO2 i.e
Xmol of C3H8 = 80.3/3
Xmol of C3H8 = 26.77 moles
Finally, we shall convert 26.77 moles of C3H8 to grams.
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Mole of C3H8 = 26.77 moles
Mass of C3H8 =..?
Mass = mole x molar mass
Mass of C3H8 = 26.77 x 44
Mass of C3H8 = 1177.88g
Therefore, 1177.88g of C3H8 are needed for the reaction
A maple tree could be studied in many fields of science. What aspects of a maple tree might be studied in chemistry?
Answers
Answer:
Chemical reactions, kinetics, organic chemistry
Explanation:
You might study the chemical reaction, learn about the differences between products and reactants, about delta H and exothermic and endothermic reactions. You may also study Kinetics by studying the rates of reactions with certain chemicals in a maple's enzymatic processes.
Another thing that you might learn about is organic chemistry. The glucose molecules, carbohydrates, lipids, nucleic acids, all have a structure based on the Carbon atom. You can learn about the specific structures of some chemicals that are involved in photosynthesis and simple hydrocarbons that are involved in photosynthetic/bio-synthetic pathways.
There's probably a lot more - but these are the most basic things I could think of.
In the activity, click on the Keq and ΔG∘ quantities to observe how they are related. Calculate ΔG∘using this relationship and the equilibrium constant (Keq) obtained in Part A at T=298K:Keq=1.24×1020Express the Gibbs free energy (ΔG∘) in joules to three significant figures.
Answers
Answer: The Gibbs free energy of the reaction is -114629.4 J
Explanation:
To calculate the Gibbs free energy of the reaction, we use the equation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy of the reaction = ?
R = Gas constant = [tex]8.314 J/K.mol[/tex]
T = temperature of the reaction = 298 K
[tex]K_{eq}[/tex] = equilibrium constant of the reaction = [tex]1.24\times 10^{20}[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o=-(8.314J/mol.K\times 298K\times \ln (1.24\times 10^{20}))\\\\\Delta G^o=-114629.4J[/tex]
Hence, the Gibbs free energy of the reaction is -114629.4 J
Benzene can be converted to 1,3,5-tribromobenzene in five reaction steps and four intermediate compounds. Select the appropriate reagent from the followings.
Br2, R2O2
CH3Cl, AlCl3
CH3COCl, AlCl3
NaNO2, HCl
HNO3, H2SO4
H3PO2
H3PO4
KMnO4
Answers
Answer:
The appropriate reagent is: H3PO2.
Explanation:
H3PO2 is in charge of eliminating the amino group by diazotization, remember that the amino group had previously achieved bromination at positions m; that is to say that it achieved in the beginning that the three bromine atoms of 1,2,4 tribromobenzene were introduced in the meta positions among themselves, which finally corresponds as part of the last reaction to the 1,3,5-tribromobenzene position.
A solution of pentane and ethanol (CH3CH2OH)that is 50.% pentane by mass is boiling at 57.2°C. The vapor is collected and cooled until it condenses to form a new solution.
Calculate the percent by mass of pentane in the new solution. Here's some data you may need:
normal boiling point density vapor pressure at
57.2°C
pentane 36.°C 0.63gmL 1439.torr
ethanol 78.°C 0.79gmL 326.torr
Be sure your answer has 2 significant digits.
dont round during math only for answer!
Note for advanced students: you may assume the solution and vapor above it are ideal.
Answers
Answer:
The correct answer is 81.52 percent.
Explanation:
Based on the given information, the boiling point of pentane is 36 degree C and the boiling point of ethanol is 78 degree C. The density of pentane and ethanol is 0.63 g/ml and 0.79 g/ml. The vapor pressure of pentane at 57.2 degree C is 1439 torr and the vapor pressure of ethanol at 57.2 degree C is 326 torr.
In the given case, 50 percent pentane by mass signifies that mass of pentane is 50 grams. Thus, the mass of ethanol will be 100-50 = 50 grams.
The moles or n can be calculated by using the formula,
n = weight/molecular mass
The molecular mass of pentane is 72.15 g per mol and the molar mass of ethanol is 46.07 g/mol.
The moles of pentane is,
= 50 g/72.15 g/mol = 0.6930 mol
The moles of ethanol is,
= 50 g/46.07 g/mol = 1.0853 mol
The mole fraction of pentane is,
= 0.6930 mol / (0.6930 + 1.0853) mol = 0.3897
The mole fraction of ethanol is,
= 1.0853 mol / (0.6930 + 1.0853) mol = 0.6103
Now the vapor pressure of solution will be,
= pressure of pentane * mole fraction of pentane + pressure of ethanol * mole fraction of ethanol
= (1439 * 0.3897) + (326 * 0.6103)
= 759.736 torr
The vapor pressure of pentane within the solution,
= vapor pressure of pentane * mole fraction of pentane
= 1439 torr * 0.3897
= 560.778 torr
The fraction of pentane is,
= 560.778 / 759.736 = 0.738
Let us assume that the total mole is 1, the mole fraction of pentane is 0.738, so the mole fraction of ethanol will become, 1-0.738 = 0.262
The mass of pentane = 0.738 * 72.15 = 53.2467
The mass of ethanol = 0.262 * 46.07 = 12.07034
The percent by mass of pentane in new solution will be,
Mass% = mass of pentane/Total mass * 100%
= 53.2467/(53.2467 + 12.07034) * 100%
= 53.2467/65.31704 * 100 %
= 81.52 %
A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Litre volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-
Answers
Answer:
[tex]M=0.213M[/tex]
Explanation:
Hello,
In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:
[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]
[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]
[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]
We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:
[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]
Finally, we compute the molarity:
[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]
Regards.
Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )
Answers
Answer:
[tex]\large \boxed{\text{-486 kJ}}[/tex]
Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
2H₂ + O₂ ⟶ 2H-O-H
Bonds: 2H-H 1O=O 4H-O
D/kJ·mol⁻¹: 436 498 464
[tex]\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.[/tex]
A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to rise? The molecules in the water move closer together. The molecules in the thermometer’s liquid spread apart. The kinetic energy of the water molecules decreases. The kinetic energy of the thermometer’s liquid molecules decreases.
Answers
Answer: The molecules in the thermometer's liquid spread apart.
Explanation:
Mercury is the only metal that remains liquid at room temperature. It has a high coefficient of expansion therefore the its level rises when exposed to a temperature range. It can detect a slight change in temperature. It has a high boiling point.
When the thermometer is placed in the water to measure the temperature, the molecules of thermometer liquid that is mercury only will spread due to high coefficient of expansion. This can be seen as rise in temperature.
Answer:
B
Explanation:
Just did the test
Which of these statements gives a correct reason as to why our body needs water?
(1 Point)
1. It provides us with energy.
2. It helps us to eliminate waste.
3. It helps regulate our body temperature.
Answers
Answer:
2. It helps us to eliminate waste
3. It helps regulate our body temperature
Explanation:
In addition to the function of bringing nutrients to the cells, water provides the elimination of substances out of the body. This occurs, for example, through urine, which is basically formed by water and toxic or excess substances dissolved.
Water also helps in regulating body temperature. This occurs when the heat becomes exaggerated, sweat is released, which has water in its composition. When in contact with the medium, the sweat evaporates on the surface of the skin, causing the body to cool.
given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43
Answers
Answer:
Less than 0.033 M:
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Explanation:
Hello,
In this case, the described equilibrium is:
[tex]2A+B\rightarrow 2Z[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]
Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]
Which has the following solution:
[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]
But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).
Regards.
When pressure is increased on the following equilibrium, where will the shift be? 3H2 + N2 2NH3
Answers
Answer:
Explanation:
it is based on le chatliers principles
the left side of reaction you have 4 moles , where as at the right hand side you have 2 moles,,,,
so when you increase the pressure the reaction will shift towards the lower moles producing reaction that is reaction move towards forward in you case.
what is the equation for "acid dissociation constant" of "carbonic acid"
Answers
Answer:
H2CO3 = 2H+ + CO3-
Explanation:
It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.
What happens in a double replacement reaction
Answers
Answer: D
Explanation: The elements in two compunds switch places
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Determine how much heat (in kJ) is produced by the decomposition of 1.71 mol of H2O2 under standard conditions.
Answers
Answer:
[tex]Q=-361.56kJ[/tex]
Explanation:
Hello,
In this case, the decomposition of hydrogen peroxide is given by:
[tex]2H_2O_2\rightarrow 2H_2O+O_2[/tex]
Which occurs in gaseous phase, therefore the enthalpy of reaction is:
[tex]\Delta _rH=2\Delta _fH_{H_2O}-2\Delta _fH_{H_2O_2}[/tex]
Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:
[tex]\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol[/tex]
Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:
[tex]Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ[/tex]
Whose sign means a released heat.
Regards.
The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answers
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
2. In a paper chromatography analysis, three pigments, A, B, and C, were dissolved in a polar solvent. A is slightly polar, B is highly polar, and C is moderately polar. List in order how these will appear on the surface of the chromatography
Answers
how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure
Answers
Answer:
1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.
Explanation:
To determine the absolute age of rocks and fossils, geologists use _____.
Answers
Answer:
The rates of decay of radioactive elements
Explanation:
The age of a rock in years is called its absolute age. Geologists find absolute ages by measuring the amount of certain radioactive elements in the rock. When rocks are formed, small amounts of radioactive elements usually get included.
