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3 years ago

What is the kcalorie value of a meal supplying 30 g of carbohydrates, 8 g of protein, 10 g of fat, and 3 g of alcohol?

Chemistry

2 answers:

hammer [34]3 years ago

7 0

Answer:

mass of carbohydrate = 30g

mass of protein = 8g

mass of fat = 10g

mass of alcohol 3g

Caloric value of carbohydrate = 4

Caloric value of protein = 4

Caloric value of fat = 9

Caloric value of alcohol = 6.9

The kcalorie of the meal = mass *caloric value of each food item

Kcalorie value =

30*4+8*4+10*9+3*6.9

Kcalorie = 120+32+90+20.7

Kcalorie = 262.7

Assoli18 [71]3 years ago

5 0

Answer:

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Explanation:

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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

A chemist weighed out 20.7 g of sodium . Calculate the number of moles of sodium she weighed out

Pachacha [2.7K]

Answer:

about 0.9 mol

Explanation:

there are 22.990 g/mol of Na

20.7/22.99 = 0.900391 mol

about 0.9 mol

8 0

3 years ago

Which is an element? ... air ...carbon dioxide ...hydrogen ...water Description

Natalka [10]

All Elements are obtained in the periodic table & water is Not an element it is a compound.

5 0

3 years ago

Read 2 more answers

How many moles of NaCl are needed to make 10.0L of a 5 M solution of Salt water?*

vesna_86 [32]

Answer: 50 mol

Explanation:

$Molarity=\frac{Moles -of -solute}{liters -of -solution}$

$M=\frac{mol}{L}$

$mol=M*L$

$mol=5*10\\mol=50mol$

3 0

3 years ago

THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.

Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

6 0

3 years ago