Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
Answer:
The products are: KCl03 and H20.
Explanation:
The reaction between HC03 (chloric acid) and KOH (potassium hydroxide) is:
HC03 + KOH ----> KCl03 (KCl03 and H20) + H20 (water)
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<em>The reaction is of the double displacement type (in this case parts of the reagents are exchanged, producing two generating new compounds).</em>
Answer is: the ratio of the effusion rate is 1.59 : 1.
1) rate of effusion of carbon monoxide gas = 1/√M(CO).
rate of effusion of carbon monoxide gas = 1/√28.
rate of effusion of carbon monoxide gas = 0.189.
2) rate of effusion of chlorine = 1/√M(Cl₂).
rate of effusion of chlorine = 1/√70.9.
rate of effusion of chlorine = 0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine =
= 0.189 : 0.119 / ÷0.119.
rate of effusion of carbon monoxide : rate of effusion of chlorine = 1.59 : 1.
Answer:
KJ8RT898TGHO7-6734354546746R476
Explanation:
