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Sever21 [200]
2 years ago
11

One serving of peanut butter contains 5.0 g of carbohydrate, 12 g of fat, and 6.0 g of

Chemistry
1 answer:
larisa [96]2 years ago
3 0

The serving of peanut butter contains 117kcal

<h3>Calculation of calories of food Nutrients</h3>

1 gram of both carbohydrate and protein contains 4 (kcal)

1 gram of fat contains 9 (kcal)

But peanut butter contains 5g of fat. The kilocalories of fat present is;

1 g = 9 kcal

5g = F

F = 5×9

F = 45 kcal

The kilocalories of carbohydrates present is;

1g = 4kcal

12g = C

C= 12×4

C = 48kcal

The kilocalories of proteins present is;

1g = 4kcal

6 g = P

P = 4× 6

P = 24 kcal

Therefore, the total kilocalories of the peanut = 45 + 48 + 24 = 117kcal

Learn more about kilocalories here:

brainly.com/question/6423812

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A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i
erastova [34]

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

7 0
3 years ago
A buffer is prepared by adding 139 mL of 0.39 M NH3 to 169 mL of 0.19 M NH4NO3. What is the pH of the final solution? (Assume th
Paha777 [63]

Answer:

pH = 9.48

Explanation:

We have first to realize that NH₃ is a weak base:

NH₃ + H₂O ⇔ NH₄⁺ + OH⁻     Kb = 1.8 x 10⁻⁵

and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

pOH = pKb + log ( [ conjugate acid ] / [  weak base ]

mol NH₃ = 0.139 L x 0.39 M = 0.054 mol

mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol

Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52

pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48

The solution is basic which agrees  with NH₃ being  a weak base.

5 0
3 years ago
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