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3 years ago

What is the molarity of a solution made by dissolving 8.60 g of a solid with a

Chemistry

1 answer:

Dima020 [189]3 years ago

8 0

Answer:

1.43 M

Explanation:

We'll begin by calculating the number of mole of the solid. This can be obtained as follow:

Mass of solid = 8.60 g

Molar mass of solid = 21.50 g/mol

Mole of solid =?

Mole = mass / molar mass

Mole of solid = 8.60 / 21.50

Mole of solid = 0.4 mole

Next, we shall convert 280 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

280 mL = 280 mL × 1 L / 1000 mL

280 mL = 0.28 L

Thus, 280 mL is equivalent to 0.28 L.

Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:

Mole of solid = 0.4 mole

Volume = 0.28 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.4 / 0.28

Molarity = 1.43 M

Thus, the molarity of the solution is 1.43 M.

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3 years ago

How many moles of HCl are in 30.00mL of a 0.1000M HCl solution? A. 0.003000mol. B. 300.0mol C. 0.03000mol D. 3.000mol

Katyanochek1 [597]

Given the volume of HCl solution = 30.00 mL

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Molarity, moles and volume are related by the equation:

Molarity = $\frac{Moles of solute}{Volume of solution (L)}$

Converting volume of HCl from mL to L:

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Calculating moles of HCl from volume in L and molarity:

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3 years ago

If you have 3 moles of a gas at a pressure of 2.5 atm and a volume of 8 liters, what is the temperature?

fomenos

Answer:

Explanation:

pv=nrt

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3 years ago

Read 2 more answers

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

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Given the following parameters

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$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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