Answer:
1.1713 moles
Explanation:
RFM of N2O5= (14*2)+(16*4)=108
Moles of N2O5= Mass/RFM= 63.25/108= 0.5856 moles
Mole ratio of N2O5:NO2 = 2:4
Therefore moles of NO2= 4/2*0.5856= 1.1713 moles
Answer:
d. Sum of product enthalpies minus the sum of reactant enthalpies
Explanation:
The standard enthalpy change of a reaction (ΔH°rxn) can be calculated using the following expression:
ΔH°rxn = ∑n(products) × ΔH°f(products) - ∑n(reactants) × ΔH°f(reactants)
where,
ni are the moles of products and reactants
ΔH°f(i) are the standard enthalpies of formation of products and reactants
Answer:
- <u><em>1.7 × 10³ kg of ore.</em></u>
Explanation:
Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.
Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:
- X × 71% × 82% = 1.0 × 10³ kg
↑ ↑ ↑ ↑
(mass of ore) (% of Al in the ore) (yield) ( Al metal to obtain)
You must just simplify, solve and compute:
- X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg
Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.
The answer is AIM NOT POSTIVE
The answer is A, good luck
