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Natali [406]
3 years ago
13

How many mL of 50% (v/v) alcohol should be mixed with 10% alcohol to obtain 50mL of 35% (v/v) alcohol? Round to the hundredths p

lace.
Chemistry
1 answer:
Furkat [3]3 years ago
8 0

Answer:

The volume of 50% alcohol used to make the mixture = 31.25 L

The volume of 10% alcohol used to make the mixture = 18.75 L

Explanation:

Let the volume of 50% alcohol used to make the mixture = x L

Let the volume of 10% alcohol used to make the mixture = y L

Total volume of the mixture = x + y = 50 L .................. (1)

For 50% alcohol:

C₁ = 50% , V₁ = x L

For 10% alcohol :

C₂ = 10% , V₂ = y L

For the resultant alcohol solution:

C₃ = 35% , V₃ = 50 L

Using  

C₁V₁ + C₂V₂ = C₃V₃

50×x + 10×y = 35×50

So,  

5x + y = 175 .................. (2)

Solving 1 and 2 we get,

<u>x = 31.25 L </u>

<u>y = 18.75 L</u>

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2 years ago
The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval
worty [1.4K]
<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

<h3><u>Explanation;</u></h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t  

The average rate of disappearance of ozone ... is found to  

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t  

rate = -(1/2)*(-7.25 × 10^–3 atm)

      = 3.625 × 10^–3 atm  

Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

              = 1.0875 x 10-2 atm over the same time interval

4 0
3 years ago
True or false?<br><br> When sodium and chlorine combine and bond, a molecule is formed.
CaHeK987 [17]

When Sodium and Chlorine come together they transfer an electron.

- Source: google


Hopefully this was clear and you understood!

3 0
3 years ago
. Ashley is identifying insects and finds a camo moth. A camo moth appears like a leaf and is able to easily blend in with its s
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5 0
3 years ago
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If one adds 0.1 mol of the weak acid hf (pk_a = 3.2) to a solution with a ph = 2, which species would be most abundant and how m
zavuch27 [327]

F^{-} is most abundant and 6310 times more than HF.

<h3>What is a strong and weak acid?</h3>

When an acid is dissolved in water, all of its molecules disintegrate, making the acid powerful.

When an acid is dissolved in water, only a small number of its molecules disintegrate, making the acid weak. Strong acids have a lower pH than weak acids.

The powerful acids include perchloric acid, chloric acid, nitric acid, sulfuric acid, hydrobromic acid, and hydroiodic acid.

Given:

Pka=3..2

pH=7

Let the volume be 1 liter

[HF]=01 M

pH=pka+log \frac{F^{-}}{HF} \\\\7=3.2+log\frac{F^{-}}{HF} \\3.8=log\frac{F^{-}}{HF}\\ \frac{F^{-}}{0.1}=10^{3.8} \\F^{-}=630.95 M

Now,

\frac{F^{-}}{HF}=\frac{630.95}{0.1}\\ =6309.57

F-:HF= 6309.57:1

Therefore, the most abundant is F^{-}and has 6310 times more than HF is F^{-}.

To know more about strong and weak acids, visit: brainly.com/question/12811944

#SPJ4

4 0
2 years ago
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