How does thermal energy affect chemical changes

Chemistry

1 answer:

solniwko [45]3 years ago

7 0

Answer:

It makes chemicals hot or cold which makes chemicals react differently

Explanation:

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Look at the four chemical reactions.

frutty [35]

Reaction 3 is different from the rest and it is because a single product is synthesized.

Option 2

<u>Explanation:</u>

The reaction given in 1,2 and 4 are dissociation reaction where a single compound is getting split into two compounds. So all those three reactions are exothermic as they are breaking the bonds leading to release of energy to the surrounding. But the reaction 3 is a formation reaction, where two chemical compounds are forming a single compound.

So in this case, new bonds are formed leading to absorption of energy from the surrounding. So this is an endothermic reaction. Thus, reaction 3 is different from the rest and it is because a single product is synthesized.

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Being able to roll the tongue is dominant over not being able to roll the tongue.

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Explanation:

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3 years ago

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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$