Answer:
There will be produced:
2.97 moles HMnO4
4.45 moles Pb(NO3)2
2.97 moles H2O
Explanation:
Step 1: Data given
Manganese(II) oxide = MnO2
lead(IV) oxide = PbO2
nitric acid = HNO3
Moles of HNO3 = 8.90 moles
Step 2: The balanced equation
2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O
Step 3: Calculate moles of reactants and products
For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water
For 8.90 moles of HNO3, there will react:
8.90 / 3 = 2.97 moles MnO2
8.90 / 2 = 4.45 moles PbO2
There will be produced:
8.90/3 = 2.97 moles HMnO4
8.90/2 = 4.45 moles Pb(NO3)2
8.90 / 3 = 2.97 moles H2O
Answer:
96.66 g of Fe₃O₄
Explanation:
In order to calculate the weight of Fe₃O₄ in 100.0 g of Fe₂O₃ we need elements in common between both substances, for instance, the mass of Fe in each one. We will use the molar mass of Fe = 55.84g/mol The conversion factors we need are:
- 159.69 g of Fe₂O₃ contain 2 × 55.84 = 111.68 g of Fe
- 231.54 g of Fe₃O₄ contain 3 × 55.84 = 167.52 g of Fe
Then, we can use proportions:
Atomic mass Ca = 40.078 a.m.u
40.078 g -------------------- 6.02x10²³ atoms
72.8 g ---------------------- ??
72.8 x ( 6.02x10²³) / 40.078 =
4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms
hope this helps!
Answer:
i is d " The carbons on either side of the double bond are pointed in the
same direction."Explanation:
