What type of radiation is carbon emitting in the following equation? 14/6C=>0/-1e+14/7N

If 27.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 32.40 mL of 0.185 M HCl, what is the concentration of the

zhannawk [14.2K]

Given information : Volume of HCl = 32.40 mL

$32.40 mL\times \frac{1 L}{1000 mL}$

Volume of HCl = 0.0324 L

Concentration of HCl = 0.185 M or 0.185 mol/L (M = mol/L)

Volume of Ca(OH)2 = 27.0 mL

$27.0 mL\times \frac{1 L}{1000 mL}$

Volume of Ca(OH)2 = 0.027 L

We need to find the concentration of Ca(OH)2.

To find the concentration of Ca(OH)2 we need moles and volume of Ca(OH)2.

$Concentration (Molarity) = \frac{(Moles of Ca(OH)2)}{(Volume of Ca(OH)2)}$

Moles of Ca(OH)2 can be calculated using stoichiometry and volume of Ca(OH)2 is already given to us.

Step 1 : Find the moles of HCl using its given volume and concentration.

$Moles = Concentration \times Volume in L$

$Moles = 0.185\frac{mol}{1L}\times 0.0324 L$

Moles of HCl = 0.005994 mol HCl

Step 2 : We need to find moles of Ca(OH)2 using mol of HCl with the help of mole ratio.

Mole ratio are the coefficient present in front of the compound in a balanced equation.

Mole ratio of Ca(OH)2 : HCl = 1:2 ( 1 coefficient of Ca(OH)2 and 2 coefficient of HCl)

$(0.005994 mol HCl)\times \frac{(1 mol Ca(OH)2)}{(2 mol HCl)}$

Moles of Ca(OH)2 = 0.002997 mol Ca(OH)2

Step 3 : Find the concentration of Ca(OH)2 using its moles and volume.

$Concentration (Molarity) = \frac{(Moles of Ca(OH)2)}{(Volume of Ca(OH)2)}$

Moles of Ca(OH)2 = 0.002997 mol and volume of Ca(OH)2 = 0.027 L

$Concentration (Molarity) = \frac{(0.002997 mol)}{(0.027 L)}$

Concentration of Ca(OH)2 = 0.111 mol/L or 0.111 M

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Hope this helps! :)</span>

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