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 percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
<span>A. Missing part is ⁴He₂
You can also calculate it by adding/subtracting for atomic mass & atomic numbers
Hope this helps!
</span>
Answer:
C
Explanation:
Because nothing is left at the end of the reaction.
Take 46g / 46 g/mol = 1 mol NO2 produced
1 mol * 1/2 = 0.5 mol O2 needed (1/2 ratio between NO2 and O2)
