It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 30 mL

Now, crossing the products:
x · 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = x g : 10 mL

Now, crossing the products:
x · 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
x = 3.6 g

So, from 10 mL mixture, 3.6 g of NaCl could be extracted.

Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
3.6/10.8 = 1/3

Therefore, it will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

5 0

3 years ago

When 280. mL of 1.50 × 10⁻⁴ M hydrochloric acid is added to 135 mL of 1.75 × 10⁻⁴ M Mg(OH)₂, the resulting solution will be?

goldfiish [28.3K]

Answer:

B is the correct answer!

Explanation:

Balanced equation is Mg(OH)2 + 2 HCl = MgCl2 + 2 H2O

HCl is the limiting reactant because it gives the smaller amount of MgCl2 I did the math. This means you will be left with Mg(OH)2 since HCl completely runs out.

If HCl is the limiting reactant, Mg(OH) 2 is the excess reactant.

Mg(OH) 2 is a base, so the solution will be basic.

6 0

3 years ago