N=m(g)/Mwt (3(1)+1(14))
1.35=m/17
m=17*1.35=22.95 g
<h3>
Answer:</h3>
23.95 °C
<h3>
Explanation:</h3>
We are given;
- Mass of the sample is 0.5 gram
- Quantity of heat released as 50.1 Joules
- Specific heat capacity is 4.184 J/g°C
We are required to calculate the change in temperature;
- Quantity of heat absorbed is given by the formula;
- Q = mass × specific heat capacity × Change in temperature
That is, Q = mcΔT
Rearranging the formula;
ΔT = Q ÷ mc
Therefore;
ΔT = 50.1 J ÷ (0.5 g × 4.184 J/g°C)
= 23.95 °C
Therefore, the expected change in temperature is 23.95 °C
This is not a question kid
Xe +f2 →Xef2
ΔXe = ΣB.P reactants - Σ B.d products
-108k.s/ mol = B. D f₂ - 2 B.D xe-f
-108 k.s/mol =155 k.s/mol - 2B.Dxe-f
263kJ/mol/2 = B. D xe-f
B.D xef = 131.5 kJ/mol
132 kJ/mol
