Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
hooc are carboxyl groups
your r or amino groups are those unique structures which have different atoms in them. your nh2 groups are your hydrogen atoms
Answer:
The minimum molecular weight of the enzyme is 29.82 g/mol
Explanation:
<u>Step 1:</u> Given data
The volume of the solution = 10 ml = 10*10^-3L
Molarity of the solution = 1.3 mg/ml
moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol
<u>Step 2:</u> Calculate the mass
Density = mass/ volume
1.3mg/mL = mass/ 10.0 mL
mass = 1.3mg/mL *10.0 mL = 13mg
<u>Step 3:</u> Calculate minimum molecular weight
Molecular weight = mass of the enzyme / number of moles
Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol
Molecular weight = 29.82 g/mole
The minimum molecular weight of the enzyme is 29.82 g/mol
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![Cu_4(OH)_6(SO_4)](https://tex.z-dn.net/?f=Cu_4%28OH%29_6%28SO_4%29)
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You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =
![\frac{2}{3} \times 2.3 \times 10^{-6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%202.3%20%5Ctimes%2010%5E%7B-6%7D%20)