Ask question

Ask question

All categories

Salsk061 [2.6K]

2 years ago

14

What is the molarity of a solution in which 175.8 grams of NaCl is dissolved in 1.5 L of water?

1 answer:

Sophie [7]2 years ago

8 0

175.8 g NaCl to moles:
(175.8 g)/(58.44 g/mol) = 3.008 mol NaCl

Molarity = moles of solute/volume of solution in liters

(3.008 mol NaCl)/(1.5 L) = 2.0 M.

The molarity of this solution would be 2.0 M.

You might be interested in

Suppose you wish to apply SSA to a triangle, in order to find an angle

emmasim [6.3K]

First of all there nothing exists something like SSA congruence criterion, It is SAS Congruence where the sides of a Triangle and the angle between them is exactly same to the other two sides and angle between them of another triangle.

So the correct option is D.

Explanation:

If the Triangle has two sides equal, so the two angles there are equal. Also the other angle be some other definite angle. So

45+x+x=180(Considering definite angle to be 45)
2x=135
x=67.5

So there is only one solution to the triangle.

8 0

3 years ago

Please help! I really don’t understand it

chubhunter [2.5K]

Answer:

C. It makes sure there is a balance inside the body

Explanation:

Hope this helps

7 0

2 years ago

Read 2 more answers

What is the specific latent heat of a fussion for a substance that takes 550 kj to melt 14 kg at 262 k?

Lina20 [59]

The latent heat is correlated with energy as follows:

Q = mL

550 * 103 = 14 * 103 * L

L = 39.285 J /g

Thus, latent heat of the substance is 39.285 j /g

7 0

3 years ago

Julie finds a snail on the sidewalk and wants to know whether or not the snail moves throughout the day. She places a

AURORKA [14]

Acting as a reference point for detecting motion

3 0

3 years ago

Read 2 more answers

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

2 years ago