Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7
MThe heat energy required to raise the temperature of 0.36Kg of copper from 22 c to 60 c is calculate using the following formula
MC delta T
m(mass)= 0.360kg in grams = 0.360 x1000 = 360 g
c(specific heat energy) = 0.0920 cal/g.c
delta T = 60- 23 = 37 c
heat energy is therefore= 360g x0.0920 cal/g.c x 37 c= 1225.44 cal
Answer:
A picture .There is high reactions in progression still the energy reaction reduced to indicate complete reaction.