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2 years ago

What volume, in milliliters, of a 0.140 M solution of KCl contains 2.65 g of the compound?

Chemistry

1 answer:

marishachu [46]2 years ago

3 0

Answer:

Calculate moles KCl: 2.55 g / 74.55 g/mol = 0.0342 moles KCl

Volume KCl = 0.0342 mol X (1 L/0.150 mol) = 0.228 L X 1000 mL/L = 228 mL

Explanation:

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Explanation:

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What is the symbol of a ion with a 22 protons, 34 neutrons, and 18 electrons

Dominik [7]

Well knowing the number of protons, you can find what the element is, which is Ti = titanium. Then to find the charge, simply subtract 22 and 18 which is +3.

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3 years ago

Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass

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Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

$\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles$

$\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol$

According to stoichiometry :

2 moles of $NaCl$ are formed from = 2 moles of $Na$

Thus 16 moles of $NaCl$ are formed from= $\frac{2}{2}\times 16=16moles$ of $Na$

Mass of $Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g$

Thus 368 grams of sodium reacted.

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2 years ago

Which has a higher average speed of molecules 95 degrees Celsius or 95 degrees Fahrenheit

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2 years ago

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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

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2 years ago