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3 years ago

CAN SOMEONE DO MY PROJECT PLEASE!!!

Chemistry

2 answers:

Tasya [4]3 years ago

8 0

Answer:

just do it

Explanation:

just do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do itjust do it

PtichkaEL [24]3 years ago

6 0

sorry I can do it your project

you have to doing it

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3 years ago

What are the possible values of ml for l=2<br>

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Explanation:

There are four types of quantum numbers;

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3 years ago

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3 years ago

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Question 11. Identify the reducing agent <br> Sn+2 + AG 0 —> Sn0 + Ag+

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Answer:

Ag 0 is the reducing agent.

Explanation:

Reducing -> gaining electrons

Oxidizing -> losing electrons

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$