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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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Answer:

FALSE

Since 0.385 < 0.526, the value for week 3 is accepted.

Explanation:

Qexp = (|Xq - Xₙ₋₁|)/w

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3. 5.6

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Since 0.385 < 0.526, the value for week 3 is accepted.

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3 years ago
What type of energy is stored for use at a later time? Bond energy Free energy Kinetic energy Potential energy
Nataly [62]
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5 0
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How many moles are in 9.8 grams of calcium?
kondor19780726 [428]
<span>when the number of moles Ca = mass of Ca / molar mass of Ca.

and we can get the molar mass of Ca, it is = 40 g/mol

and we have already the mass of Ca (given) = 9.8 g

so, by substitution: the moles Ca = 9.8 g / 40 g/mol

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White raven [17]

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

Temperature (T) = 353 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

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PV = nRT

1 × V = 1.27 × 0.0821 × 353

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5 0
3 years ago
Calculate the mass in grams in nine molecules of CH3COOH? Please show how you got your answer.
WINSTONCH [101]
M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>

m 9CH₃COOH: 60u×9 = <u>540u</u>

<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
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</u>
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7 0
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