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Katen [24]
2 years ago
8

Find the mass in 2.6 mol of lithium bromide. (Hint:  You need to write the formula)

Chemistry
1 answer:
marissa [1.9K]2 years ago
6 0
Lithium Bromide : LiBr.

Molar mass of lithium bromide : 86.845 g/mol.

To find the mass of 2.6 mol of lithium bromide, simply multiply in order to cancel out the moles : 2.6 mol * 86.845 g / 1 mol = 225.8 grams.
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A sample of fluorine gas exerts a pressure of 900 mmHg. When the pressure is changed to 1140 mmHg, the volume is 250 mL. What wa
UNO [17]
Let's use Boyle's Law here. P1*V1 = P2*V2
Given: (assuming that there are decimals at the end for Sig Figs)
P1 = 900.mmHg
P2 = 1140.mmHg
V1 = ???
V2 = 250.mL

900.mmHg* ??? = 1140.mmHg * 250.mL
??? = 1.27*250.mL
??? = 318.mL

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8 0
2 years ago
The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this
Semmy [17]

Answer:

The molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

,=\frac{0.07}{100}\times 21.5}\\=0.0151g of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M

The chemical equation for the ionization of copper (II) sulfate follows:

CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = 4.73\times 10^{-5}M

Hence, the molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

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The answer is B.) 10^5

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