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4vir4ik [10]
3 years ago
6

How will adding NaCl affect the freezing point of a solution?

Chemistry
1 answer:
zaharov [31]3 years ago
7 0
Answer is: adding NaCl will lower the freezing point of a solution. A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water). The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
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Question 5 of 10
pychu [463]
The answer is D. S2O6
4 0
3 years ago
Is work done whenever you hold a heavy object for a long time?
AnnZ [28]

No, work is not done whenever you hold a heavy object for a long time

<h3>What is work done ?</h3>

The result of a force's displacement and its component of force exerted by the object in the direction of displacement is what is known as the force's work. When we push a block with some force, the body moves quickly and work is completed.

  • No work, as that term is used here, is done until the object is moved in some way and a component of the force travels along the path that the object is moved. Because there is no displacement when holding a heavy object still, energy is not transferred to it.

Learn more about Work done here:

brainly.com/question/25573309

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7 0
1 year ago
19. If it takes more energy to break the bonds of the reactants than is released when the bonds of
frozen [14]

Answer:

If it takes more energy to break the original bonds than is released when the new bonds are formed, then the net energy of the reaction is negative. This means that energy must be pumped into the system to keep the reaction going. Such reactions are known as endothermic.

Explanation:

5 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
Saturn is about 1 429 000 km from the sun. how many meters is saturn from the sun?
stira [4]

Answer:

1.493 trillion

Explanation:

8 0
3 years ago
Read 2 more answers
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