Answer:
10.5 × 10^5 M
Explanation:
E°cell = E°cathode - E°anode
E°cell = 0.85 - (-0.74) = 1.59 V
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
1.499 = 1.59 - 0.0592/6 log [Cr^3+]/6.35x10^-4
1.499 - 1.59 = - 0.0592/6 log [Cr^3+]/6.35x10^-4
-0.091 = -0.00987 log [Cr^3+]/6.35x10^-4
-0.091/ -0.00987 = log [Cr^3+]/6.35x10^-4
9.22 = log [Cr^3+]/6.35x10^-4
Antilog (9.22) = [Cr^3+]/6.35x10^-4
1.66 × 10^9 = [Cr^3+]/6.35x10^-4
[Cr^3+] = 1.66 × 10^9 × 6.35x10^-4
[Cr^3+] = 10.5 × 10^5 M
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 0.125 mL
Atomic number of an element correlates to the number of protons an atom of that element will have. An atom of this element would have 63 protons.
Answer is: <span>requirement of valine is 0,0222 mol for a100 kilograms adult.
</span>m(C₅H₁₁NO₂) = 2,60 g.
M(C₅H₁₁NO₂) = 5 · 12 + 11 · 1 + 1 · 14 + 2 · 16 · g/mol.
M(C₅H₁₁NO₂) = 117 g/mol.
n(C₅H₁₁NO₂) = m(C₅H₁₁NO₂) ÷ M(C₅H₁₁NO₂).
n(C₅H₁₁NO₂) = 2,60 g ÷ 117 g/mol.
n(C₅H₁₁NO₂) = 0,0222 mol.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.
First convert the 2 temperatures to degrees K:
4 + 273 ---> 277 and 22---> 295
So using Charles' Law
V1 / T1 = V2 / T2
0.50 / 295 = V2 / 277
solving for V2 gives a volume of 0.469 L to nearest milliliter
