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Y_Kistochka [10]
2 years ago
12

How many formula units are found in 2.10 gram of Mgo? (No units needed.)

Chemistry
1 answer:
disa [49]2 years ago
8 0

Answer:

84.63924

Explanation:

one moles equals 40.30440

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3. Which of the following substances is NOT an electrolyte? (1 point)
iren [92.7K]

Answer: electrolytes hydrate you and give you energy.

Explanation: please add more information to this question. I can tell you this though from my answer it should help you define what is and isn't an electrolyte. I hope that this is helpful to you.

7 0
2 years ago
An object exerts a reaction force when it _____.
VashaNatasha [74]
Touch or pushed........
7 0
3 years ago
Read 2 more answers
Which equations are in standard form? Check all that apply
aliya0001 [1]

Answer:

2x+3y=-6

4x+3y=12

3x-y=5

5x+3y=1

Explanation:

standard form: ax + by = C

4 0
3 years ago
the vapor pressure of a naqueous solution is found to be 24.9 mmgh at 25C. what is the mole fraction of solute in this solution?
Gekata [30.6K]

Answer:

Mole fraction of solute is 0.0462

Explanation:

To solve this we use the colligative property of lowering vapor pressure.

First of all, we search for vapor pressure of pure water at 25°C  = 23.8 Torr

Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.

Formula for lowering vapor pressure is:

ΔP = P° . Xm

Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)

Xm = mole fraction

24.9 mmHg - 23.8 mmHg = 23mmHg . Xm

Xm = (24.9 mmHg - 23.8 mmHg) /  23mmHg

Xm = 0.0462

8 0
2 years ago
Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid (FCH2CO
PSYCHO15rus [73]

[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of FCH_{2}COOH

FCH_{2}COOH(aq) + H_{2}O (l)    FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = [H_{3}O^{+}] =[FCH_{2}COO^{-}] = 0.00770 M

pK_{a} = 2.59K_{a} = 10^{-2.59}   = 0.00257 M

K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}

0.00257 = \frac{0.00005929}{x - 0.00770}

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

Therefore, initial concentration of the weak acid is <u>0.031 M</u>

4 0
3 years ago
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