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Natasha2012 [34]
3 years ago
11

How many molecules are in 12.5g of water (H2O)?

Chemistry
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

.694 moles

Explanation: do given over one so 12.5 over 1 then multiply that by 1 over 18.01528( atomic mass of water) and then you get ans answear that you sig fig but i did that for you

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10 points please please help
Inga [223]

Your answer would be respiration. This would be your answer because, plants take in carbon dioxide, and animals, including humans, breath oxygen. I hope this had helped you out.

4 0
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)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

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

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
2 years ago
Chemistry table practice
Sonbull [250]

Explanation:

I don't understand your question....

thank you

5 0
3 years ago
Ammonia (nh3) is widely used as a fertilizer and in many household cleaners. how much ammonia is produced when 6.64 mol of hydro
olga55 [171]
N₂ + 3H₂ ⇒ 2NH₃

doesnt matterN₂ + 6.64H₂ ⇒ 2NH₃

(6.64H₂/3H₂) x (2NH₃) =4.4266667

rounded to sig figs= 4.43

5 0
2 years ago
Entropy can be thought of as the degree of disorganization of a system.<br> TRUE<br> FALSE
Mice21 [21]

Answer:

True

Explanation:

The entropy of a system denoted by S is a thermodynamic function that increases in value when there are more ways to arrange the particles in the system. Some spontaneous chemical processes are entropy driven. An increase in entropy is said to drive the dissolution of ionic salts along with the evaporation of water are related to the spreading out of energy.

The entropy of a system can be taken as a measure of disorder of a system. In a spontaneous chemical process, the entropy of the universe is said to increase. ΔSunivu>0. Making the answer true.

7 0
2 years ago
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