Answer:
a
. eight tenths of her cookies
Explanation:
Let the total number of Lakesha's cookies be represented by x.
So that;
She gave three tenths to Bailey = 
of x
= 
She gave five tenths to Helen = 
of x
= 
Fraction of Lakesha's cookies given away = +
= 
= 
Thus, the fraction of cookies given away by Lakesha is 
.
Answer:
- <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>
Explanation:
You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.
To compare, the two volumes must be on the same system of units.
Quarts is a measure of volume equivalent to 1/4 of gallon.
One gallon is approximately 3.785 liters.
3.785 liter = 3.785 liter × 1,000 ml/liter
Then, to convert 1 quart to ml use the unit cancellation method:
- (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml
Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
the energy gained by proteins and carbohydrates differs from the energy gained by fats.
proteins and carbohydrates both give 4 kcal per gram
fats give 9 kcal per gram
mass of proteins - 2 g
energy given by proteins - 2 g x 4 kcal/g = 8 cal
mass of carbohydrates - 20 g
energy given by carbohydrates - 20 g x 4 kcal/g = 80 cal
mass of fat - 1 g
energy given by fat - 1 g x 9 kcal/g = 9 cal
total energy = 8 + 80 + 9 = 97 kcal
energy = 97 kcal
