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Margaret [11]
3 years ago
11

Which of the following is the best explanation for the fact that crushing increases the rate at which a solid dissolves in a liq

uid?
Select one:
a. The temperature of the solution is increased when the solid is crushed.
b. The surface area of the solute is increased when it is crushed.
c. Crushing increases the amount of solute present.
d. Crushing makes stirring the solution easier.​
Chemistry
1 answer:
Oksana_A [137]3 years ago
7 0
It’s D cause i know I know I know I know
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When oil and water are mixed vigorously, the oil is broken up into tiny droplets that are dispersed throughout the water. this d
Bumek [7]
Emulsion or heterogeneous mixture
4 0
2 years ago
2N2 + 3H2 → 2NH3
denis23 [38]

There are two N≡N bonds and three H–H  bonds are in reactants.

Given:

The reaction between nitrogen gas and hydrogen gas.

2N_2+3H_2\rightarrow 2NH_3

To find:

Bonds on the reactant side

Solution:

2N_2+3H_2\rightarrow 2NH_3

Reactants in the reaction = N_2, H_2

The bond between nitrogen atoms in single N_2 molecule  = N≡N (triple bond)

Then in two N_2  molecules = 2 N≡N (triple bonds)

The bond between hydrogen atoms in single H_2 molecule = H-H (single bond)

Then in three H_2  molecules = 3  H-H (single bonds)

Product in the reaction =NH_3

The bonds between nitrogen and hydrogen atoms in single NH_3 molecule = 3 N-H (single bond)

Then in two NH_3  molecules = 6  N-H (single bonds)

So, there are two N≡N bonds and three H–H bonds are in reactants.

Learn more about reactants and products here:

brainly.com/question/21517037?referrer=searchResults

brainly.com/question/20602904?referrer=searchResults

3 0
2 years ago
Classify the following change as physical or chemical: A match lights when struck.
mr_godi [17]
I think the answer is:
B. Chemical Change.
8 0
3 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
2 years ago
Helooo i need help please :)
Tatiana [17]
It is a ,d and c just make sure tho
6 0
2 years ago
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