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frozen [14]
2 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]2 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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A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23
Anna35 [415]

Answer:

0.8696\ \text{J/g}^{\circ}\text{C}

Explanation:

m_m = Mass of metal = 19 g

c_m = Specific heat of the metal

\Delta T_m = Temperature difference of the metal = 99-23=76^{\circ}\text{C}

V = Volume of water = 150 mL = 150\ \text{cm}^3

\rho = Density of water = 1\ \text{g/cm}^3

c_w = Specific heat of the water = 4.186 J/g°C

\Delta T_w = Temperature difference of the water = 23-21=2^{\circ}\text{C}

Mass of water

m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}

Heat lost will be equal to the heat gained so we get

m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}

The specific heat of the metal is 0.8696\ \text{J/g}^{\circ}\text{C}.

4 0
2 years ago
Please answer, with explanation. Thanks!​
nadya68 [22]

Answer:

Explanation:

a = 40.1 g of Ca

Number of moles = mass / molar mass

Number of moles = 40.1 g/ 40.1 g/mol

Number of moles = 1 mol

b = 11.5 g Na

Number of moles = mass / molar mass

Number of moles = 11.5 g/ 23 g/mol

Number of moles = 0.5 mol

c = 5.87 g Ni

Number of moles = mass / molar mass

Number of moles = 5.87 g/ 58.7 g/mol

Number of moles = 0.1 mol

d = 150 g of S

Number of moles = mass / molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

e = 2.65 g Fe

Number of moles = mass / molar mass

Number of moles = 2.65 g/ 55.85 g/mol

Number of moles = 0.05 mol

f = 0.00750 g Ag

Number of moles = mass / molar mass

Number of moles = 0.00750 g/ 107.9 g/mol

Number of moles = 6.95 × 10⁻⁵ mol

g = 2.25 × 10²⁵ atoms Zn

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms × 2.25 × 10²⁵ atoms

0.17  × 2.25 × 10²⁵ moles

38.25 moles

h = 50 atoms of Ba

1 mole = 6.022 × 10²³ atoms

1 mol / 6.022 × 10²³ atoms ×50 atoms

0.17 × 10²³ × 50 moles

8.5 × 10²³ moles

6 0
3 years ago
2. (Exercise 4.1.6) A liquid adhesive consists of a polymer dissolved in a solvent. The amount of polymer in the solution is imp
Lilit [14]

Answer:

800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

Explanation:

3000 lb of 13% solution is required .

Total adhesive in weight = 3000 x .13 = 390 lb of adhesive

Available = 500 lb of 10% solution = 50 lb of adhesive

Rest = 390 - 50 = 340 lb required .

rest mass of solution = 3000 - 500 = 2500 lb

mass of adhesive required = 340 lb

Let the mass  of 20% required be V

mass of adhesive = .20 V

.20 V = 340

V = 1700

rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent

So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

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