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2 years ago

Diamond has a density of 3.26 g/cm^3. What is the mass of a diamond that has a volume of 25 cm^3?

Chemistry

1 answer:

givi [52]2 years ago

4 0

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 25 × 3.26

We have the final answer as

Hope this helps you

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Find the density of an object that has a mass of 28 grams and a volume of 2 points<br> 38.6 cm^3. *

Shtirlitz [24]

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{28}{38.6} \\ =0.72538...$

We have the final answer as

Hope this helps you

4 0

2 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

2 years ago

After Dalton's proposed atomic theory, several atomic models were developed. Which model was the first propose the existence of

Lostsunrise [7]

Answer: Rutherford's model was the first propose of the existence of small, dense nucleus's.

Explanation: The atom, as described by Ernest Rutherford, has a tiny, massive core called the nucleus. The nucleus has a positive charge. Electrons are particles with a negative charge. Electrons orbit the nucleus. The empty space between the nucleus and the electrons takes up most of the volume of the atom.

7 0

3 years ago

What is the molarity of a solution preparedby dissolving 10.0 g of acetone, C3H6O, in enough water to make 125 mL of solution?

pentagon [3]

Answer:

1.379 M

Explanation:

The computation of the molarity of a solution is as follows;

As we know that

Molarity(M) = number of moles of solute ÷ volume of solution ( in L )

Given that

Mass = 10 grams

Molar mass of acetone(C3H6O) = 58 g/mol

moles = mass ÷ molar mass

= 10 g ÷ 58 g/mol

= 0.172414 mol

Now the molarity of the solution is

Volume of solution = 125 mL = 0.125 L

1 L = 1000 mL

so,

125 mL = (125 mL × ( 1 L ÷ 1000 mL) = 0.250 L)

Now the molarity is

= 0.172414 mol ÷ 0.125 L

= 1.379 M

7 0

2 years ago

Which of the following is a consequence of ocean acidification?

olga2289 [7]

Answer:

Explanation:

because that in or out I don't know

3 0

2 years ago