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olganol [36]
3 years ago
6

What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?

Chemistry
1 answer:
Viefleur [7K]3 years ago
6 0
Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³ 
y =  \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}
\boxed{y \approx 0.04\:mol}


Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

x = 40*0.04
\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark

Answer:
The mass is 1.6 grams
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A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centr
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<h3><u>Answer;</u></h3>

<em><u> = 48,828.125 mi/hr²</u></em>

<h3><u>Explanation and solution</u>;</h3>
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4 0
3 years ago
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If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
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Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

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Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

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