Answer:
Explanation:
Given that:

From above:

To predict the effect of the addition of Br₂(g);
The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr
The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.
Answer:
The correct answer is: d. The pKa of the chosen buffer should be close to the optimal pH for the biochemical reaction.
Explanation:
The buffer resist or maintain the change in pH in case of Acid or basic addition to the solution. The buffer capacity should be within one or two pH units when compared to the optimal pH.
Thus it is important to select a buffer with pKa close to the optimum pH of the reaction because the ability for the buffer to maintain the pH is is great at the pH close to pKa.
Answer:
There is 5.56 g of gold for every 1 g of chlorine
Explanation:
The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction 
You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

The proportion is the equal relationship that exists between two reasons and is represented by: 
This reads a is a b as c is a d.
To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

Solving for the mass of gold gives:

mass of gold= 5.56 grams
So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.