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4 years ago

2. As the sun heats the surface of the oceans, two layers of water result. What separates the layers?

Chemistry

2 answers:

worty [1.4K]4 years ago

7 0

Carbon dioxide I think lol

zubka84 [21]4 years ago

6 0

The correct answer is Thermocline.

The ocean is majorly divided into 3 main layers on the basis of temperature. The sun heats up the upper layer of water of the ocean and so it is called a warm layer.

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Can some one help me in this I hope this will be easy for you :)

kramer

Answer:

c) i) Mg²⁺ ii) O²⁻

Explanation:

I don't know the answer to Q7. because you don't show the diagram

5 0

3 years ago

Electron dot of NH3

aliya0001 [1]

Answer:

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Explanation:

4 0

3 years ago

5K + KNO3 ------ 3K2O + N

masya89 [10]

Answer:

174,957.143 grams of potassium and 89,228.478 grams of potassium nitrate will be needed.

Explanation:

$5K +KNO_3\rightarrow 3K_2O + N$

Mass of nitrogen = 27 lbs = 12,247 g

1 lbs = 453.592 g

Moles of nitrogen = $\frac{12,247 g}{14 g/mol}=874.786 mol$

According to reaction, 1 mole of nitrogen is produced from 5 moles of potassium and 1 mole of potassium nitrate.

Then 874.786 mol of nitrogen will be obtained from :

$\frac{5}{1}\times 874.786 mol=4,373.928 mol$ of potassium.

Then 874.786 mol of nitrogen will be obtained from :

$\frac{1}{1}\times 874.786 mol=874.789 mol$ of potassium nitrate.

Mass of 4,373.928 moles of potassium:

$4,373.928 mol\times 40 g/mol=174,957.143 g$ of potassium

Mass of 874.789 moles of potassium nitrate:

$874.789 mol\times 102 g/mol=89,228.478 g$ of potassium nitrate

3 0

3 years ago

When 3.18g of copper(ii)oxide was carefully heated in a stream of dry hydrogen, 2.54g of copper and 0.72g of water was formed. D

Klio2033 [76]

Answer:

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7 0

3 years ago

For the aqueous reaction dihydroxyacetone phosphate is the reactant and glyceraldehyde 3 phosphate is the product. dihydroxyacet

lbvjy [14]

<u>Answer:</u> The Gibbs free energy of the reaction is -445 J/mol.

<u>Explanation:</u>

The chemical equation for the conversion follows:

$\text{Dihydroxyacetone phosphate}\rightleftharpoons \text{Glyceraldehyde-3-phosphate}$

The expression for $K_{eq}$ of above equation is:

$K_{eq}=\frac{\text{[Glyceraldehyde-3-phosphate]}}{\text{[Dihydroxyacetone phosphate]}}$

We are given:

[Glyceraldehyde-3-phosphate] = 0.00400 M

[Dihydroxyacetone phosphate] = 0.100 M

Putting values in above equation, we get:

$K_{eq}=\frac{0.004}{0.100}=0.04$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G=\Delta G^o+RT\ln K_1$

where,

$\Delta G^o$ = Standard Gibbs free energy = 7.53 kJ/mol = 7530 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

Putting values in above equation, we get:

$\Delta G=7530J/mol+(8.3145J/Kmol)\times 298K\times \ln (0.04)\\\\\Delta G=-445J/mol$

Hence, the Gibbs free energy of the reaction is -445 J/mol.

3 0

4 years ago