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Archy [21]
3 years ago
5

Which condition would lower the salinity of ocean water? A. high rates of evaporation B. freezing of glaciers C. extremely hot w

eather D. a river flowing into the sea
Chemistry
2 answers:
Paraphin [41]3 years ago
5 0

It is a I have done this in a quiz and got it right


oksano4ka [1.4K]3 years ago
5 0

\boxed{{\text{D}}{\text{. a river flowing into the sea}}} would lower the salinity of ocean water.

Further Explanation:

Water that comprises oceans and seas covers more than 70 % of the surface of the earth. Ocean water is not just pure water; it also contains a variety of substances in it. It mainly contains sodium chloride in it and also includes some other chemicals like sulfate, potassium, calcium, and magnesium. When it rains, minerals present in rocks get dissolved in some portion of water which then flows into rivers or seas. But evaporation of this water takes place and during this process, minerals are left behind, making the ocean water salty. The salinity of water depends on the number of minerals dissolved in it.

When evaporation of ocean water occurs, it removes water from it and salt is left behind so it results in an increase in the salinity of ocean water. The freezing of glaciers does not have any impact on the salinity of ocean water.

When the weather is extremely hot, it results in more evaporation of water, leaving large amounts of salt behind so it increases the salinity of ocean water.

The river consists of fresh water that has no salt in it. So when a river flows into the sea the amount of salt in it decreases due to the addition of fresh water to it. Therefore a river flowing into the sea would decrease or lower the salinity of ocean water.

Learn more:

  1. Identify the phase change in which crystal lattice is formed: brainly.com/question/1503216
  2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer details:

Grade: Senior School

Chapter: Phase transition

Subject: Chemistry

Keywords: ocean water, evaporation, fresh water, ocean water, salinity, glaciers,  

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marta [7]

Molality of the solution is defined as the number of moles of a substance dissolved divided by the mass of the solvent:

Molality = number of moles / solvent mass

From the concentration of 39% (by mass) of HCl in water, we construct the following  reasoning:

in 100 g solution we have 39 g hydrochloric acid (HCl)

number of moles  = mass / molecular weight

number of moles of HCl = 39 / 36.5 = 1.07 moles

solvent (water) mass = solution mass - hydrochloric acid mass

solvent (water) mass = 100 - 39 = 61 g

Now we can determine the molality:

molality = 1.07 moles / 61 g = 0.018

8 0
3 years ago
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i
alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

Explanation:

The pH of the solution = 2.46

pH=-\log[H^+]

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HA\rightleftharpoons H^++A^-

Initially

0.0144         0      0

At equilibrium

(0.0144-x)       x       x

The expression if an dissociation constant is given by :

K_a=\frac{[A^-][H^+]}{[HA]}

K_a=\frac{x\times x}{(0.0144-x)}

x=[H^+]=0.003467 M

K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}

K_a=1.099\times 10^{-3}

The value of dissociation constant of the monoprotic acid is 1.099\times 10^{-3}.

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I hope it helps, Regards.
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