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Bezzdna [24]
3 years ago
9

You are given a 250 ml sample of HCI and asked to find out what its concentration is. You have a 0 118 M Ca(OH)2 solution and it

takes 13.7 ml to
neutralize the acid sample. What is the concentration of the HCI? Show explanation Please help this is urgent
Chemistry
1 answer:
anygoal [31]3 years ago
4 0

Answer:

The concentration of the acid, HCl is 0.013 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nₐ) = 2

The mole ratio of base, Ca(OH)₂ (n₆) = 1

Finally, we shall determine the molarity of the HCl. This can be obtained as follow:

Volume of acid, HCl (Vₐ) = 250 mL

Molarity of base, Ca(OH)₂ (M₆) = 0.118 M

Volume of base, Ca(OH)₂ (V₆) = 13.7 mL

Molarity of acid, HCl (Mₐ) =?

MₐVₐ / M₆V₆ = nₐ/n₆

Mₐ × 250 / 0.118 × 13.7 = 2/1

Mₐ × 250 / 1.6166 = 2

Cross multiply

Mₐ × 250 = 1.6166 × 2

Mₐ × 250 = 3.2332

Divide both side by side 250

Mₐ = 3.2332 / 250

Mₐ = 0.013 M

Thus, the concentration of the acid, HCl is 0.013 M

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\huge\blue{answer}

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Explanation:

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3 years ago
Which equivalence factor should you use to convert from 38.2 grams of K to moles of K?
UNO [17]

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Read 2 more answers
If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: 2KMnO4 + 5Hg2Cl2 + 16HCl -&gt
IceJOKER [234]

Answer:

The limiting reactant is Hg2Cl2.

Explanation:

Step 1: Data given

Mass of each reactant = 5.0 grams

KMnO4 MM=158 g/mol

Hg2Cl2 MM=472.1 g/mol

HCl MM=36.5 g/mol

HgCl2 MM=271.5 g/mol

MnCl2 MM=125.8 g/mol

KCl MM=74.6 g/mol

H2O MM=18 g/mol)

Step 2: The balanced equation

2KMnO4 + 5Hg2Cl2 + 16HCl -> 10HgCl2 + 2MnCl2 + 2KCl + 8H2O

Step 3: Calculate moles

KMnO4 = 5.00 grams / 158 g/mol = 0.0316 mol

Hg2CL2 = 5.00 grams / 472.1 g/mol = 0.0106 mol

HCl = 5.00 grams / 36.5 g/mol = 0.137 mol

Step 3: Calculate limiting reactant

For 2 moles of KMno4 we need 5 moles of Hg2Cl2 and 16 moles of HCl

Hg2Cl2 has the smallest amount of moles.

For 5 moles Hg2Cl2 ( 0.0106 mol) we need 0.0106 / (5/2) = 0.00424 mol KMnO4

For 5 moles Hg2Cl2 we need (16/5) *0.0106 = 0.03392 moles of HCl

So the limiting reactant is Hg2Cl2.

Step 4: Calculate moles of product produced:

2*0.0106 = 0.0212 moles of HgCl2

(2/5) * 0.0106 = 0.00424 moles of MnCl2 and 0.00424 moles of KCl

(8/5) * 0.0106 = 0.01696 moles H2O

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3 years ago
What is the graph of the orange line and is it a positive or negative connotation
dimulka [17.4K]

Answer:

Positive

Explanation:

4 0
3 years ago
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