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3 years ago

If the [H+] in a solution is 1 × 10–1 mol/L, what is the [OH–]? Show your work.

1 answer:

3 0

Hello!

The basic equations to solve this is
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
------------------------------------------------------------------------------------------------------
Find pH

pH = -log(1 * 10^-1)
pH = 1
------------------------------------------------------------------------------------------------------
Find pOH
1 + pOH = 14

pOH = 13
------------------------------------------------------------------------------------------------------
Find OH-
[OH-] = 10^(-pOH)
[OH-] = 1 * 10^-13mo/L

The answer is $[OH-] = 1 * 10^{-13} mol/L$

Hope this helps!

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As long as there is some residual liquid present after equilibrium is reached, the vapor pressure of a liquid at any given tempe

stepan [7]

Answer:

True

Explanation:

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Examples of intensive properties include temperature, density, vapor pressure and viscosity.

Assuming that there is some residual liquid left after equilibrium is reached, no matter how much liquid is present, at any given temperature, the vapor pressure will be the same because it is an intensive property.

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3 years ago

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

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Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: $V_{1}$ = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

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As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$

Substitute the values into above formula as follows.

$[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M$

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

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3 years ago

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Answer:

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3 years ago

5. Which of the following elements will have a charge of 4+ or 4- as an ion?

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Answer:

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Explanation:

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