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2 years ago

Given the following information:

Chemistry

1 answer:

bonufazy [111]2 years ago

6 0

Answer:

Ya'll should really practice more because things like these are easy

Explanation:

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

Montano1993 [528]

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:

$m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g$

8 0

3 years ago

A 0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+]? Calculate BrCH2COO^ - ? Calculate BrCH2COOH

Natali [406]

<u>Answer:</u> The concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

<u>Explanation:</u>

We are given:

Molarity of bromoacetic acid = 0.100 M

Percent of ionization = 13.2 %

The chemical equation for the ionization of bromoacetic acid follows:

$BrCH_2COOH\rightarrow BrCH_2COO^-+H^+$

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion

Molarity of hydrogen ion = 13.2 % of 0.100 = $\frac{13.2}{100}\times 0.100=0.0132M$

Molarity of bromoacetate ion = molarity of hydrogen ion = 0.0132 M

Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance

Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M

Hence, the concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

8 0

3 years ago

Match the substance with its chemical formula.

ch4aika [34]

H 3 0 + is hydronium ion. OH- is hydroxide ion. H+ is hydrogen ion

4 0

3 years ago

Read 2 more answers

im making a poster for chemistry. the topic is acid and base. i have to make a creative title to go with the poster. any ideas?

REY [17]

The correct answer for the question that is being presented above is this one: Right now, I haven't seen the poster but there are things that i have in mind. It can be "The Game of Acids and Bases," or you can another one like, "Unity Amidst Diversity."

3 0

3 years ago

AJUTOR VA ROG AM TEST SI MAI AM 10 MINUTE AM NEVOIE DE SUBPUNCTUL C SI D

BabaBlast [244]

Explanation:

desk I have a lot of work and I will send it in the one

7 0

3 years ago