Answer:
the equator is closer to the sun
Answer : The mass of
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of
.

Now we have to calculate the moles of
.
The balanced chemical reaction will be,
produced from 1 mole of 
So, 0.005 mole of
produced from 0.005 mole of 
Now we have to calculate the mass of 


Therefore, the mass of
needed are, 1.515 grams.
Endothermic or C, if you would like a reason why, do not hestitate to ask
Answer:
A) [H3PO4] will increase, [KH2PO4] will decrease, and pH will slightly decrease.
Explanation:
A buffer is a solution which resists changes to its pH when a small amount of acid or base is added to it.
Buffers consist of a weak acid (HA) and its conjugate base (A–) or a weak base and its conjugate acid. Weak acids and bases do not completely dissociate in water, and instead exist in solution as an equilibrium of dissociated and undissociated species. When a small quantity of a strong acid is added to a buffer solution, the conjugate base, A-, reacts with the hydrogen ions from the added acid to form the weak acid and a salt thereby removing the extra hydrogen ions from the solution and keeping the pH of the solution fairly constant. On the other hand, if a small quantity of a strong base is added to the buffer solution, the weak acid dissociates further to release hydrogen ions which then react with the hydroxide ions of the added base to form water and the conjugate base.
For example, if a small amount of strong acid is added to a buffer solution that is 0.700 M H3PO4 and 0.700 M KH2PO4, the following reaction is obtained:
KH₂PO₄ + H+ ----> K+ + H₃PO₄
Therefore, [H₃PO₄] will increase, [KH₂PO₄] will decrease, and pH will slightly decrease.:
According to the balanced chemical equation:
4 HPO₃ + 12 C → 2 H₂ + 12 CO + P₄
4 moles of HPO₃ ---gives---> 12 moles of CO
2.73 moles of HPO₃ ---gives---> ? moles of CO
so number of moles of CO =

= 8.19 moles of CO
Number of molecules of CO = number of moles * Avogadro's number
= 8.19 * (6.022 * 10²³) = 4.93 * 10²⁴ molecules