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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

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Read 2 more answers

If E=mc^2,solve for both m and c. Also, if m=80 and c=0.40 what is the value of E?

Pie

Answer:

$m = \frac{E}{c^2}$

$c = \sqrt{\frac{E}{m} }$

$E = 12.8 J$

Explanation:

Part 1: Solving for m

We are given that:

E = mc²

To solve for m, we will need to isolate the m on one side of the equation

This means that we will simply divide both sides by c²

$m = \frac{E}{c^2}$

Part 2: Solving for c

We are given that:

E = mc²

To solve for c, we will need to isolate the m on one side of the equation

This means that first we will divide both sides by m and then take square root for both sides to get the value of c

$c^2 = \frac{E}{m}\\ \\ c=\sqrt{\frac{E}{m}}$

Part 3: Solving for E

We are given that:

m = 80 and c = 0.4

To get the value of E, we will simply substitute in the given equation:

E = mc²

E = (80) × (0.4)²

E = 12.8 J

Hope this helps :)

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3 years ago