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frozen [14]
3 years ago
11

You determine that Lauren's EER is 2200 kcal/day. You recommend and intake of 60% CHO, 15% protein, and 25% fat. Approximately h

ow many grams of carbohydrate can Lauren have per day?
Chemistry
1 answer:
const2013 [10]3 years ago
6 0

Answer:

The answer to your question is 330 g of CHO

Explanation:

Data

Calories needed = 2200 kcal/day

CHO = 60%

Proteins = 15%

Fats = 25%

Grams of carbohydrates needed = ?

Process

1.- Calculate the number of calories  in 60% of 2200 kcal

                      2200 kcal ----------------  100%

                          x            ---------------      60%

                       x = (60 x 2200) / 100

                       x = 1320 kcal

2.- Calculate the grams of CHO

                       1 g of CHO ---------------- 4 kcal

                       x                 ----------------  1320 kcal

                       x = (1320 x 1) / 4

                       x = 1320/4

                       x = 330 g of CHO

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Joy mixes one cup of sugar and one cup of lemon juice into three cups of water. The solvent in this recipe is the?
kobusy [5.1K]

Answer: Water is the solvent in this recipe.

Explanation: A solvent is " a molecule that has the ability to dissolve other molecules". Lemon juice and sugar are solutes.

5 0
3 years ago
A nonmetal
mars1129 [50]

^{16}_{\phantom{1}8}\text{O}.

Oxygen-16 is the atom in question.

  • Atomic number: 8.
  • Mass number: 16.
<h3>Explanation</h3>

The superscript of the ion says "2-". That means that the ion here carries a charge of -2.

  • The charge is negative, meaning that there are more electrons (which are negative) than protons (which are positive) in that ion.
  • The size of the charge is 2. There are two more electrons than protons in that ion.

There are 10 electrons in total in that ion. There are two more electrons than protons. That means that there are 10 - 2 = 8 protons in that ion.

The atomic number of an atom is the same as its number of protons. The atomic number of X is 8.

The atomic number determines the element. Atomic number 8 is oxygen. Thus element X is oxygen.

Mass number is the sum of number of protons and neutrons in an atom. 8 + 8 = 16 for this atom.

5 0
3 years ago
For the reaction 6 Li + N2 → 2 Li3N , what is the maximum amount of Li3N (34.8297 g/mol) which could be formed from 14.18 mol (6
Yuki888 [10]
Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g
4 0
3 years ago
NH₄NO₃ → N₂O + 2H₂O When 45.70 g of NH₄NO₃ decomposes, what mass of each product is formed?
Anna007 [38]

Answer: 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles

The balanced chemical equation is:

NH_4NO_3\rightarrow N_2O+2H_2O  

According to stoichiometry :

1 mole of NH_4NO_3 produce = 1 mole of N_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{1}{1}\times 0.571=0.571moles  of N_2O  

Mass of N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g

1 mole of NH_4NO_3 produce = 2 moles of H_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{2}{1}\times 0.571=1.142moles  of H_2O  

Mass of H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g

Thus 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

5 0
3 years ago
What is the density of a gas at 242.5K and 0.7311atm. The molar mass of this gas is 70.90g/mol
Pie

Answer:

0.384g/l

Explanation: the density version of the ideal gas law is pm=drt

in which p= pressure, m=molar mass,d=, density, r= to a constant which is 0.08206, and t=temperature so just input the values

PM=DRT. so to find d the formula would be D=RT\PM

D=<u>0.08206*242.5</u>

        0.7311*70.90

D=0.384g/l

4 0
3 years ago
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