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3 years ago

Can some answer this please?

Chemistry

2 answers:

Mumz [18]3 years ago

7 0

Answer:

i think its C

Explanation:

marysya [2.9K]3 years ago

4 0

Answer:

Explanation:

it's b because I just went over that frome my class and got it correct

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2 C2H6 + 7 026 H2O + 4 CO2

Kay [80]

Answer:

Oxygen is limiting reactant

Explanation:

Based on the chemical reaction:

2C2H6 + 7O2 → 6H2O + 4CO2

<em>2 mole of ethane reacts with 7 moles of oxygen</em>

<em />

For a complete reaction of 5.25 moles of ethane are required:

5.25 moles Ethane * (7mol Oxygen / 2mol Ethane) = 18.38 moles of oxygen

As there are just 15.0 moles of oxygen

<h3>Oxygen is limiting reactant</h3>

3 0

3 years ago

Why is it important to use chemicals of high quality (purity) and exact(correct) concentrations in experiments?

Darya [45]

It is important to use chemicals of high quality and correct concentrations in experiment so that we can obtain the predictable result in an experiment

Purity of substances/chemicals is very important when performing experiments in the laboratories, because it plays an important role in the determination of the chemical properties of substances which will have a direct effect on the result obtained from the experiment. likewise using the correct concentration of reagents

During experiments to obtain the best/predictable result in an experiment it is of best practice to make use of chemicals of high quality and also at the exact(needed) concentrations, that way the predictable result of the experiment will be obtained.

Learn more : brainly.com/question/8684258

6 0

3 years ago

Read 2 more answers

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 $(\Delta S^o)$ .

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

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