Answer:
answer A
Explanation:
A) the quantity of usable energy declines with each transformation → True . Since the entropy increases , the amount of energy that can not be converted to useful energy increases and since the total amount of energy is conserved, the quantity of useful energy decreases.
B) energy can be neither created nor destroyed → False in the context of entropy , since the energy is conserved regardless of the changes in entropy (First law → conservation of energy vs second law → increase of entropy)
C) life should be impossible → False . Since the second law states that the entropy of the <u>universe </u>increases with time . Then the system (life) can experience a decrease in entropy at the expense of a larger increase in entropy of the surroundings ( so the net increase is positive)
D) it is not possible to observe an increase in molecular organisation → False . Same as C. A system can experience a decrease in entropy at the expense of a larger increase in entropy of the surroundings ( so the net increase is positive)
A proton is the same as an H+ ion, and Arrhenius acids are the ones that release H+ in solution, so the answer is A
This is a combination reaction because two molecules becomes one
A + B ---> AB
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%
m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water
%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
