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TEA [102]
3 years ago
5

Define a function ComputeGasVolume that returns the volume of a gas given parameters pressure, temperature, and moles. Use the g

as equation PV = nRT, where P is pressure in Pascals, V is volume in cubic meters, n is number of moles, R is the gas constant 8.3144621 ( J / (mol*K)), and T is temperature in Kelvin.
Chemistry
1 answer:
andre [41]3 years ago
7 0

Answer: A function can be defined as a relation in which one thing is dependent on another for its value.

Explanation: Given R = 8.314J/mol*k

PV = nRT

V = nRT/ P

V = 8.314RT / P (cm^3)

Volume of gas =(( 8.314 * R* T) / P ) cm3

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A sample of helium gas has a volume of 620. Ml at a temp of 500 K. If we decrease the temperature to 100K while keeping the pres
Ne4ueva [31]

Answer:

A sample of helium gas has a volume of 620mL at a temperature of 500 K. If we ... to 100 K while keeping the pressure constant, what will the new volume be?

Explanation:

3 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
Chemistry is: a) the mystical search for the elixir of life. b) speculation about the nature of matter. c) the study of matter a
Yuki888 [10]

Answer:

c) the study of matter and the changes it undergoes

Explanation:

Chemistry -

It is the sub topic of science , dealing with the study of matter , the properties of matter , the type of interaction between the particles of matter , the reason for the particle of matter to combine and separate in order to form new substance .

The basic concepts of chemistry are applicable on the day to day activities.

Hence, from the given options , the correct statement is c) the study of matter and the changes it undergoes.

7 0
3 years ago
NEED HELP ASAP PLEASE
frutty [35]
C. NO2 is covalent. Hope this helps!
6 0
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Read 2 more answers
If you start with 10ml of 0.75 m cu(no3)2 how much cu (s) in grams should be recovered in step #7
LenaWriter [7]
<span>0.48 grams. Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced. To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So 0.0075 mol * 63.546 g/mol = 0.476595 g. Round the results to 2 significant figures, giving 0.48 grams.</span>
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3 years ago
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