Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
= 
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝) 



percentage% (∝) = 
= 0.704%
For Part B; where Concentration of B =
M



percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C= 



percentage% (∝) = 0.02608 × 100%
= 2.60%
Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2
O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving
0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2
Answer: The
value is 0.664
Explanation:
Distance travelled by solvent front = (7.7-1.45)cm = 6.25 cm
Distance travelled by unknown = (5.6-1.45) cm = 4.15 cm
The retention factor or the
value is defined as the ratio of distance traveled by the unknown to the distance traveled by the solvent front.
Thus the
value is 0.664
Answer:
i am not sure but its 2 ican"t qry
Explanation:
Answer:
Plants consume carbon through transpiration
Explanation:
In transpiration, plants lose water vapor through the stomata in their leaves. No carbon is involved in transpiration, which has an outbound direction. Nothing can be consumed through the stomata when vapor is going out of the plant. It´s like trying to get in through the exit.