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4 years ago

if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?

Chemistry

1 answer:

ipn [44]4 years ago

6 0

Answer:

M of HI = 5.4 M.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HI = (XMV) Ca(OH)₂.

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = 5.4 M.

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vesna_86 [32]

Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

$2AgClO_3\rightarrow 2AgCl+3O_2$

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in $\frac{0.466}{22.4}=0.0208$ mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from $\frac{2}{3}\times 0.0208=0.01387$ moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

$\text{Mass of }AgClO_3=\text{Moles of }AgClO_3\times \text{Molar mass of }AgClO_3$

Molar mass of silver chlorate = 191.32 g/mole

$\text{Mass of }AgClO_3=0.01387mole\times 191.32g/mole=2.654g$

Therefore, the mass of silver chlorate will be 2.654 grams.

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3 years ago

3. If a sound is louder or a light is brighter, what does this mean about the amplitude of

hodyreva [135]

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According to the law of definite proportions, if a sample of a compound contains 24.00 grams of carbon and 4.00 grams of hydroge

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2 years ago

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Sergio [31]

Answer:

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