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garik1379 [7]
3 years ago
12

Given: h2o(aq) + h2o(l) h3o+(aq) + oh–(aq) δh°rxn > 0 when the temperature of a sample of pure water is raised above 25°c,

Chemistry
1 answer:
sesenic [268]3 years ago
7 0
When Δ H > 0 that means this is an endothermic reaction, and we can assume heat as one of the reactants. SO when the temperature increases so the reaction will move rightward to decrease the reactants and increase the products and as the products increase it will make the Kw increase also as
 Kw = [Products] /[reactants].

So our answer will be: the value of Kw will increase.
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Return the first n>0 elements of list l. assume the list elements are numbered from 1 to n
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3 years ago
Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast
baherus [9]

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

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