Hello there!
Hydrogen has one proton because the number of protons in an atom is equal to the atomic number of the element. S
ince hydrogen's atomic number is 1, it has one proton.
I hope that's help !
Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
Answer:
Explanation:
Hello,
In this case, we use the Avogadro's number to compute the molecules of C2F4 whose molar mass is 100 g/mol contained in a 485-kg sample as shown below:
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Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
