The answer is 12.5 mL.
Solution:
<u>M, 6.00M V₁ =?</u>
M2 = 3.00M, V₂ = 25.0mL
<u>M1V1 = M2 V2/M1</u>
V1 = M2V2 M,
V1 = 25.0X3.00/6.00
∴ V1 = 12.5 mL
The initial concentration of the resulting solution is 2 molar which is the volume multiplied by the concentration. The volume of the starting solution is therefore 2,500 ml × 2.25 mol. Calculate the number of moles of HCl dissolved in a given volume of acid by reacting an acid with CaCO3.
This is essentially a double exchange reaction with the decomposition of one of the products. To determine the amount of stock solution required divide the number of moles of glucose by the molarity of the stock solution.
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Answer:

Explanation:
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In this case, according to the gas laws, we would be able to assume HCl can be modeled as an ideal gas for this calculation purpose; in such a way, we use the following equation to compute the temperature:

In such a way, we plug in moles, volume and pressure to obtain:

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It will stay at a constant temperature while it is boiling which is 212°F
Answer:
6.69%
Explanation:
Given that:
Mass of the fertilizer = 0.568 g
The mass of HCl used in titration (45.2 mL of 0.192 M)
= 
= 0.313 g HCl
The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)
= 
= 0.0058919 mole of NaOH
From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process
Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g
= 0.2151 g HCl
From above ; the total amount of HCl used = 0.313 g
The total amount that is used for complete neutralization = 0.2151 g
∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g
= 0.0979 g
We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g
Now; the amount of HCl neutralized by 0.0979 HCl = 
= 0.0456 g
Therefore, the mass of nitrogen present in the fertilizer is:
= 
= 0.038 g
∴ Mass percentage of Nitrogen in the fertilizer =
%
= 6.69%