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2 years ago

10

If there is a graduated cylinder containing 20mL of water and a small rock is gently placed inside it, and the water level insid

e the graduated cylinder rises to 25mL...what is the volume of the small rock?

2 answers:

LekaFEV [45]2 years ago

8 0

Answer:

5

Explanation:

just subtract 20-25 and 5 is left over which is how much the water when up with the rock.

Vanyuwa [196]2 years ago

7 0

Answer:

5

Explanation:

quick maths

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Option B

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3 years ago

If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc

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Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

$Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)$

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

$c_{HCl}=\frac{n_{HCl}}{V_{HCl}}$

Rearranging for moles of HCl, n:

$n_{HCl}=c_{HCl}V_{HCl}$

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

$n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}$

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

$n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}$

Using the last two equations, we see that:

$\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}$

Substitute the given data, as well as the molar mass of calcium hydroxide:

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