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zheka24 [161]
3 years ago
7

Would you be able to fit a mole of rice in a 30.0m*8.5m*3.5m room??

Chemistry
1 answer:
Julli [10]3 years ago
8 0

Answer:

No.

Explanation:

1. Volume of a rice grain

There is no standard for the size of a rice grain, so let's make an arbitrary assumption.

An average rice grain behaves as if it were a rectangular solid with dimensions 7 mm × 2 mm × 2 mm.

The volume of one rice grain is

V = lwh = 7 mm × 2 mm × 2 mm = 28 mm³

Convert to cubic metres :

V = \text{28 mm}^{3} \times \left (\dfrac{\text{1 m}}{\text{1000 mm}}\right )^{3} = \mathbf{2.8 \times 10^{-8}}\textbf{ m}^{\mathbf{3}}

2. Volume of the room

V = lwh = 30.0 m × 8.5 m × 3.5 m = 890 m³

3. Volume of a mole of rice  

V= 6.022 \times 10^{23}\text{ grains} \times \dfrac{2.8 \times 10^{-8}\text{ m}^{3}}{\text{1 grain }} = \mathbf{1.7 \times 10^{16}}\textbf{ m}^{\mathbf{3}}

4. Conclusion

The room is not big enough to hold a mole of rice.

If you work it out, you will find that it takes about

20 000 000 000 000 (twenty trillion) rooms to hold a mole of rice.

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anastassius [24]

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present =\frac{20}{171.34}

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

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From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

6 0
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