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Archy [21]
4 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]4 years ago
5 0

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


oksano4ka [1.4K]4 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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Answer:

11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.

Explanation:

Concentration of sodium stearate acid : c

Moles of sodium stearate = \frac{3.96 g}{306 g/mol}=0.01294 mol

Volume of the solution = 10.0 mL = 0.010 L

c=\frac{0.01294 mol}{0.010 L}=1.294 M

C_{17}H_{35}COONa\rightleftharpoons C_{17}H_{35}COO^-+Na^+

[C_{17}H_{35}COO^-]=c=1.294 M

C_{17}H_{35}COO^-+H_2O\rightleftharpoons C_{17}H_{35}COOH +OH^-

initially c

c           0    0

At equilibrium

(c-x)       x    x

Dissociation constant of an acid = K_a=1.3\times 10^{-5}

Expression of a dissociation constant of an acid is given by:  

K_a=\frac{[C_{17}H_{35}COOH][OH^-]}{[C_{17}H_{35}COO^-]}

K_a=\frac{(x)^2\times x}{(c -x)}

1.3\times 10^{-5}=\frac{x^2}{1.294-x}

Solving for x;

x = 0.0041 M

[OH^-]=0.0041 M

The pOH of the solution:

pOH=-\log[OH^-]=-\log[0.0041 M]=2.39

pH = 14 -pOH

pH = 14 - 2.39 = 11.61

11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.

7 0
4 years ago
Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th
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Answer: from the Zn anode to the Cu cathode


Justification:


1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)


2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).


3) Then, you can already tell that electrons go from Zn to Cu.


4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.


So you get that the electrons flow from the anode (Zn) to the cathode (Cu).


Always oxidation occurs at the anode, and reduction occurs at the cathode.

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