Answer:
= 25.05°C
Explanation:
Given:
the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.
mass = 0.905g of dimethylphthalate
molar mass = 194.18g dimethylphthalate
number of moles of dimethylphthalate = ???
= 21.5°C
= 6.15 kJ/°C
= ???
since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;
0.905g of dimethylphthalate × 
number of moles of dimethylphthalate = 0.000466 moles
Heat released = moles of dimethylphthalate × heat of combustion
= 0.000466 moles × 4685 kJ
= 21.84 kJ
∴ Heat absorbed by the calorimeter = 
21.84 kJ =6.15 kJ/°C 
21.84 KJ = 
21.84 KJ = 
- 132.225 kJ
21.84 KJ + 132.225 kJ =
154.065 kJ =
= 
=25.05°C
Ur ans is option d
addition
Chemicals that’s and ghebreziqbiher
Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
Answer:
2578.99 years
Explanation:
Given that:
100 g of the wood is emitting 1120 β-particles per minute
Also,
1 g of the wood is emitting 11.20 β-particles per minute
Given, Decay rate = 15.3 % per minute per gram
So,
Concentration left can be calculated as:-
C left = ![[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5Cfrac%7B11.20%5C%20per%5C%20minute%7D%7B15.3%5C%20per%5C%20minute%5C%20per%5C%20gram%7D%5Ctimes%20%5BA_0%5D%3D%200.7320%5BA_0%5D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Also, Half life of carbon-14 = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.000120968 year⁻¹
Time =?
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
So,
![\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.000120968%5Ctimes%20t%7D)
![\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B0.7320%5BA_0%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-0.000120968%5Ctimes%20t%7D)
<u>t = 2578.99 years</u>
