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netineya [11]
11 months ago
13

O The Arrhenius Model of acids and bases applies toward substances that are nonaqueous.O The Arrhenius Model of acids and bases

was developed before the Bronsted-Lowry Model.O The Bronsted-Lowry Model can apply to bases that do not contain hydroxide ions.O The Bronsted-Lowry Módel applies to a wider range of acid-base phenomena than does the Arrhenius MoO none of the above
Chemistry
1 answer:
Alex11 months ago
4 0

ANSWER

Option A

EXPLANATION

Arrhenius defines an acid as a substance that will produce hydrogen ions as the only positive ion when dissolved in water.

He also defines a base as a substance that contains a hydroxide ion (OH^-).

Hence, the Arrhenius Model of acids and bases applies that is applied to the aqueous solution.

The correct answer is option A

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The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t
Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
How do you know that volume of a diamond is 15.1g
marshall27 [118]

Answer:

Density = mass / volume,  

        therefore volume = mass / density. Note that 1 mL = 1 cubic centimeter.

Explanation:

Volume = 15.1g / 3.52g/mL = 4.3mL = 4.3 cubic centimeters.

then a  diamond has a density of 3.52 g/mL.

4 0
3 years ago
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

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23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

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from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

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from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

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