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Ugo [173]
3 years ago
8

What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K

Chemistry
1 answer:
Ivan3 years ago
3 0

Answer:

1.67 atm.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = ??? atm).

V is the volume of the gas in L (V = 5.0 L).

n is the no. of moles of the gas in mol (n = 0.5 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 203 K).

∴ P = nRT/V = (0.5 mol)(0.0821 L.atm/mol.K)(203 K)/(5.0 L) = 1.67 atm.

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What is the final volume of the solution when 9.80 mL of a 1.56 m solution to need to be diluted to 1.27 m?
musickatia [10]

hAnswer:

Explanation:

8 0
3 years ago
An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t
Elenna [48]

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol

Thus, the concentration of ethylamine in solution is:

[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M

Now, we can also infer that some salt is formed, and has the following concentration:

[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0

Finally, the pH turns out to be:

pH=14-pOH=14-3\\\\pH=11

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

4 0
2 years ago
When forming an ion, iodine will do which of the following?
Mariana [72]

Answer:

B

Explanation:

Iodine has a negative charge, so it will gain electrons.

4 0
2 years ago
Use the given heats of formation to calculate the enthalpy change for this reaction. B2O3(g) + 3COCl2(g) →2BCl3(g) + 3CO2(g) ΔHo
UkoKoshka [18]

Answer:

the enthalpy change for this reaction is -57.7 kJ/mol

Explanation:

Given:

HB₂O₃ = -1272.8 kJ/mol

HCOCl₂ = -218.8 kJ/mol

HBCl₃ = -403.8 kJ/mol

HCO₂ = -393.5 kJ/mol

Those are all standard enthalpies

Question: Calculate the enthalpy change for this reaction, ΔHreaction = ?

The enthalpy of the reaction is calculated using the standard enthalpies of formation of both products and reagents. To understand better, the reaction is as follows

B₂O₃ + 3COCl₂ → 2BCl₃ + 3CO₂

Where the compounds on the left are the reactants and the compounds on the right are the products

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

delta(H)_{reaction} =((2*(-403.2)+(3*(-393.5))-((1*(-1272.8)+(3*(-218.8))=-57.7kJ/mol

Please be careful with the signs.

4 0
3 years ago
Help with chemistry, 50 points
expeople1 [14]

Answer:

Thank you

I didn't wait because there is someone who will take it and there is someone who answer a wrong answer

Explanation:

4 0
2 years ago
Read 2 more answers
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