The other dude is rude wrong
The correct answer is Dmitri Mendeleev
In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,
Answer:
Explanation:
<u>1) Data:</u>
a) m = 18 kg
b) T₁ = 285 K
c) T₂ = 318 K
d) Q = 267.3 kJ
e) S = ?
<u>2) Principles and equations</u>
The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.
The mathematical relation between the specific heat and the heat energy absorbed is:
Where,
- Q is the heat absorbed,
- S is the specific heat, and
- ΔT is the temperature increase (T₂ - T₁)
<u>3) Solution:</u>
<u>a) Substitute the data into the equation:</u>
- 267.3 kJ = 18 kg × S × (318 K - 285 K)
<u>b) Solve for S and compute:</u>
- S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)
The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.
<u>c) Convert to J /( kg . k):</u>
- 0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)
Now we can see that the option A is is the answer, assuming the units.
Answer:
- The abundance of 107Ag is 51.5%.
- The abundance of 109Ag is 48.5%.
Explanation:
The <em>average atomic mass</em> of silver can be expressed as:
107.87 = 106.90 * A1 + 108.90 * A2
Where A1 is the abundance of 107Ag and A2 of 109Ag.
Assuming those two isotopes are the only one stables, we can use the equation:
A1 + A2 = 1.0
So now we have a system of two equations with two unknowns, and what's left is algebra.
First we<u> use the second equation to express A1 in terms of A2</u>:
A1 = 1.0 - A2
We <u>replace A1 in the first equation</u>:
107.87 = 106.90 * A1 + 108.90 * A2
107.87 = 106.90 * (1.0-A2) + 108.90 * A2
107.87 = 106.90 - 106.90*A2 + 108.90*A2
107.87 = 106.90 + 2*A2
2*A2 = 0.97
A2 = 0.485
So the abundance of 109Ag is (0.485*100%) 48.5%.
We <u>use the value of A2 to calculate A1 in the second equation</u>:
A1 + A2 = 1.0
A1 + 0.485 = 1.0
A1 = 0.515
So the abundance of 107Ag is 51.5%.