Answer:
5) donating H to a solution when they have been depleted and accepting H when they are in excess.
Explanation:
A buffer solution is a solution that undergoes a negligible change in pH in addition of moderate quantities of acid or alkali. In other words, a buffer solution is one that resists a change in pH on addition or dilution of small amounts of acids or alkalis.
So, from the given question:
Buffers are substances that help resist shifts in pH by donating H to a solution when they have been depleted and accepting H when they are in excess.
N(H₂O)=1g÷18g/mol=0,055mol
N(H₂O)=0,055mol · 6·10²³ 1/mol (Avogadro number)= 3,33·10²² molecules.
<u>Given:</u>
Isotopes of silver with atomic masses:
Ag - 107 and Ag - 109
% Abundance of Ag-107 = 51.35 %
% Abundance of Ag-109 = 48.65 %
<u>To determine:</u>
The average atomic mass of Ag
<u>Explanation:</u>
The average atomic mass can be calculated using the formula-
Average Atomic Mass = f1M1 + f2M2 + .......+fnMn
where:
f = fraction representing the abundance of that isotope
M = atomic mass of that isotope
In the case of silver
Average atomic mass = (51.35/100)*107 + (48.65/100)*109
= 54.9445 + 53.0285 = 107.973 amu
Ans: the average atomic mass for silver is 107.973 amu
Answer:
A. [isocitrate]/[citrate] = 0.724
B. [citrate] = 24.1 mM
Explanation:
Using the equation, ∆G'° = -RTlnK'eq
Where, ∆G'° = 0.8 KJ/mol = 800 J/mol; R is molar gas constant = 8.315 J/mol; T is standard temperature = 298 K; Keq is equilibrium constant = [isocitrate]/[citrate]
Making Keq subject of formula in the above equation;
Keq = e^(-∆G'°/RT)
= e^ {-800/(8.315*298)}
= e^(-0.323)
Keq = 0.724
Therefore, [isocitrate]/[citrate] = 0.
724
B. Keq = [isocitrate]/[citrate]
Where Keq = 0.724, [isocitrate] = 0.03mM.
[citrate] = Keq/[isocitrate]
= 0.724/0.03
[citrate] = 24.1 mM
The empirical formula, <span>C<span>H2</span></span>, has a relative molecular mass of
<span>1×<span>(12.01)</span>+2×<span>(1.01)</span>=14.04</span>
This means that the empirical formula must be multiplied by a factor to bring up its molecular weight to 70. This factor can be calculated as the ratio of the relative masses of the molecular and empirical formulas
<span><span>7014.04</span>=4.98≈5</span>
Remember that subscripts in molecular formulas must be in whole numbers, hence the rounding-off. Finally, the molecular formula is
<span><span>C<span>1×5</span></span><span>H<span>2×5</span></span>=<span>C5</span><span>H<span>10</span></span></span>
