Answer: Hmmmmm that's crazy....
There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.
Fraction remaining (FR) = 0.5n
n = number of half lives that have elapsed
In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).
0.0156 = 0.5n
log 0.0156 = n log 0.5
-1.81 = -0.301 n
n = 6.0 half lives have elapsed
Explanation:
Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in 
of ethane gas
And, as we know that
1 mole of ethane molecule contains 
molecules of ethane
2.869 moles of ethane molecule contains 
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Answer:
Compound A is succinic anhydride and B is methyl succinate (the monomethyl ester).
Structural Diagram is attached.
Explanation:
A.
Succinic anhydride appears as colorless needles or white crystalline solid. ... Succinic anhydride, also called dihydro-2, 5-furandione, is an organic compound with the molecular formula C4H4O3. It is the acid anhydride of succinic acid.
B.
Monomethyl succinate is a dicarboxylic acid monoester that is succinic acid in which one of the carboxy groups has been converted to its methyl ester. It is a dicarboxylic acid monoester and a hemisuccinate. ... They have the general structure RC(=O)OR', where R=fatty aliphatic tail or organyl group and R'=methyl group.
<span>The answer to this question would be: B.They are different.
A compound will have different characteristics from their element. </span>The interaction of different elements in a compound makes it so. They might have similar or opposite characteristic but it didn't always happen so those answers are not best.
Answer:
E_a = 103.626 × 10³ KJ/mol
Explanation:
Formula to solve this is given by;
Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))
Where;
k2 is rate constant at second temperature
k1 is rate constant at first temperature
R is universal gas constant
T1 is first temperature
T2 is second temperature
We are given;
k1 = 2.8 × 10^(-3) /s
k2 = 4.8 × 10^(-4) /s
R = 8.314 J/mol.k
T1 = 60°C = 333.15 K
T2 = 45°C = 318.15 K
Thus;
Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))
We now have;
-0.76592 = -0.00000739121E_a
E_a = -0.76592/-0.00000739121
E_a = 103.626 × 10³ KJ/mol
