Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)
Therefore, the molal freezing point depression constant of X is
Answer:
The correct answer is 1.
Explanation:
The metabolism of homocysteine produces a sulfur amino acid that is normally formed from methionine during the fulfillment of its function as a donor of methyl groups. Metabolic fate such as remethylation and transsulfuration, involving the enzymatic forms of the vitamins folacin, B12, and B6, gives rise to homocysteine and mixed disulfides including so-called protein-linked homocysteine, the main form circulating in plasma. B6 deficiency would have a direct impact on the metabolism of homocysteine to cysteine.
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Answer:
6,8 g
Explanation:
c = 4.18 J/(g * °C) = 4180 J / (kg * °C)
= 25 °C
= 36,4 °C
Q = 325 J
The formula is: Q = c * m * (
)
m = 
Calculating:
m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g
Answer:
cold
Warm air lifted over a moving cold air mass will produce a _____ front.
