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olasank [31]
3 years ago
8

What mass of F2 is needed to produce 180 g of PF3 if the reaction has a 78.1% yield?

Chemistry
1 answer:
nexus9112 [7]3 years ago
6 0

Answer:

91.26 g

Explanation:

Given data:

Mass of PF₃ = 180 g

Mass of F₂ required = ?

Solution:

Chemical equation:

P₄ + 6F₂   → 4PF₃

Moles of PF₃:

Number of moles = mass/ molar mass

Number of moles = 180 g/ 88 g/mol

Number of moles = 2.05 mol

Now we will compare the moles of PF₃ with F₂.

                        PF₃            :           F₂

                          4               :           6

                          2.05         :           6/4×2.05 = 3.075

Mass of  F₂:

Mass of F₂ = moles × molar mass

Mass of F₂ = 3.075 mol × 38 g/mol

Mass of F₂ =  116.85 g

If reaction yield is 78.1%:

116.85 /100 ×78.1 = 91.26 g

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5 0
3 years ago
A plant living in the environment shown in the picture would have to be adapted not to loose to much water through its___
Ray Of Light [21]

Answer:

Many plants have thorns on their stems or leaves. What is the MOST likely explanation for the evolution of thorns?

A) Thorns help plants produce more food from photosynthesis.

B) Thorns are an example of a mutation that arises in the genetic code of plants.

C) Thorns help plants to conserve resources like water and soil nutrients that may be used by other organisms.

D) Thorns are an adaptation that some plants have evolved in order to discourage herbivores from eating the plant.

2)

Explanation:

5 0
2 years ago
Read 2 more answers
How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?
zhenek [66]

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\  =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\

2) Determine the limitting reagent

\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation:

6 0
2 years ago
Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How ma
Art [367]

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

3 0
3 years ago
Both 1,2−dihydronaphthalene and 1,4−dihydronaphthalene may be selectively hydrogenated to 1,2,3,4−tetrahydronaphthalene. One of
pochemuha

Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.

In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

8 0
2 years ago
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