Chemistry: What volume of 0.5M HCl can be prepared from 1L of 12M HCl?

07.03.2021

What volume of 0.5M HCl can be prepared from 1L of 12M HCl?

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Chemistry

A student is asked to prepare 400.0 mL of a 0.25M HCl solution. The students find a bottle labeled 2.0M HCl. How many mL of the 2.0M solution are needed to prepare the 0.25M solution?

Step-by-step answer

P Answered by PhD

The answer to your question is He needs 50 ml of 2.0 M HCl.

Explanation:

Data

Volume 2 = 400 ml

[HCl] 2 = 0.25 M

[HCl] 1 = 2.0 M

Volume 1 = ?

To solve this problem use the dilution formula

Formula

V₁C₁ = V₂C₂

Solve for V₁

V₁ = V₂C₂ / C₁

Substitution

V₁ = (400)(0.25) / 2

Simplification

V₁ = 100 / 2

Result

V₁ = 50 ml

Chemistry

A student is asked to prepare 400.0 mL of a 0.25M HCl solution. The students find a bottle labeled 2.0M HCl. How many mL of the 2.0M solution are needed to prepare the 0.25M solution?

Step-by-step answer

P Answered by PhD

The answer to your question is He needs 50 ml of 2.0 M HCl.

Explanation:

Data

Volume 2 = 400 ml

[HCl] 2 = 0.25 M

[HCl] 1 = 2.0 M

Volume 1 = ?

To solve this problem use the dilution formula

Formula

V₁C₁ = V₂C₂

Solve for V₁

V₁ = V₂C₂ / C₁

Substitution

V₁ = (400)(0.25) / 2

Simplification

V₁ = 100 / 2

Result

V₁ = 50 ml

Chemistry

What volume of 1.25m hcl would be required to prepare 180 ml of a 0.500m hcl solution?

Step-by-step answer

P Answered by Master

The volume required of 1.25M HCL to prepare 180ml of a 0.500M Hcl solution is 72ml

Chemistry

How many milliliters of 6.45m hcl solution should be used to prepare 4.50 l of 0.600m hcl

Step-by-step answer

P Answered by Master

418.6 mL of stock solution of HCl will be needed to prepare the required amount of solution.

Explanation:

To calculate the volume of stock solution required to prepare the HCl solution, we use the following equation:

M_1V_1=M_2V_2

where,

$M_1\text{ and }V_1$ are the molarity and volume of Stock HCl solution

$M_2\text{ and }V_2$ are the molarity and volume of required HCl solution.

We are given:

Conversion factor: 1L = 1000mL

$M_1=6.45M\\V_1=?mL\\M_2=0.6M\\V_2=4.5L=4500mL$

Putting values in above equation, we get:

$6.45\times V_1=0.6\times 4500\\\\V_1=418.6mL$

Hence, 418.6 mL of stock solution of HCl will be needed to prepare the required amount of solution.

Chemistry

How many milliliters of 6.45m hcl solution should be used to prepare 4.50 l of 0.600m hcl

Step-by-step answer

P Answered by Master

418.6 mL of stock solution of HCl will be needed to prepare the required amount of solution.

Explanation:

To calculate the volume of stock solution required to prepare the HCl solution, we use the following equation:

M_1V_1=M_2V_2

where,

$M_1\text{ and }V_1$ are the molarity and volume of Stock HCl solution

$M_2\text{ and }V_2$ are the molarity and volume of required HCl solution.

We are given:

Conversion factor: 1L = 1000mL

$M_1=6.45M\\V_1=?mL\\M_2=0.6M\\V_2=4.5L=4500mL$

Putting values in above equation, we get:

$6.45\times V_1=0.6\times 4500\\\\V_1=418.6mL$

Hence, 418.6 mL of stock solution of HCl will be needed to prepare the required amount of solution.

Chemistry

Question 5 a solution is prepared at 25°c that is initially 0.35m in chlorous acid hclo2, a weak acid with =ka×1.110−2, and 0.29m in sodium chlorite naclo2. calculate the ph of the solution. round your answer to 2 decimal places.

Step-by-step answer

P Answered by Master

Answer : The pH of the solution is, 1.88

Explanation : Given,

$K_a=1.1\times 10^{-2}$

Concentration of HClO_2 = 0.35 M

Concentration of NaClO_2 = 0.29 M

First we have to calculate the value of pK_a .

The expression used for the calculation of pK_a is,

$pK_a=-\log [K_a]$

Now put the value of K_a in this expression, we get:

$pK_a=-\log (1.1\times 1-^{-2})$

$pK_a=2-\log (1.1)$

pK_a=1.96

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}$

Now put all the given values in this expression, we get:

$pH=1.96+\log (\frac{0.29}{0.35})$

pH=1.88

Therefore, the pH of the solution is, 1.88

Chemistry

Step-by-step answer

P Answered by Specialist

Answer : The pH of the solution is, 1.88

Explanation : Given,

$K_a=1.1\times 10^{-2}$

Concentration of HClO_2 = 0.35 M

Concentration of NaClO_2 = 0.29 M

First we have to calculate the value of pK_a .

The expression used for the calculation of pK_a is,

$pK_a=-\log [K_a]$

Now put the value of K_a in this expression, we get:

$pK_a=-\log (1.1\times 1-^{-2})$

$pK_a=2-\log (1.1)$

pK_a=1.96

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaClO_2]}{[HClO_2]}$

Now put all the given values in this expression, we get:

$pH=1.96+\log (\frac{0.29}{0.35})$

pH=1.88

Therefore, the pH of the solution is, 1.88

Chemistry

What volume of 2.0M HCl is needed to prepare 0.50L of a 0.75M solution? Please show your work

Step-by-step answer

P Answered by Master

V = 0.5 L

Explanation:

Given that,

The volume of HCl, V = 2 M

We need to find the volume of 2.0M HCl is needed to prepare 0.50L of a 0.75M solution.

Let n be the number of moles of HCl.

$n=c\times V$

Where

c is molarity

$n=2\times 0.5\\\\n=1\ mol$

Let V be the volume of the solution. So,

$V=\dfrac{n}{c}\\\\V=\dfrac{1}{2}\\\\V=0.5\ L$

So, the required volume of the solution is equal to 0.5 L.

Chemistry

How many moles of HF are in 30.mL of 0.15MHF(aq) ?

Step-by-step answer

P Answered by Master

Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarity * L
mol = 0.15 * 0.03 = 0.0045 mol
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarit

Chemistry

What elements combined with Strontium, St, in a 1:1 ratio?

Step-by-step answer

P Answered by PhD

Answer: chalcogens.
Explanation: Strontium is an alkaline earth metal, it always exhibits a degree of oxidation in its compounds +2.
Chalcogens are a group of 6 chemical elements (oxygen O, sulfur S, selenium se, tellurium te, polonium Po) that have an oxidation state of -2 => Chalcogens will combine with strontium in a ratio of 1:1.

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