I think the answer might be 51.14% since the formula had to equal a 100% just add 41.86 and 6.98 and subtract the sum to 100.
Answer:
Explanation:
The relation between equilibrium constant and Ecell is given below .
E⁰cell = (RT / nF ) lnK , F is faraday constant T is 273 + 25 = 298 K
E⁰cell = 1.46 - 1.21 = .25 V
n = 2
Putting the values
.25 = (8.314 x 298 lnK) / (2 x 96485 )
lnK = 19.47
K = 2.85 x 10⁸
2 )
Change in free energy Δ G
Δ G ⁰ = nE⁰ F
n = 4
E⁰ = .4 + .83 = 1.23 V
Δ G ⁰= 4 x 1.23 x 96485
= 474706 J / mol
3 )
E⁰cell = (RT / nF ) lnK
n = 2
1.78 = 8.314 x 298 lnK / 2 x 96485
lnK = 138.638
K = 1.62 x 10⁶⁰
Answer:
C5H6N2O3
Explanation:
First the empirical formulas
C= 41.8÷12= 2.48
H= 4.7÷1= 4.7
O= 37.3÷ 16= 2.33
N= 16.3÷14 = 1.16
Divide by the smallest
C= 3.48/1.16=3
H= 4.7/1.16=4.1
O= 2.33/1.16=2
N= 1.16/1.16=1
Therefore empirical formula = C3H4NO2
To calculate molecular formula for osmotic pressure,
π= cRT
Or
π=cgRT/M where cg is in g/liter & T is temperature in Kelvin. Thus
π= (7.480*0.0821*300)/ M
M= 184.23/1.43
M= 128.83
To find molecular Formula
Molecular Mass= (empirical mass)n
128.83= (C3H4NO2)n
128.83= 86n
n= 1.5
Therefore the molecular formula
(C3H4NO2)1.5
= C4.5H6N1.5O3
Approximately
C5H6N2O3
Answer
During the process of polymerization,
<h2><em>
monomers</em></h2>
combine by sharing electrons. This process forms a
<h2><em>
polymer</em></h2>
which is made of repeating subunits. The resulting material is used in a variety of ways.
Explanation:
Answer:
Half life of phosphorous-32 = 14 days
Explanation:
Given data:
Total mass of phosphorous-32 = 2.0 g
After 42 days mass left = 0.25 g
Half life of phosphorous-32 = ?
Solution:
First of all we will calculate the number of half lives passed.
At time zero = 2.0 g
At first half life = 2.0 g/2 = 1.0 g
At 2nd half life = 1.0 g/2 = 0.5 g
At 3rd half life = 0.5 g/2 = 0.25 g
Half life:
Half life = T elapsed / half lives
Half life = 42 days/ 3
Half life = 14 days
