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MrRa [10]
3 years ago
11

Concentrated sulfuric acid contains very little water, only 5.0% by mass. it has a density of 1.84g/ml. what is the molarity of

this acid
Chemistry
1 answer:
kvasek [131]3 years ago
5 0
When m for water is only 5 % so, m for the acid is (m%) = 95%

and D = 1.84 g/ml, assume 1 L of solution therefore,
1- D= m / v
m = D x v =  1.84 g/ml x 1 L x 10^3 ml /L
                 = 1840 g solution
2- m% = m acid / m sol x 100 %
m acid = 1840 g H2So4 x 0.95
            = 1750 g H2So4
3- n = 1750 g H2So4 x 1 mol H2So4 x 98.09 H2So4
   = 17.8 mol H2So4
4- So the molarity of { H2So4} = n / v = 17.8 mol H2So4 / 1 L
                               = 17.8 M
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Answer/Explanation:

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7 0
3 years ago
What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid?
Shkiper50 [21]
Precipitation and gas evolution
7 0
3 years ago
Read 2 more answers
Help with this please
Len [333]

Answer:

12

Explanation:

There are 4 sulfur atoms in SO4

4×3=12

This means that it turns into 3×(SO4)

=3SO4

6 0
3 years ago
Consider the following reaction:
Sidana [21]
Your limiting is CuCI2 and the excess is KI (from what i’ve heard from my tc to find it just use the moles or look at the grams)Do you want me to do the qn and give u the ans or?

Explanation:You have more grams of KI than CuCI2
irl example : I need 200g of flour to bake 1 muffin and 100g of butter.But I have 300g of butter and only 200g of flour.This means I can only bake up to 1 muffin since I got excess grams of butter.But to use up all my 300g of butter I need 400g more of flour.Making my butter the excess while my flour the limiting since I have less of it and it also determines how much muffin would I get at the end of the bake.

im sorry if that example sounds clowny T-T
6 0
2 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
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