Answer:
17.04%
Explanation:
Actual Value = 173.1
Measured Value = 143.6
Percent error is obtained using the equation;
Percent error = (Measured - Actual) / Actual ]* 100
Percent error = [ (143.6 - 173.1) / 173.1 ] * 100
The absolute value of (Measured - Actual) is taken,
Percent Error = [29.5 / 173.1 ] * 100
Percent Error = 0.1704 * 100 = 17.04%
Answer:
4.9 x 10⁻⁶.
Explanation:
<em>2H₂O(g) ⇄ 2H₂(g) + O₂(g),</em>
'
P of H₂ = 0.002 atm, P of O₂ = 0.006 atm, and P of H₂O = 0.07 atm.
∴ The equilibrium constant (Kp) = (P of H₂)²(P of O₂)/(P of H₂O)² = (0.002 atm)²(0.006 atm)/(0.07 atm)² = 4.9 x 10⁻⁶.
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
Answer: covalent bond
Explanation: Yeah that's the answer
The pressure would increase. When the temperature change form cold to hot, the gas will find ways to escape from containment. Thus, if it cannot escape that pressure will keep on increasing as the temperature rises.
