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Ahat [919]
2 years ago
3

Given the reaction: p4(l) + 6 cl2(g) → 4 pcl3(l) if the percent yield is 82%, what mass of p4 is required to obtain 2.30 g pcl3

(cl2 in excess)?
Chemistry
1 answer:
kozerog [31]2 years ago
6 0

The  mass  of P4   required to obtain 2.30  PCl3  0.694  g


  calculation

p4 + 6 Cl2→  4 pcl3

step 1:calculate the moles  of  PCl3  

moles=  mass/molar mass

from periodic table  the  molar mass   of PCl3  =  31 + (3 x35.5) = 137.5  g/mol

 moles = 2.30  / 137.5 =0.0167   moles

step 2: use the  mole  ratio to find the  moles of P4

from equation  above the  mole ratio  of p4 :PCl3   is  1:3  therefore  the  moles  of p4

    = 0.0167   x1/3  =0.0056 moles

step 3:  find the  mass of P4

mass = moles x   molar mass

from periodic table the   molar mass  of P4 =  31 x4= 124  g/mol

mass =  0.0056  x 124 = 0.694  g

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Which of the following reactions is the least energetic? Question 18 options: ATP + H2O → ADP + Pi ATP + H2O → AMP + PPi AMP + H
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Answer:

The correct answer is AMP+H2O→ Adenosine + pi

Explanation:

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2 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

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\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

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1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

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