Answer:
the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.
Explanation:
3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)
∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants
∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]
∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K
∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.
For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,
then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.
Answer:
96.09 g/mol
Explanation:
You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.
If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).
Also, please take note that I will be using the unit g/mol for all the weights. Thus,
Step 1
N = 14.01 g/mol
H = 1.008 g/mol
O = 16.00 g/mol
C = 12.01 g/mol
Since your compound is
(
N
H
4
)
2
C
O
3
, you need to multiply the atomic weights by their subscripts. Therefore,
Step 2
N = 14.01 g/mol × 2 =
28.02 g/mol
H = 1.008 g/mol × (4×2) =
8.064 g/mol
O = 16.00 g/mol × 3 =
48.00 g/mol
C = 12.01 g/mol × 1 =
12.00 g/mol
To get the mass of the substance, we need to add all the weights from Step 2.
Step 3
molar mass of
(
NH
4
)
2
CO
3
=
(28.02 + 8.064 + 48.00 + 12.01) g/mol
=
96.09 g/mol
this is a google search and a example i hope is helps to solve
Answer:
Oxygen.
Explanation:
Hello,
Yeast fermentation in the food industry must be carried out under anaerobic conditions as long as when oxygen is present respiration occurs rather than fermentation.
Best regards.
This method relies on the exothermic reaction of lithium hydroxide with carbon dioxide gas to create lithium carbonate (Li2CO3) solid and water (H2O).
might be wrong but....oh well
scandium is a transition metal and shows a valency of 3.
