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harkovskaia [24]
3 years ago
9

How many moles of silver (Ag) are present in a sample of 3.8 × 1024 atoms Ag?

Chemistry
2 answers:
const2013 [10]3 years ago
4 0

Answer : The correct option is, (C) 6.3 mole Ag

Explanation : Given,

Number of atoms of Ag = 3.8\times 10^{24}

As we know that, 1 mole contains 6.022\times 10^{23} number of atoms.

As, 6.022\times 10^{23} number Ag atoms present in 1 mole of Ag

So, 3.8\times 10^{24} number Ag atoms present in \frac{3.8\times 10^{24}}{6.022\times 10^{23}}=6.3 mole of Ag

Therefore, the number of moles of Ag present are, 6.3 moles

xz_007 [3.2K]3 years ago
3 0

1 mol=6.022 x 10²³ atoms.

1 mol-----------------------------6.022 * 10²³ atoms
   x--------------------------------3.8 * 10²⁴ atoms

x=(1 mol * 3.8 * 10²⁴ atoms) / 6.022*10²³ atoms=6.3 moles of Ag,

Answer: C. 6.3 mol Ag
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if the mass of 191 grams NaCl reacted with 74 frams of calcium hydroxide and 80 grams of sodium hydroxide is produced, what mass
nadya68 [22]
<h3>Answer:</h3>

110.98 g/mol

<h3>Explanation:</h3>

The reaction between NaCl and Ca(OH)₂ is given by the equation;

2NaCl(aq) + Ca(OH)₂(s) → 2NaOH(aq) + CaCl₂(aq)

We are required to determine the mass of CaCl₂ produced,

We will use the following simple steps;

Step 1: Moles of NaCl and Ca(OH)₂ given

Number of moles = Mass ÷ Molar mass

Moles of NaCl

Mass of NaCl = 191 g

Molar mass NaCl = 58.44 g/mol

Number of moles = 191 g ÷ 58.44 g/mol

                             = 3.268 moles

                             = 3.27 Moles

Moles of Ca(OH)₂

Mass of Ca(OH)₂ = 74 g

Molar mass of Ca(OH)₂ = 74.093 g/mol

Number of moles = 74 g ÷ 74.093 g/mol

                             = 0.998 mole

                              = 1.0 mole

However, from the equation  2 moles of NaCl requires 1 mole of Ca(OH)₂

Therefore, from the amount of reactants available NaCl was in excess and Ca(OH)₂ is the limiting reactant .

Step 2: Moles of CaCl₂ produced

From the equation

1 mole of Ca(OH)₂ reacts with NaCl to produce 1 mole of CaCl₂

Therefore; the mole ratio of Ca(OH)₂ to CaCl₂ is 1: 1

Thus;

Moles of CaCl₂ produced is 1.0 moles

Step 3: Mass of CaCl₂ produced

Moles of CaCl₂ = 1.0 mole

Molar mass CaCl₂ = 110.98 g/mol

But; mass = number of moles × Molar mass

Therefore;

Mass of CaCl₂ = 1.0 mole × 110.98 g/mol

                       = 110.98 g CaCl₂

3 0
3 years ago
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