<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:
In the option(A) moles of HCl left are 0.100 moles which is wrong, making the option incorrect.
Explanation:
Moles of HCl = n
Molarity of HCl = 1.0M
Volume of HCl solution = 30.0 mL = 0.030 L (1 mL = 0.001L)
Moles of Fe = 
According to recation , 1mol of Fe reacts with 2 mol HCl. Then 0.01 mole of Fe will recat with :
of HCl
This means that HCl uis in excess , hence excessive reagent.
Moles of HCl left unreacted :
= 0.030 mol - 0.020 mol = 0.010 mol
But in the option moles of HCl left are 0.100 moles which is wrong, making the option incorrect.
C
0.70
I hope this is help, I’m so so sorry if I’m incorrect
Enzymes
i got this from google
Enzymes<span> are soluble protein molecules that can speed up chemical reactions in cells. These reactions include respiration, photosynthesis and making new proteins. For this reason </span>enzymes<span> are sometimes called biological catalysts</span>
The equilibrium constant of the reaction is 282. Option D
<h3>What is equilibrium constant?</h3>
The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.
Concentration of bromine = 0.600 mol /1.000-L = 0.600 M
Concentration of iodine = 1.600 mol/1.000-L = 1.600M
In this case, we must set up the ICE table as shown;
Br2(g) + I2(g) ↔ 2IBr(g)
I 0.6 1.6 0
C -x -x +2x
E 0.6 - x 1.6 - x 1.190
If 2x = 1.190
x = 1.190/2
x = 0.595
The concentrations at equilibrium are;
[Br2] = 0.6 - 0.595 = 0.005
[I2] = 1.6 - 0.595 = 1.005
Hence;
Kc = [IBr]^2/[Br2] [I2]
Kc = ( 1.190)^2/(0.005) (1.005)
Kc = 282
Learn more about equilibrium constant:brainly.com/question/15118952
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