Answer:
The heat produced is -15,1kJ
Explanation:
For the reaction:
2SO₂+O₂ → 2SO₃
The enthalpy of reaction is:
ΔHr = 2ΔHf SO₃ - 2ΔHf SO₂
As ΔHf SO₃ = -395,7kJ and ΔHf SO₂ = -296,8kJ
<em>ΔHr = -197,8kJ</em>
Using n=PV/RT, the moles of reaction are:
= <em>0,153 moles of reaction</em>
As 2 moles of reaction produce -197,8kJ of heat, 0,153moles produce:
0,153mol×
= <em>-15,1kJ</em>
<em></em>
I hope it helps!
Answer:
I think it's
there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium
The periodic table<span>, and its respective </span>melting<span> and </span>boiling points<span>. ... </span>Chemistry.2<span> The student </span>will <span>investigate and understand that the placement of elements ... </span>Families/groups<span> ... As </span>you<span> analyze </span>your <span>graph, try to </span>answer<span> the </span>following questions<span>: ... </span>period<span>. How </span>would you describe<span> the </span>trend<span> in </span>boiling point<span> as the atomic number ...</span>
Answer:
The answer is Sodium Sulfate = Na2SO4
Explanation:
Molar mass of sulfate = 1 (S) + 4 (O) = 1 (32) + 4 (16) = 32 + 64 = 96
Molar mass of sodium sulfate = 2 (23) + 96 = 46 + 96 = 142
% of Sulfate = (96/142)*100 = 67.6%
Percent mistake in Studen A,
(I) % mistake = (67.6 - 68.6)/67.6 = 1.48
(ii) % mistake = (67.6 - 66.2)/67.6 = 2.07
(iii) % mistake = (67.6 - 67.1)/67.6 = 0.74
For understudy B
(I) % mistake = (67.6 - 66.7)/67.6 = 1.33
(ii) % mistake = (67.6 - 66.6)/67.6 = 1.48
(iii) % mistake = (67.6 - 66.5)/67.6 = 1.63
Sutdent An is some how exact.
Understudy B is exact however not precise.
