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VARVARA [1.3K]
3 years ago
8

Sodium metal and water react to form hydrogen and sodium hydroxide. If 11.96 g of sodium react with water to form 0.52 g of hydr

ogen and 20.80 g of sodium hydroxide, what mass of water was involved in the reaction
Chemistry
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

9.36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction.

2Na + 2H2O —> 2NaOH + H2

Next, we shall determine the mass of Na and H2O that reacted from the balanced equation.

This is illustrated below:

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

46g of Na reacted with 36g of H2O.

Now, we can determine the mass of H2O involved in reaction as follow:

From the balanced equation above,

46g of Na reacted with 36g of H2O.

Therefore, 11.96g of Na will react with = (11.96 x 36)/46 = 9.36g of H2O.

Therefore, 9.36g of H2O were used for the reaction.

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Answer:

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Explanation:

Given that:

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r = \dfrac{a}{2\sqrt{2}}

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Making "a" the subject of the formula:

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to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

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mass of unit cell = mass of atom × numbers of atoms per unit cell

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So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

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