Chemistry, 02.04.2021 22:50 Anthony1442

Food Energy Math For this assignment you will have to calculate the food energy required to power various devices. You will be required to select 5 foods you have at your home and calculate how many of each item will be required to perform each mechanical or electrical process. Remember to calculate the per item calories; a serving of 2 chocolate sandwich cookies is 80 calories, one sandwich cookie is 40 calories.
For example, a coffee maker will require about 400 Watts of energy to run for an hour. 1 Kilocalorie is equal to 1.163 Watts, so the coffeemaker will require 343.9 Kilocalories to run for this length of time. If it could be converted to electricity how many fastfood cheeseburgers, (417 calories) would that require? Only 1 cheeseburger
Conversions:
1 Kcal- 1.163 Watts per hour
31,500 Kcal – 1 Gallon of gasoline
Food Calories per item

1. 60 Watt light bulb- 60 Watts Per Hour
Food / Calories Number required

2. Modern Video Game Console – 120 Watts Per Hour
Food / Calories Number required

3. Drive a car for 30 miles – 1 Gallon of gasoline
Food / Calories Number required

4. Run emergency generator for 1 hour - .75 Gallon of gasoline
Food / Calories Number required

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Answer from: Quest

c-14 is carbon-14, or radiocarbon. it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.

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Its trait is different and it the organism survive

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Answersublimation;

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the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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