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mash [69]
2 years ago
10

Gas Law

Chemistry
1 answer:
Ratling [72]2 years ago
3 0

69 mmHg pressure is exerted by oxygen if 40% of the gas pressure is exerted by the oxygen gas.

<h3>What is pressure?</h3>

Pressure is defined to be the amount of force exerted per area

40% of pressure is exerted by oxygen gas.

So, 60% of pressure will be exerted by carbon dioxide gas.

60% of 115 mmHg = 69 mmHg

Hence, 69 mmHg pressure is exerted by oxygen if 40% of the gas pressure is exerted by the oxygen gas.

Learn more about pressure here:

brainly.com/question/15175692

#SPJ1

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4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T
andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

q=c\times \Delta T

where,

c = heat capacity of calorimeter = 1620 J/K

\Delta T = change in temperature = 0.150^oC=0.150K   (Change remains same)

Putting values in above equation, we get:

q=1620J/K\times 0.15K=243J

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}

We are given:

Moles of monoclinic sulfur = 1 mole

  • To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,  

q = amount of heat released = -243 J

n = number of moles = 1 mole

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol

Hence, the enthalpy change of the reaction is -243 J/mol

8 0
3 years ago
Moises performs a chemical reaction according to the reaction
taurus [48]

Answer:

Explanation:

mass of the reactant = mass of the product

15.31 + 1.50 = 16.81g

4 0
3 years ago
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
2 years ago
Based on the information in the table, which of the following arranges the bonds in order of decreasing
tatuchka [14]

Answer:

what grade so i can help

Explanation:

3 0
2 years ago
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Noble gas notation for platinum
SOVA2 [1]

Pt

Hope this helps!!!

4 0
3 years ago
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