All categories

4 years ago

If a 32.5 mL of a 15.1 M HCl solution was diluted to 3.25 L with water, what is the final concentration?

Chemistry

1 answer:

user100 [1]4 years ago

8 0

Answer:

The final concentration is 0,151 M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We convert the unit of volume in L into ml: 3,25 x 1000= 3250 ml

We use the formula for dilutions:

C1 x V1 = C2 x V2

C2= (C1 xV1)/V2

C2= (32, 5 ml x 15, 1 M)/ 3250 ml

C2=0,151 M

You might be interested in

Robert decides it would be fun to jump off of the back of his canoe while fishing in the lake one day. After he jumps out of the

Lady_Fox [76]

This is an example of Newton’s third law of motion, which states that for every action in nature there is an equal and opposite reaction. So when Robert jumped off his kayak, he pushed it back as he used it to launch himself, with his feet.

3 0

4 years ago

Read 2 more answers

How many unpaired electrons would you expect on iron in iron ii sulfide. enter an integer?

zimovet [89]

What are ur answers of the question

5 0

3 years ago

Read 2 more answers

) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole% O2, 79%

LiRa [457]

Answer:

Here's what I get.

Explanation:

1. Write the chemical equation

$\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O$

Assume that we start with 4 L of HCl

2. Calculate the theoretical volume of oxygen

$\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}$

3. Add 35% excess

$\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}$

4. Calculate the theoretical volume of nitrogen

$\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}$

4. Calculate volumes of reactant used up

Only 85 % of the HCl is converted.

We can summarize the volumes in an ICE table

4HCl + O₂ + N₂ → 2Cl₂ + 2H₂O

I/L: 4 1.35 5.08 0 0

C/L: -0.85(4) -0.85(1) 0 +0.85(2) +0.85(2)

E/L: 0.60 0.50 5.08 1.70 1.70

5. Calculate the mole fractions of each gas in the product stream

Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L

$\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H$_{2}${O}}} = 0.1044\times 1.70 = 0.177\\\\$

8 0

3 years ago

HELP ASAP 20 POINTS

Colt1911 [192]

Solubility of a compound in water can be referred to as the amount of the compound that can be dissolved in 1 L of the solvent (water) at any given temperature. Solubility of a compound can be expressed in the units of g/L or mg/L.

Given that the solubility of calcium carbonate in water = 14 mg/L

We have to calculate the volume of water that can dissolve 11 g of calcium carbonate.

Converting 11 g calcium carbonate to mg:

$11g *\frac{1000mg}{1g} =11,000 mg$

Volume of water that would dissolve 11000 mg calcium carbonate

= $11000 mg CaCO_{3}*\frac{1L}{14 mg}$

=785.7 L

Rounding the volume 785.7 L to two significant figures, we get 790 L water.

Therefore, we would need 790 L water to completely dissolve 11 g of calcium carbonate.

4 0

3 years ago

Two water solutions that have a density of

Colt1911 [192]

<h2>Answer : 627 J</h2><h2>Brainliest</h2>

5 0

3 years ago