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3 years ago

Which TWO statements are true about the changing salt concentrations of ocean water?

Chemistry

1 answer:

marissa [1.9K]3 years ago

3 0

Answer:

The answer is 3

Explanation:

The ocean gets saltier when more water is evaporated

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In an exothermic reaction, an increase in temperature favors the formation of products.

pentagon [3]

False, in an exothermic reaction, an increase in temperature does not favor the formation of products. Instead, it favors the backward reaction. An exothermic reaction is a reaction where energy is transferred from the system out to the environment.

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4 years ago

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What could you predict about a methane molecule that would most likely be made by using both the quantum model of the atom and J

Svetlanka [38]

By using the quantum model of the atom and John Dalton's model you can predict that the methane molecule will most likely be made up of Hydrogen and Carbon.

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3 years ago

Consider the reaction 4HCl(g) + O2(g)2H2O(g) + 2Cl2(g) Using standard thermodynamic data at 298K, calculate the entropy change f

Ulleksa [173]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$4HCl(g)+O_2(g)\rightarrow 2H_2O(g)+2Cl_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cl_2(g))})+(2\times \Delta S^o_{(H_2O(g))})]-[(4\times \Delta S^o_{(HCl(g))})+(1\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(O_2(g))}=205.14J/K.mol\\\Delta So_{(HCl(g))}=186.91J/K.mol\\\Delta S^o_{(Cl_2(g))}=223.07J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (223.07))+(2\times (188.82))]-[(4\times (186.91))+(1\times (205.14))]\\\\\Delta S^o_{rxn}=-129J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-129) J/K = 129 J/K

We are given:

Moles of HCl gas reacted = 2.27 moles

By Stoichiometry of the reaction:

When 4 mole of HCl gas is reacted, the entropy change of the surrounding will be 129 J/K

So, when 2.27 moles of HCl gas is reacted, the entropy change of the surrounding will be = $\frac{129}{4}\times 2.27=73.21J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of HCl gas is reacted is 73.21 J/K

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3 years ago

Write the balanced Ka and Kb reactions for HSO3- in water. Be sure to include the physical states of each species involved in th

Hunter-Best [27]

Answer:

Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]

Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]

Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

And kb, base dissociation constant is:

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3 years ago

While following the Dumas method, a student assumes that the contents of the container are the same temperature as the water bat

Nataliya [291]

Answer:

Explanation:

the Molar mass will be smaller as the content of the container is not directly proportional to the temperature of the water bath.

5 0

3 years ago