Phosphate group , pentose sugar , nitrogen base nucleotide
Answer:
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Hope that helps :)
Answer:
4.86 moles of HCl
Explanation:
1. First write the balanced chemical equations involved in the process:
2. Calculate what amount of 
is formed in the first reaction.
As the yield of each reaction is 90.0%, the amount of produced is the following:
3. Calculate the amount of HCl produced.
The total amount of HCl produced with a 90.0% yield is:
The acid dissociation constant is defined as Ka = [H+][A-]/[HA] where [H+], [A-] and [HA] are the concentrations of protons, conjugate base, and acid in solution respectively. Assuming this is a weak acid as the pH is quite high for a 1.35 M solution, we can assume that the change in [HA] is negligible and therefore [HA] = 1.35 M.
To calculate [H+] we can use the relationship pH = -log[H+], rearranging to give: [H+] = 10^(-pH) = 10^(-2.93) = 1.17 x 10^(-3).
Since the acid is relatively concentrated we can assume therefore that [H+] = [A-] as for each proton dissociated, a conjugate base is formed.
Therefore, we can calculate Ka as:
Ka = [H+]^2/[HA] = (1.17 x 10^-3 M)^2/1.35 = 1.01 x 10^-6 M
