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3 years ago

How much heat is released when you condense 93.9 g of water vapor? Show work please ❤️

Chemistry

1 answer:

sammy [17]3 years ago

7 0

Answer:

211.9 J

Explanation:

The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.

The amount of heat released can be represented by the formula:

$Q=mL_e$

where $Q$ = heat energy, $m$ = mass of water and $L$ = latent heat of evaporation.

The latent heat of evaporation for water is $L=2257 kJ/kg$ and the mass of the water is $m=93.9 g=0.0939 kg$ .

The amount of heat released in this process is:

$Q=mL_e = (0.0939kg)(2257 kJ/kg)=$ 211.9 J

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The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b

____ [38]

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

$\Delta S_{vap}=\frac{\Delta H_{vap}}{T}$

We can solve for the temperature as follows:

$T=\frac{\Delta H_{vap}}{\Delta S_{vap}}$

Thus, with the proper units, we obtain:

$T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C$

Hence, answer is approximately 100 °C.

Best regards.

6 0

3 years ago

The table shows a chemical equation and the dimensional analysis used by a student to calculate the number of moles of Ba3N2 req

UNO [17]

This problem is providing the chemical reaction whereby barium nitride reacts with water to produce barium hydroxide and ammonia, so the number of moles of barium nitride are required in order to produce 8.3 moles of ammonia. It asks for us to evaluate the student's setup, so we conclude the answer is C. "1 mol of NH3 should be replaced with 2 mol of NH3", according to:

<h3>Mole ratios:</h3>

In chemistry, stoichiometric calculations are used to figure out the moles or mass of a substance, given information about another one in the reaction. In this case, for the given chemical equation:

$Ba_3N_2 + 6H_2O \rightarrow 3Ba(OH)_2 + 2NH_3$

We evidence a 1:2 mole ratio of barium nitride to ammonia, for that reason, the student's setup:

$8.3 mol NH_3 * \frac{1 mol Ba_3N_2 }{1 mol NH_3}$

Is incorrect, because the ammonia must be accompanied by a 2 rather than the 1 it is given there:

$8.3 mol NH_3 * \frac{1 mol Ba_3N_2 }{2mol NH_3}$

Thereby, the correct answer is C. "1 mol of NH3 should be replaced with 2 mol of NH3"

Learn more about mole ratios: brainly.com/question/15288923

5 0

2 years ago

How many mols are in 15 of mg (0H) 2

SpyIntel [72]

The answer is 0.0171468704904. We assume you are converting between moles Mg(OH)2 and gram. This compound is also known as Magnesium Hydroxide. 1 mole is equal to 1 moles Mg(OH)2, or 58.31968 grams.

5 0

3 years ago

Calculate the percent composition of the following: A. VO3 B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

4 0

3 years ago

Determine the empirical formula for a compound containing 85.6 g of carbon and 14.4 g of hydrogen. Please show work

photoshop1234 [79]

Answer: $C_{1}H_{2}$

Explanation:

take gram & convert to mol (using the mm of the element) ,

85.6g C $\frac{1 mol }{12.01g}$ = 7.12 14.4g H $\frac{1 mol}{1.01g}$ =14.2 then take the smallest number & divide 7.12/7.12 =1 14.2/7.12= 2. since the numbers are in whole numbers the answer is $C_{1}H_{2}$ .

4 0

2 years ago