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Marysya12 [62]
3 years ago
5

1. How many grams would 8.1 x 1021 molecules of sucrose (C12H22011) weigh?

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
7 0

Answer:

Mass = 4.6 g

Explanation:

Given data:

Number of molecules of sucrose = 8.1 ×10²¹ molecules

Mass of sucrose = ?

Solution:

First of all we will calculate the number of moles by using Avogadro number.

1 mole × 8.1 ×10²¹ molecules / 6.022×10²³ molecules

1.35 × 10⁻² mol

Mass of sucrose:

Mass = number of moles × molar mass

Molar mass = 342.3 g/mol

Mass = 1.35 × 10⁻² mol ×342.3 g/mol

Mass = 462.1  × 10⁻² g

Mass = 4.6 g

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7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m
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The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

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Mole ratio of the base, NaOH (n_{B}) = 2

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH (M_{B}) = 0.505 M

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Molarity of the acid, H₂SO₄ (M_{A}) = 0.505 M

<h3>Volume of the base, NaOH (V_{B}) =? </h3>

\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}

Cross multiply

0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4

Divide both side by 0.505

V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

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