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Yuri [45]
3 years ago
6

How would you prepare 10 ml of a 0.25% m/v hcl solution if 1% m/v hcl was available? how much 1% m/v hcl is needed? how much dis

tilled water is used?
Chemistry
1 answer:
Fed [463]3 years ago
8 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.

M1V1 = M2V2

1% x V1 = 0.25% x 10 mL

V1 =2.5 mL

Therefore, you will need to have 2.5 mL of the 1% HCl solution and 7.5 mL of distilled water. In mixing the two liquids, you should remember that the order of mixing would be acid to water. So, you use a 10 mL volumetric flask . Put small amount of distilled water and add the 2.5 mL of HCl solution. Lastly, dilute with distilled water up to the 10 mL mark.

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The number of protons determines an element's atomic number (Z) and distinguishes one element from another. For example, carbon's atomic number (Z) is 6 because it has 6 protons. The number of neutrons can vary to produce isotopes, which are atoms of the same element that have different numbers of neutrons.

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Please help me I need help with this questions I’m very confused fused as to what the answer is please
zepelin [54]
Answer: This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
3 0
3 years ago
What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?
Morgarella [4.7K]

This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5%​ as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

\% H=\frac{m_H*1}{M_{HF}}*100\%

Where m_H stands for the atomic mass of hydrogen and M_{HF} for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%

Learn more about percent compositions: brainly.com/question/12247957

5 0
2 years ago
Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
bija089 [108]

<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=K_1\times K_2

We are given:  

K_1=K_b

K_2=\frac{1}{K_w}

Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

7 0
3 years ago
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