Answer:
Check explanation
Explanation:
From the question, the parameters given are 64.7g of benzene,C6H6; a starting temperature of 41.9°C and bringing it to 33.2°C.
Molar mass of benzene,C6H6= 78.11236 g/mol.
Things to know: heat capacity of benzene, C6H6= 1.63 J/g.K, the heat of fusion = 9.87 kj/mol.
STEP ONE(1): ENERGY USED IN MELTING BENZENE SOLID.
Using the formula below;
Energy used in melting the solid(in JOULES) = (mass of benzene/molar mass of benzene) × heat of Fusion.
=(64.7 g of C6H6/ 78.11236(g per mol) of C6H6) × 9.87 kJ per mol.
= 8.175 J.
= 0.008175 kJ.
STEP TWO (2): ENERGY OF HEATING THE LIQUID.
It can be calculated from the formula below;
Energy= heat capacity (J/g.K) × mass of benzene× (∆T).
= 1.63 J/g.K × 64.7 × (41.9-33.2).
= 917.5J.
= 0.9175 kJ.
Energy required to boil benzene= Energy required to melt the bezene + energy required for boiling.
= 0.008175+ 0.9175.
= 0.93kJ
Approximately, 1 kJ
Answer:
2N2 + O2 ⇒ 2N2O
Explanation:
Since I cannot see the product's of the second chemical equation I will only solve the first one.
In order to balance a chemical equation you need to make sure that the atoms on both sides are equal.
In this case we have...
N2 + O2 = N2O
N = 2
O = 2
N = 2
O = 2
Change the coefficient:
N2O = 2N20
2 × 2 = 4
1 × 2 = 2
2N2 + O2 ⇒ 2N2O
Hope this helps.
Answer:
For example, N2O4 is referred to as dinitrogen tetroxide, not dinitrogen tetraoxide, and CO is called carbon monoxide, not carbon monooxide.
...
Binary molecular (covalent) compounds.
Prefixes used in chemical nomenclature
prefix number of atoms
tetra- 4
penta- 5
hexa- 6
Explanation:
Answer:
0.84M
Explanation:
Hello,
At first, the equilibrium constant should be computed because the whole situation is at the same temperature so it is suitable for the new condition, thus:
![K_{eq}=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \\K_{eq}=\frac{0.6^2}{0.2*0.2}\\ K_{eq}=9](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNO%5D%5E2_%7Beq%7D%7D%7B%5BN_2%5D_%7Beq%7D%5BO_2%5D_%7Beq%7D%7D%20%5C%5CK_%7Beq%7D%3D%5Cfrac%7B0.6%5E2%7D%7B0.2%2A0.2%7D%5C%5C%20K_%7Beq%7D%3D9)
Now, the new equilibrium condition, taking into account the change x, becomes:
![9=\frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}}\\9=\frac{[0.9+2x]^2}{[0.2-x][0.2-x]}](https://tex.z-dn.net/?f=9%3D%5Cfrac%7B%5BNO%5D%5E2_%7Beq%7D%7D%7B%5BN_2%5D_%7Beq%7D%5BO_2%5D_%7Beq%7D%7D%5C%5C9%3D%5Cfrac%7B%5B0.9%2B2x%5D%5E2%7D%7B%5B0.2-x%5D%5B0.2-x%5D%7D)
Nevertheless, since the addition of NO implies that the equilibrium is leftward shifted, we should change the equilibrium constant the other way around:
![\frac{1}{9} =\frac{[N_2]_{eq}[O_2]_{eq}}{[NO]^2_{eq}}\\\frac{1}{9} =\frac{[0.2+x][0.2+x]}{[0.9-2x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D%20%3D%5Cfrac%7B%5BN_2%5D_%7Beq%7D%5BO_2%5D_%7Beq%7D%7D%7B%5BNO%5D%5E2_%7Beq%7D%7D%5C%5C%5Cfrac%7B1%7D%7B9%7D%20%3D%5Cfrac%7B%5B0.2%2Bx%5D%5B0.2%2Bx%5D%7D%7B%5B0.9-2x%5D%5E2%7D)
Thus, we arrange the equation as:
Finally, the new concentration is:
![[NO]_{eq}=0.9-0.06=0.84M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D0.9-0.06%3D0.84M)
Best regards.
