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mr Goodwill [35]
3 years ago
14

Sulfur is a nonmetalic mineral that has a density of 2 g/cm3. The volume of a sample of sulfur was measured to be 5.0 cm3. what

is the mass of the sample
Chemistry
1 answer:
taurus [48]3 years ago
8 0

Answer:

<h2>10 g</h2>

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 = 2 × 5

We have the final answer as

<h3>10 g</h3>

Hope this helps you

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

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7 0
3 years ago
Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

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