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hjlf
2 years ago
7

For lunch, a patient consumed 3 oz of skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk (see Table 3.

9). How many kilocalories did the patient obtain from the lunch?
Chemistry
1 answer:
Volgvan2 years ago
3 0

Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk

Energy content of 3 oz skinless chicken is = 110 kcal

Energy content of 3 oz broccoli = 30 kcal

Energy content of 1 medium apple = 60 kcal

Energy content of 1 cup non-fat milk = 90 kcal

So the kilocalories of energy patient obtained from lunch

                   = 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal


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Vadim26 [7]

Answer:

pH 8.89

Explanation:

English Translation

If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.

Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹

Assuming all of the salts involved all ionize completely

MgCl₂ ionizes to give Mg²⁺ and Cl⁻

MgCl₂ ⇌ Mg²⁺ + 2Cl⁻

1 mole of MgCl₂ gives 1 moles of Mg²⁺

Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M

Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻

Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²

(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²

[OH⁻]² = (6×10⁻¹¹)

[OH⁻] = √(6×10⁻¹¹)

[OH⁻] = 0.000007746 M

p(OH) = - log [OH⁻] = - log (0.000007746)

pOH = 5.11

pH + pOH = 14

pH = 14 - pOH = 14 - 5.11 = 8.89

Hope this Helps!!!

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