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3 years ago

Calculate the mass in atomic mass units of 0.25 mol of carbon-12 atoms

Chemistry

2 answers:

liraira [26]3 years ago

8 0

Each C-12 atom weights 12 amu
1mol = 6.02*10^23 atoms
6.02*10^23 * 0.25 * 12 = 1.806*10^24

Semmy [17]3 years ago

4 0

Answer : The mass of carbon-12 atoms is, 3 amu

Solution : Given,

Moles of carbon-12 atoms = 0.25 moles

The molar mass of carbon-12 atoms = 12 g/mole = 12 amu

(1 amu = 1 gram/mole)

Now we have to calculate the mass of carbon-12 atoms.

$\text{Mass of carbon-12}=\text{Moles of carbon-12}\times \text{Molar mass of carbon-12}$

Now put all the given values in this formula, we get the mass of carbon-12 atoms.

$\text{Mass of carbon-12}=(0.25moles)\times (12amu)=3amu$

Therefore, the mass of carbon-12 atoms is, 3 amu

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Determine the concentration of a solution made by dissolving 1.40grams NaCL in enough water to make 30.0mL of solution.

KiRa [710]

Answer: 0.8M

Explanation:

Given that,

Amount of moles of NaCl (n) = ?

Mass of NaCl in grams = 1.40 g

For molar mass of NaCl, use the molar masses:

Sodium, Na = 23g;

Chlorine, Cl = 35.5g

NaCl = (23g + 35.5g)

= 58.5g/mol

Since, amount of moles = mass in grams / molar mass

n = 1.40g / 58.5g/mol

n = 0.024 mole

Now, given that:

Amount of moles of NaCl (n) = 0.024

Volume of NaCl solution (v) = 30.0mL

[Convert 30.0mL to liters

If 1000 mL = 1L

30.0mL = 30.0/1000 = 0.03L]

Concentration of NaCl solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

c = 0.024 mole / 0.03 L

c = 0.8 M (0.8M means concentration is in moles per litres)

Thus, the concentration of the solution is 0.8M

3 0

3 years ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.