If 3.8 kcal is released by combustion of each gram of glucose, how many kilocalories are released by the combustion of 1.50 mol
1 answer:
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<u>Answer:</u>
P2 = 778.05 mm Hg = 1.02 atm
<u>Explanation:</u>
We are to find the final pressure (expressed in atm) of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.
For this, we would use the equation:
where P1 = 724 mm hg
V1 = 3.05 L
T1 = 298 K
P2 = ?
V2 = 2.6 L
T2 = 173 K
Substituting the given values in the equation to get:
P2 = 778.05 mm Hg = 1.02 atm
Sodium-22 remain : 1.13 g
<h3>Further explanation </h3>The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually, radioactive elements have an unstable atomic nucleus.
General formulas used in decay:
T = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
half-life = t 1/2=2.6 years
T=15.6 years
No=72.5 g
Answer:
The mass percentage of calcium carbonated reacted is 2.5%.
Explanation:
The reaction is:
Thus the Kp of the equilibrium will be:
Kp = partial pressure of carbon dioxide [as the other are solid]
Moles of calcium carbonate initially present =
Let us apply ICE table to the equilibrium given:
Initial 0.2 0 0
Change -x +x +x
Equilibrium 0.2-x x x
Kp = partial pressure of carbon dioxide
Kp = Kc(RT)ⁿ
where n = difference in the number of moles of gaseous products and reactants
for given reaction n = 1
R = gas constant = 8.314 J /mol K
T = temperature = 800 ⁰C = 1073 K
Putting values
Kc =
Kc =
On calculating
x = 0.005
where x = the moles of calcium carbonate dissociated or reacted.
Percentage of the moles or mass reacted = %