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malfutka [58]
2 years ago
8

How much heat energy is required to boil 66.7 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol.

Chemistry
1 answer:
pochemuha2 years ago
6 0

Answer:

91.7 kJ

Explanation:

Step 1: Given data

  • Mass of ammonia (m): 66.7 g
  • Molar heat of vaporization of ammonia (ΔH°vap): 23.4 kJ/mol

Step 2: Calculate the moles (n) corresponding to 66.7 g of ammonia

The molar mass of ammonia is 17.03 g/mol.

66.7 g × 1 mol/17.03 g = 3.92 mol

Step 3: Calculate the heat (Q) required to boil 3.92 moles of ammonia

We will use the following expression.

Q = ΔH°vap × n

Q = 23.4 kJ/mol × 3.92 mol = 91.7 kJ

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<span>C4H4 The compound in question has an equal ratio of hydrogen and carbon. The atomic weight of carbon is roughly 12 and the atomic weight of hydrogen is roughly 1. The mass of the compound in question is roughly 52. 52/13=4 C4H4</span>
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Which best describes all new rocks that form near an oceanic ridge?
Alexxandr [17]

Answer:

a have stripped pattern

Explanation:

did it on edunuity and studied on quizlet

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Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine
Amiraneli [1.4K]

Answer:

% yield =  82.5%

Explanation:

HgO + 2Cl₂ →  HgCl₂ +  Cl₂O

Our reactants are:

  • HgO and Cl₂

Our products are:

  • HgCl₂ +  Cl₂O

We do not have information about moles of reactants, but we do know the theoretical yield and the grams of product, in this case Cl₂O, we have produced.

Percent yield = (Yield produced / Theoretical yield) . 100

Theoretical yield is the mass of product which is produced by sufficent reactant. We replace data:

% yield = (0.71 g/0.86g) . 100 = 82.5%

7 0
3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
6. What is the molecular formula for this compound? The compound's empirical formula and
igomit [66]

Explanation:

A compound's empirical formula tells you the smallest whole number ratio ,The molar mass tells you what the total mass of one mole

6 0
2 years ago
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