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3 years ago

What volume of a 2.00 m kcl solution is required to prepare 500. ml of a 0.100 m kcl solution?

Chemistry

1 answer:

den301095 [7]3 years ago

3 0

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2 M x V1 = 0.1 M x .5 L

V1 = 0.025 L or 25 mL of the 2 M KCl solution is needed

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Which SI unit would be most appropriate for expressing the mass of this animal

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Kilograms, I think.

Explanation:

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Which element would you expect to behave the most like carbon (C)?

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3 years ago

Read 2 more answers

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50Ti (5.3%). What is the average atomic

NNADVOKAT [17]

Hey there!:

Isotopes : abundance :

46 Ti 8.0%

47 Ti 7.8 %

48 Ti 73.4 %

49 Ti 5.5 %

50 Ti 5.3 %

Weighted average = ∑ Wa * % / 100

Therefore:

( 46 * 8.0) + (47 * 7.8 ) + (48 * 73.4 ) + ( 49 * 5.5 ) + ( 50*5.3 ) / 100 =

4792.3 / 100

= 47.923 a.m.u

Hope that helps!

7 0

3 years ago