Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
Answer:
A population is a group of organisms belonging to the same species that live in the same area and interact with one another.
Explanation:
Levels of Ecological Organization
Level # 2. Population:
Level # 3. Community:
Level # 4. Ecosystem:
Level # 5. Biome:
Level # 6. Biosphere:
The balanced chemical reaction is
<span>2al + 3cl2 = 2alcl3
To determine the maximum amount of product, we need to determine which is the limiting reactant. Then, use the initial amount of that reactant to calculate the amount of the product that would be produced. We do as follows:
7 mol Al (3 mol Cl2 / 2 mol Al) = 10.5 mol Cl2
8 mol Cl2 ( 2 mol Al / 3 mol Cl2) = 5.3 mol Al
Thus, it is Cl2 that is the limiting reactant.
8 mol Cl2 ( 2 mol AlCl3 / 3 mol Cl2) = 5.3 moles of AlCl3 is produced</span>
Answer:
Explanation:
2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.
