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Pepsi [2]
3 years ago
15

Urea is a common fertilizer with the formula CO(NH2)2 that is sometimes used as a chemical de-icer for icy road surfaces in the

winter months. Adding a solute to a solvent causes the freezing point to be lowered and allows the solvent to remain liquid at lower temperatures. The value of Kf for water is 1.86°C/m, and its normal freezing point is 0.00°C. Urea is a nonvolatile nonelectrolyte solute. What molality of urea is required to lower the freezing point of ice water by 1.13°C?
Chemistry
1 answer:
nordsb [41]3 years ago
4 0

Answer:

The answer to your question is  molality = 0.61

Explanation:

Freezing point is the temperature at which a liquid turns into a solid if a solute is added to a solution, the freezing point changes.

Data

Kf = 1.86 °C/m

molality = ?

ΔTc = 1.13°C

Formula

ΔTc = kcm

Solve for m

m = ΔTc/kc

Substitution

m = 1.13 / 1.86

Simplification and result

m = 0.61

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3 years ago
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5 0
3 years ago
Mature elephants consume an average of 600 pounds of food per day. If an elephant
MAXImum [283]

Answer:

\large \boxed{\text{4 600 000 kg or 4600 Mg}}

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its  

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}

3. Convert days to pounds of feed

\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}

4. Convert pounds to kilograms and megagrams

\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

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3 years ago
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