<span>when the number of moles Ca = mass of Ca / molar mass of Ca.
and we can get the molar mass of Ca, it is = 40 g/mol
and we have already the mass of Ca (given) = 9.8 g
so, by substitution: the moles Ca = 9.8 g / 40 g/mol
= 0.245 moles</span>
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the calculated value is Ea is 18.2 KJ and A is 12.27.
According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.
At 500K, K=0.02s−1
At 700K, k=0.07s −1
The Arrhenius equation can be used to calculate Ea and A.
RT=k=Ae Ea
lnk=lnA+(RT−Ea)
At 500 K,
ln0.02=lnA+500R−Ea
500R Ea (1) At 700K lnA=ln (0.02) + 500R
lnA = ln (0.07) + 700REa (2)
Adding (1) to (2)
700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.
=ln [0.02/0 .07]
Ea= 2/35×100×8.314×1.2528
Ea =18227.6J
Ea =18.2KJ
Changing the value of E an in (1),
lnA=0.02) + 500×8.314/18227.6
= (−3.9120) +4.3848
lnA=0.4728
logA=1.0889
A=antilog (1.0889)
A=12.27
Consequently, Ea is 18.2 KJ and A is 12.27.
Learn more about Arrhenius equation here-
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Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, 
Explanation:
The reaction equation for given reaction is as follows.
Here, 1 mole of reacts with 3 moles of 
.
As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide 
is 152 g/mol.
Number of moles is the mass of substance divided by its molar mass. So, moles of is calculated as follows.
Now, moles of .given by 0.5 mol of is calculated as follows.
As molar mass of is 2.016 g/mol. Therefore, mass of is calculated as follows.
Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, .
