All categories

3 years ago

When 250 ml of water is added to 35 ml of 0.2 M HCl

2 answers:

8 0

Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl

butalik [34]3 years ago

4 0

0.028 is going to be your answer

You might be interested in

PLEASE HELP FAST!!!! I'll give brainliest to the best answer!!

MArishka [77]

Answer:

A concentrated acid is an acid which is in either pure form or has a high concentration. Laboratory type sulfuric acid (about 98% by weight) is a concentrated (and strong) acid. A dilute acid is that in which the concentration of the water mixed in the acid is higher than the concentration of the acid itself.

Explanation:

Concentrated acid - Those acids which are pure or have very high concentration in water are called as concentrated acids. For example concentrated Hydrochloric acid (HCl) and concentrated Sulphuric acid are examples of concentrated acids.

7 0

2 years ago

Read 2 more answers

A lithium atom starts from rest at a position x = xo and falls toward zinc atom with an acceleration that depends on their separ

rusak2 [61]

Answer:

Please refer to the attachment for answers.

Explanation:

Please refer to the attachment for explanation

3 0

3 years ago

When using water displament to find the volume of an irregular shaped object we use the unit...

Gekata [30.6K]

Yeah it would develop fire

5 0

2 years ago

Is it true or false I need help.

dsp73

Answer:

true

Explanation:

4 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

3 years ago