<span>The pH is given by the Henderson - Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(</span><span>1.3 x 10^-5) + log(0.50/0.40)
pH = 4.98
The answer to this question is 4.98.
</span>
Answer:
A) 54.04%
B) 13-karat
Explanation:
A) From the problem we have
<em>1)</em> Mg + Ms = 9.40 g
<em>2)</em> Vg + Vs = 0.675 cm³
Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.
We can rewrite the first equation using the density values:
<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40
So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:
We <u>express Vg in terms of Vs</u>:
We <u>replace the value of Vg in equation 3</u>:
- Vg * 19.3 + Vs * 10.5 = 9.40
- (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
- 13.0275 - 19.3Vs + 10.5Vs = 9.40
Now we <u>calculate Vg</u>:
- Vg + 0.412 cm³ = 0.675 cm³
We <u>calculate Mg from Vg</u>:
- 0.263 cm³ * 19.3 g/cm³ = 5.08 g
We calculate the mass percentage of gold:
- 5.08 / 9.40 * 100% = 54.04%
B)
We multiply 24 by the percentage fraction:
- 24 * 54.04/100 = 12.97-karat ≅ 13-karat
The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O
M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5
moles of HCl = 5.5/37.5=0.14
M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)
M(C2H6O)= 2x12+ 6 + 16=46,
moles of C2H6O=200g / 46 =<span>4.35 </span><span> moles
</span>
the sum of the moles is 0.14 + <span>4.35 </span> = 4.501 moles
finally, <span>The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031
</span>
because it can be found by 0.14 / 4.501= 0.031
Answer:
A. The person weighs 56 pounds more on Mars than on the moon
Explanation:
- If the abundance of the first isotope is 68.037%, then the abundance of the second isotope is 100%-68.037%.
Substituting into the atomic mass formula,