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Maru [420]
3 years ago
9

How many moles are there in 2.30 x 1024 atoms of silver?

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).

Step 1

Use dimensional analysis to convert the number of atoms to moles.

1 mole atoms = 6.022 × 10²³ atoms

n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag

Step 2

Convert the moles of Ag to mass.

mass (m) = moles (n) × molar mass (M)

n(Ag) = 3.8193 mol Ag

M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag

m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures

The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.

Explanation:

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The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
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Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.56 mol of m
pishuonlain [190]
<h3>Answer:</h3>

43.27 g Mg

<h3>Explanation:</h3>

The balanced equation for the reaction between magnesium metal and hydrochloric acid is;

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

From the equation;

1 mole of magnesium reacts with 2 moles of HCl

We are given;

3.56 moles of Mg and 3.56 moles of HCl

Using the mole ratio;

3.56 moles of Mg would react with 7.12 moles of HCl, and

3.56 moles of HCl would react with 1.78 moles of Mg

Therefore;

The amount of magnesium was in excess;

Moles of Mg left = 3.56 moles - 1.78 moles

                         = 1.78 moles

But; 1 mole of Mg = 24.305 g/mol

Therefore;

Mass of magnesium left = 1.78 moles × 24.305 g/mol

                                        = 43.2629 g

                                        = 43.27 g

Thus, the mass of magnesium that remained after the reaction is 43.27 g

4 0
3 years ago
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Answer:

I'd say it's a mixture, because you are putting different ingredients together to make something new; aka this salad.

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Answer: less mass than the water

Explanation:

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Marizza181 [45]
The correct answer should be D :)
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