To solve this problem, let us first calculate for the rate
constant k using the half life formula:
t1/2 = ln 2 / k
where t1/2 = half life period = 24,000 years, therefore k
is:
k = ln 2 / 24,000
k = 2.89 x 10^-5 / yr
Now we use the rate equation:
A = Ao e^(-k t)
where,
A = mass of Plutonium-239 after number of years
Ao = initial mass of Plutonium-239
t = number of years
A. t = 12,000 years, find A
A = 100g e^(- 2.89 x 10^-5 * 12,000)
A = 70.7 g
B. t = 24,000 years, find A
A = 100g e^(- 2.89 x 10^-5 * 24,000)
A = 50 g
C. t = 96,000 years, find A
A = 100g e^(- 2.89 x 10^-5 * 96,000)
<span>A = 6.24 g</span>
From what i can gather it looks like d
<span>3 elements
Nitrogen
Hydrogen
Oxygen
2 nitrogen 4 hydrogen and 3 oxygens
there are only 3 different elements</span>
(C) 0.1 mole of NaCl dissolved in 1,000. mL of water
<u>Explanation:</u>
The conductivity of 0.1 mole of NaCl dissolved in 1000 mL of water will be greatest as the number of ions in 0.1 mole of NaCl will be more than 0.001, 0.05 and 0.005 moles of NaCl. Greater the number of ions in the solution, greater will be the conductivity. Specific Conductivity decreases with a decrease in concentration. Since the number of ions per unit volume that carry current in a solution decrease on dilution. Hence, concentration and conductivity are directly proportional to each other.
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
