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2 years ago

14

A 390 mL solutions of 4.6 M HCl is left out over the weekend and 74 mL of solvent evaporates. What is the new concentration of H

Cl?

1 answer:

storchak [24]2 years ago

8 0

Answer:

5.7 M

Explanation:

You have .390 L of 4.6 M HCl.

.390 L * 4.6 mol/L = 1.794 mol HCl

.390-.074=.316 L solvent remaining

1.794 mol/.316 L=5.7 M HCl

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What volume would 75.0g of oxygen gas occupy

seropon [69]

Answer:

Explanation: It is already known that 1 mole of the gas( or 32g of O2) is equivalent to 22.4 Litres of the oxygen gas. So, 8g is equivalent to = (22.4/32) × 8 = 5.6 L of the gas.

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Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For nickel:</u>

Given mass of nickel = 14.8 g

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Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

<u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

<u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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A medication 60 mg is ordered. the vial is labeled 1 gm/ 10 ml. How many ml would you give?

timofeeve [1]

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4 0

3 years ago

What is the mole fraction of NaCl in a mixture containing 7.21 moles NaCl, 9.37 moles KCL, and 3.42

GrogVix [38]

Answer : The mole fraction of NaCl in a mixture is, 0.360

Explanation : Given,

Moles of $NaCl$ = 7.21 mole

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Now we have to calculate the mole fraction of $NaCl$ .

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Now put all the given values in this formula, we get:

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rewona [7]

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