Answer:
A scientist measures the standard enthalpy change for the following reaction to be -139.5 kj :
h2(g) + c2h4(g)c2h6(g)
based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of c2h4(g) is _____ kj/mol
Explanation:
Hydrogen ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K)
H2 (g)
0
0
130.7
Carbon ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K)
C2H6 (g)
-84.7
-32.8
229.6
<u>Ans: P = 22.5% and Cl = 77.5%</u>
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<u>Given:</u>
Total Mass of sample containing P and Cl = 50.51 g
Mass of P produced = 11.39 g
<u>To determine:</u>
Mass % of P and Cl
<u>Explanation:</u>
Mass % of a given element in a total mass is generally expressed as:
Mass % of element = [mass of element/total mass]*100
Here,
Total mass = mass of P + mass of Cl
mass of Cl = Total - mass of P = 50.51-11.39 = 39.12 g
% P = [11.39/50.51]*100 = 22.5%
%Cl = [39.12/50.51]*100 = 77.5%
Brainest please thank you
Answer:
4.12 moles
Explanation:
We can solve this problem with the Ideal Gases Law.
P . V = n . R . T
In our first case we have:
P = 2.3 atm
V = 32.8 L
n = 2.98 moles
T → 35°C + 273 = 308K
Let's replace data for the second case:
2.3 atm . 45.3L = n . 0.082 . 308K
n = (2.3 atm . 45.3L) / (0.082 L.atm/mol.K . 308K)
n = 4.12 moles