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strojnjashka [21]
3 years ago
9

A sample of marble has a volume of 8 cm3 and a density of 2.75 g/cm3. What is its mass?

Chemistry
2 answers:
Firlakuza [10]3 years ago
7 0
Using the formula Density = Mass / Volume,

Mass = Density x Volume
Mass = 2.75 x 8 = 22g
Arte-miy333 [17]3 years ago
6 0

Answer: The correct answer is 22 g.

Explanation:

The expression for the relation between mass and density is as follows;

Density=\frac{m}{V}

Here, m is the mass of the object and V is the volume of the object.

It is given in the problem that A sample of marble has a volume of 8 cm^{3} and a density of  2.75 \frac{g}{cm^{3}}.

Calculate the mass of a sample marble.

m= (Density)(V)

Put V=8 cm^{3} and Density=2.75 \frac{g}{cm^{3}}.

m= (2.75)(8)

m=22 g

Therefore, the mass of a sample of a marble is 22 g.

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A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is
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Answer:

[H^{+}] = 0.761 \frac{mol}{L}

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

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Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L)  \\ n_{H^{+} } from HNO_{3}  = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

n_{H^{+} } from HCl = (5.00)(0.093)

n_{H^{+} } from HCl = 0.465 mol

n_{H^{+} } from HNO_{3}  = (8.00)(0.037)

n_{H^{+} } from HNO_{3}  = 0.296 mol

n_{H^{+}(total) } = 0.296 + 0.465

n_{H^{+}(total) } = 0.761 mol

For molar concentration of hydrogen ions:

[H^{+}]  = \frac{n_{H^{+}}(mol)}{V(L)}

[H^{+}] = \frac{0.761}{1.00}

[H^{+}] = 0.761 \frac{mol}{L}

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

K_{w} = [H^{+} ][OH^{-} ]

[OH^{-}]=\frac{Kw}{[H^{+}] }

[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

The pH of the solution can be measured by the following formula:

pH = -log[H^{+} ]

pH = -log(0.761)

pH = 0.119

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