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zvonat [6]
2 years ago
10

according to the recipe for soap in the background 1 gallon of lye solution was required for 2 pounds of fat. how many liters ar

e there in 1.00 gallon? how many grams are there in 2.00 pounds?​
Chemistry
1 answer:
yawa3891 [41]2 years ago
3 0

Answer:

3.785 L of lye solution needed  for 907.185 g of fat

Explanation:

The computation of the number of grams in 2.00 pounds is shown below:

As we know that

One gallon of lye solution needed 2 fats pounds

Also,

1 gallon = 3.785 L

And,

1 lb = 453.592 g

So,

For 2 lb, it is

= 2 × 453.592

= 907.185 g

Therefore  

3.785 L of lye solution needed  for 907.185 g of fat

hence, the same is to be considered

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Answer:

you have to divide it

Explanation:

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How much energy (joules) is needed to heat 250 grams of copper from 22 °C to 99 °C? The specific heat capacity (C) of copper is
Naddik [55]

Answer:

7430.5 Joules (7.4*10^4 Joules)

Explanation:

Q=mc∆T

where Q is energy in Joules.

Now m=250 g

c= 0.386 J/g°C

∆T = 99 - 22 = 77 °C

plugging the values in gives

Q=250*0.386*77=7430.5 Joules

(7.4*10^4 Joules, if 2 significant figures)

5 0
2 years ago
What is the mole fraction of each component if 3.9 g of benzene (C6H6) is dissolved in 4.6 g of toluene (C7H8)
Savatey [412]

Answer:

Step 1 of 6

(a)

The mass of benzene is  , so calculate the moles of benzene as follows:



The mass of toluene is, so calculate the moles of toluene as follows:



Now, calculate the mole fraction as follows:





Therefore, the mole fraction of benzene and toluene is  and  respectively.

Step 2 of 6

(b)

The formula to calculate the partial pressure is as follows:



Here,  is the partial pressure of benzene,  is the vapour pressure of pure benzene and  is the mole fraction of benzene.

Vapour pressure of pure benzene at  is.

Substitute the values in the equation as follows:



Therefore, the partial pressure is  .

Step 3 of 6

(c)

Vapor pressure of the solution at 1 atm is  .

When the total pressure of the vapour pressure of the mixture is  at a temperature, then, the solution boils. It corresponds to the boiling point of the solution.

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is less than the atmospheric pressure, the solution will not boil at  .

Calculate the total pressure of the solution at  as follows:



Since, the total pressure is greater than the atmospheric pressure, the solution will boil at  .

Therefore, the boiling point of the solution is  .

Step 4 of 6

(d)

Mole fraction of benzene at  is calculated as follows:



Mole fraction of toluene at  is calculated as follows:



Therefore, the mole fractions of benzene and toluene are  and  respectively.

Step 5 of 6

(e)

Vapor pressure of benzene at  is  .

Partial pressure of benzene is calculated as follows:



Vapor pressure of toluene at  is  .

Partial pressure of toluene is calculated as follows:



Step 6 of 6

Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Weight composition of the vapour that is in equilibrium with the solution is calculated as follows:



Explanation:

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2 years ago
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Determine the number of atoms of O in 92.3 moles of Cr3(PO4)2
Irina18 [472]

739 is the number of atoms of O in 92.3 moles of Cr3(PO4)2.

Explanation:

Molecular formula given is  = Cr3(PO4)2

number of moles of the compound is 92.3 moles

number of 0 atoms in 92.3 moles =?

From the chemical formula 1 mole of the compound has 8 atoms of oxygen

So, it can be written as

1 mole Cr3(PO4)2 has 8 atoms of oxygen

92.3 moles of Cr3(PO4)2 has x atoms of oxygen

\frac{8}{1} = \frac{x}{92.3}

x = 8 x 92.3

x = 738.4 atoms

There will be 739 oxygen atoms in the 92.3 moles of Cr3(PO4)2.

4 0
3 years ago
At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

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Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0
2 years ago
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