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borishaifa [10]

3 years ago

9

Jenny wants to test the electrical conductivity of two substances dissolved in water. She is preparing the containers for the ex

periment. Which factor is MOST important for her to control?

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

Volume of the solutions

This is the most important factor for her to control.

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Answer: a

Explanation:

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givi [52]

Answer: The enthalpy of formation of $SO_3$ is -396 kJ/mol

Explanation:

Calculating the enthalpy of formation of $SO_3$

The chemical equation for the combustion of propane follows:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

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

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ$

Putting values in above equation, we get:

$-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol$

The enthalpy of formation of $SO_3$ is -396 kJ/mol

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What’s the answer for number 2?

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The answer to this is surface wave i think

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