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Misha Larkins [42]
2 years ago
11

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

hin a cylinder. If the original pressure was 1.00 atm, what would the final pressure be?
Chemistry
1 answer:
Triss [41]2 years ago
4 0

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

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Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
A standard solution of 0.243 m naoh was used to determine the concentration of a hydrochloric acid solution. if 46.33 ml of naoh
MrRissso [65]
The reaction of NaOH and HCl is as follows: NaOH+HCl⇒NaCl+H2O.  So the ratio between the two reactants is NaOH:HCl=1:1.  According to the question, the moles of NaOH needed to neutralize the acid is 0.243M*0.04633L=0.01126mole. Because is the above ratio (1:1), the moles of acid that are being neutralized is also 0.01126mole.  So the concentration of the acid is 0.01126mole/0.010L = 1.126 M.
7 0
2 years ago
What will occur if solution A containing 400 mosmol/L nonpenetrating solute is separated by a biological membrane from solution
Natali [406]

Answer:

The volume of solution B will increase

Explanation:

If a solute is non-penetrating it means it cannot pass through a biological membrane. In this case we have two non-penetrating solutions with different osmolarity separated by biological membrane. Solvent from solution with lower osmolarity will tend to pass the membrane in order to equalize the solute concentrations on the two sides of the membrane. This process is called osmosis and it is spontaneous. Solvent moves through the membrane from a less concentrated solution (400 mosmol/L) into a more concentrated one (600 mosmol/L). Because of that, the volume of solution B will increase.

7 0
2 years ago
When a strong acid is titrated with a strong base using phenolphthalein as an indicator, the color changes abruptly at the endpo
kupik [55]

Answer:

the acid has been saturated with the base.

Explanation:

6 0
2 years ago
Read 2 more answers
Calculate the pOH of a solution that contains 3.9 x 10-5 M H3O+ at 25°C.
exis [7]

Answer:

Option b. 9.59

Explanation:

First, let us calculate the pH. This is illustrated below:

[H3O+] = 3.9 x 10-5 M

pH = —Log [H3O+]

pH = —Log [3.9 x 10-5]

pH = 4.41

Recall: pH + pOH = 14

4.41 + pOH = 14

Collect like terms

pOH = 14 — 4.41

pOH = 9.59

6 0
2 years ago
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