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2 years ago

You have 3 L of a 5M solution of calcium chloride. How many L would you be able to make of a 0.5M solution?

Chemistry

1 answer:

san4es73 [151]2 years ago

3 0

Answer:

30.0 L.

Explanation:

To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution = (MV)after dilution</em>

M before dilution = 5.0 M, V before dilution = 3.0 L.

M after dilution = 0.5 M, V after dilution = ??? L.

∵ (MV)before dilution = (MV)after dilution

∴ (5.0 M)(3.0 L) = (0.5 M)(V after dilution)

<em>∴ V after dilution = (5.0 M)(3.0 L)/(0.5 M) = 30.0 L.</em>

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2 years ago

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vfiekz [6]

6.02 x10^23 atom

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3 years ago

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3 years ago

: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be

Jet001 [13]

Answer: New concentration of $HC_{6}H_{5}O^{2-}_{7}$ is 0.23 M.

Explanation:

The given data is as follows.

Moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 mol

Moles of $H_{2}C_{6}H_{5}O_{7}$ = 0.08 mol

Therefore, moles of $OH^{-}$ added are as follows.

Moles of $OH^{-}$ = $0.125 \times \frac{25}{1000}$

= 0.003125 mol

Now, new moles of $HC_{6}H_{5}O^{2-}_{7}$ = 0.06 + 0.003125

= 0.063125

Therefore, new concentration of $HC_{6}H_{5}O^{2-}_{7}$ will be calculated as follows.

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2 years ago

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AlexFokin [52]

Answer:

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Explanation:

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3 years ago