<span>C4H10 + 6.5 O2 ----> 4CO2 + 5H2O
2C4H10 + 13 O2 ----> 8CO2 + 10H2O
1. Count the C on the left (4), put a 4 where the C on the right.
2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O
3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left.
4. Now multiply everything on the equation by two so you have nice integer numbers.
5. check you have the same amount of everything on each side.
Example C: left 8, right 8, etc.
I hope this helps. :)</span><span>
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Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:

Rate factor in the presence of catalyst:

Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:

![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;

Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;



The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.
Answer: IF3 (Iodine Trifluoride)
This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.
Hey there!
The molar mass of nitrogen is 14.007.
Convert grams to moles:
21.2 ÷ 14.007 = 1.51
The sample containing 21.2g of nitrogen contains 1.51 moles of nitrogen.
Hope this helps!
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