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4 years ago

you are given a sample containing 0.70 moles of a substance. How many atoms are in the sample if the substance is zinc?

1 answer:

7 0

1 mole ------------- 6.02 x 10²³ atoms
0.70 moles -------- ( atoms zinc )

atoms platinum = 0.70 x ( 6.02 x 10²³ ) / 1

<span> = </span>4.2 x 10²³ atoms zinc

<span>hope this helps!</span>

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How many grams are in 3.14 moles of PI₃?

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Answer:

$\boxed {\boxed {\sf 1290 \ g \ PI_3}}$

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

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Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

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<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

$\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

We want to convert 3.14 moles to grams, so we multiply by that value.

$3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

The units of moles of PI₃ cancel.

$3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }$

$1292.698 \ g\ PI_3$

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

1292.698

The 2 in the ones place tells us to leave the 9.

$1290 \ g \ PI_3$

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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

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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: $c_{Na_2CrO_4} = 0.0400 M$

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

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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