Answer: The heat of combustion per mole for acetylene is 227.7 kJ/mol.
Explanation:
The combustion equation of acetylene is as follows.
Formula to calculate enthalpy of formation for a reaction is as follows.
![\Delta H^{o}_{rxn} = \sum \Delta H_{products} - \sum \Delta H_{reactants}\\\Delta H^{o}_{rxn} = [2\Delta H^{o}_{f}(CO_{2}) + \Delta H^{o}_{f} (H_{2}O)] - [\Delta H^{o}_{f}(C_{2}H_{2}) + \frac{5}{2} \Delta H^{o}_{f} O_{2}]\\-1299.5 = 2(-393.5) + (-285.8) - \Delta H^{o}_{f} (C_{2}H_{2})\\\Delta H^{o}_{f} (C_{2}H_{2}) = 227.7 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%20%3D%20%5Csum%20%5CDelta%20H_%7Bproducts%7D%20-%20%5Csum%20%5CDelta%20H_%7Breactants%7D%5C%5C%5CDelta%20H%5E%7Bo%7D_%7Brxn%7D%20%3D%20%5B2%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28CO_%7B2%7D%29%20%2B%20%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%20%28H_%7B2%7DO%29%5D%20-%20%5B%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%28C_%7B2%7DH_%7B2%7D%29%20%2B%20%5Cfrac%7B5%7D%7B2%7D%20%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%20O_%7B2%7D%5D%5C%5C-1299.5%20%3D%202%28-393.5%29%20%2B%20%28-285.8%29%20-%20%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%20%28C_%7B2%7DH_%7B2%7D%29%5C%5C%5CDelta%20H%5E%7Bo%7D_%7Bf%7D%20%28C_%7B2%7DH_%7B2%7D%29%20%3D%20227.7%20kJ%2Fmol)
Thus, we can conclude that heat of combustion per mole for acetylene is 227.7 kJ/mol.
Answer:
1/32 of the original sample
Explanation:
We have to use the formula
N/No = (1/2)^t/t1/2
N= amount of radioactive sample left after n number of half lives
No= original amount of radioactive sample present
t= time taken for the amount of radioactive samples to reduce to N
t1/2= half-life of the radioactive sample
We have been told that t= five half lives. This implies that t= 5(t1/2)
N/No = (1/2)^5(t1/2)/t1/2
Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;
N/No= (1/2)^5
N/No = 1/32
Hence the fraction left is 1/32 of the original sample.
MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
Answer:
190.2 grams
Explanation:
each arrow points to the amount it weighs, just add it all up
