Answer:
9.6 mol AgCl2
Explanation:
You have to use Avogadro's number: 6.023 x 10^23
5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2
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:
D
Explanation:
heterogeneous mixture you can see the different substances that go into it.
homogeneous mixture you can't see the difference, the mixture looks like one substance.
Answer:
The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3
Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3
Given: 0.600 moleCl2; 0.500 mole Al
Required: Excess reactant___; Number of moles of AlCl3 produced__
Solution: Use dimensional analysis using the mole-mole rel
0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al
0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2
Based on the given:
0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)
0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)
Answer: Excess reactant is Al; Limiting reactant is Cl2
The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3
Answer:

Explanation:
To solve this problem, we can use the Combined Gas Laws:

Data:
p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C
p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C
Calculations:
(a) Convert the temperatures to kelvins
T₁ = ( 1 + 273.15) K = 274.15 K
T₂ = (82 + 273.15) K = 355.15 K
(b) Calculate the new pressure
