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2 years ago

Which of the following is a basic solution? HCl dissolved in water b. household ammonia c. vinegar d. pure water

Chemistry

1 answer:

PSYCHO15rus [73]2 years ago

7 0

Answer:

B. Household ammonia.

Explanation:

NH₃ is a base, so the solution always will be basic.

NH₃ takes the proton from the water. In conclusion we have free OH⁻ in medium, that's why the solution is basic.

NH₃ + H₂O → NH₄⁺ + OH⁻

- HCl + H₂O → H₃O⁺ + Cl⁻

This reaction makes an acid solution, cause the H₃O⁺

- Vinegar is a compound made of acetic acid.

CH₃COOH + H₂O ⇄ H₃O⁺ + CH₃COO⁻

This reaction also makes an acid solution, cause the H₃O⁺

- Pure water makes neutral solution. It is not acid, neither basic.

2H₂O ⇄ H₃O⁺ + OH⁻

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Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M

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CH₃Br+NaOH⟶CH₃OH+NaBr

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rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x .12 x .12

k = .55555

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rate of reaction = .555555 x [.24] [ .12 ]

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3 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

Best regards.

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3 years ago