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arlik [135]
3 years ago
15

What volume of 6 m hcl is needed to prepare 150 ml of 0.4 m hcl?

Chemistry
1 answer:
defon3 years ago
4 0

Answer : The volume of 6 M HCl needed is, 10 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of diluted HCl.

M_2\text{ and }V_2 are the final molarity and volume of HCl.

We are given:

M_1=6M\\V_1=?\\M_2=0.4M\\V_2=150mL

Now put all the given values in above equation, we get:

6M\times V_1=0.4M\times 150mL\\\\V_1=10mL

Hence, the volume of 6 M HCl needed is, 10 mL

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The unique characteristic of the amino acid cysteine is _____.
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At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations
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This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, x, has taken place:

[HF]=[HF]_0-2x

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }

[HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M

Finally, the initial concentration of HF is calculated as follows:

[HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M

Learn more:

  • brainly.com/question/13043707
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