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deff fn [24]
3 years ago
12

500 mL of water is added to 400 mL of 0.35 M HCl. Find the concentration of the diluted solution.

Chemistry
1 answer:
jarptica [38.1K]3 years ago
5 0

The concentration of diluted solution is 0.16 M

<u>Explanation:</u>

As, the number of moles of diluted solution and concentrated solution will be same.

So, the equation used to calculate concentration will be:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated HCl solution

M_2\text{ and }V_2 are the molarity and volume of diluted HCl solution

We are given:

M_1=0.35M\\V_1=400mL\\M_2=?M\\V_2=(500+400)mL=900mL

Putting values in above equation, we get:

0.35\times 400=M_2\times 900\\\\M_2=\frac{0.35\times 400}{900}=0.16M

Hence, the concentration of diluted solution is 0.16 M

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Read 2 more answers
A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24
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Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

5 0
3 years ago
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