Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 = or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>
<em />
As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
<em>Where X is reaction coefficient.</em>
<em />
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X²
]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
<h3>[HI] = 0.264M</h3>
Answer:
i have no clue whats going on here but imma act like i do.....
Explanation:
The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate ( ) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate ( ) = 120 + 3(95)
molar mass of calcium phosphate ( ) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g
