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2 years ago

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Homogenized whole milk contains 4% butterfat by volume. how many milliliters of fat are there in a glass (250 ml) of milk?

1 answer:

solong [7]2 years ago

7 0

Volume of one glass of milk = 250 mL

Percent by volume of butterfat in homogenized milk = 4%

This means 4mL butterfat is present per 100mL milk

Volume of fat present in 250 mL of homogenized milk:

$250mL milk*\frac{4mL butterfat}{100mL milk}= 10mL butterfat$

Therefore, 250mL homogenized milk which is 4 % butterfat contains 10mL butterfat.

If the density of butterfat is 0.8g/mL, the mass of butterfat present in 250mL milk will be:

$10mL butterfat*\frac{0.8g butterfat}{1mL butterfat} =8g butterfat$

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Chlorinated hydrocarbon pesticides, such as DDT, were introduced during World War II to control insects in agricultural crops. A

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2 years ago

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-

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<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u> $Ni(s)\rightarrow Ni^{2+}+2e^-$

<u>Reduction half reaction (cathode):</u> $Ni^{2+}+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

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2 years ago

In a certain compound of iron and oxygen, FexOy, it is found that a sample of this compound weighing 6.285 g contains 4.396 g of

saveliy_v [14]

Answer:

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Explanation:

The given compound consists of Fe and O.

So, mass of oxygen in sample = (mass of sample) - (mass of Fe in sample)

= (6.285 g ) - (4.396 g)

= 1.889 g

Molar mass of O = 16 g/mol and molar mass of Fe = 55.845 g/mol

So, ratio of number of moles of Fe and O (Fe : O)

= $\frac{4.396}{55.845}:\frac{1.889}{16}$

= 0.0787 : 0.118

= $1:\frac{0.118}{0.0787}$

= 1 : 1.5

= 2 : 3

So, formula of oxide is $Fe_{2}O_{3}$

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2 years ago

. The Henry's law constant for helium gas in water at 30 ∘C is 3.7×10−4M/atm; the constant for N2 at 30 ∘C is 6.0×10−4M/atm. a.

ludmilkaskok [199]

Explanation:

1) Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$M_{g}=K_H\times p_{g}$

where,

$K_H$ = Henry's constant

$p_{He}$ = partial pressure of gas

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Putting values in above equation, we get:

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Putting values in above equation, we get:

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2)

a) Mass of solute or methanol , m= 14.7 g

Mass of solvent or water , m'= 186 g

Mass of the solution = M = m + m' = 14.7 g + 186 g = 200.7 g

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$=\frac{14.7 g}{200.7 g}\times 100=7.32\%$

The mass percent of methanol is 7.32%.

b) Molality is defined as moles of solute per kilograms of solvent.

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Moles of methanol = $\frac{14.7 g}{32 g/mol}=0.4594 mol$

Mass of solvent = 186 g = 0.186 kg

$m=\frac{0.4594 mol}{0.186 kg}=2.4610 mol/kg$

The molality of methanol is 2.4610 mol/kg.

4 0

2 years ago