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Alisiya [41]
3 years ago
8

How many molecules of PF5 are found in 39.5 grams of PF5?

Chemistry
1 answer:
djyliett [7]3 years ago
7 0

Answer:

1.9 \times 10^{23} molecules of PF_5 are found in 39.5 grams of PF_5.

Explanation:

Atomic weights : P= 31, F= 19,

molar mass of = 1 atomic weight of P+ 5 atomic weight of

 F= 31+5 \times 19

= 31+95

 =126 g/mole

moles in 39.5 gm of

= \frac{Mass}{Molar mass}

 = \frac{39.5}{126}moles

 =0.3134 moles

1 mole of any substance contains

0.3131 moles contains 0.3134  

 = 1.9 \times 10^{23} molecules

Therefore, 1.9 \times 10^{23} molecules of PF_5 are found in 39.5 grams of PF_5.

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The mass of a single gold atom is 3.27X10^-22 grams. How many gold Adams with there be in 57.8 mg of gold.
vichka [17]

Answer:

18 * 10^19 atoms

Explanation:

We must first convert 57.8 mg to grams.

If 1000 mg = 1g

  57.8 mg = 57.8/1000 = 57.8 * 10^-3 g

Now;

If 1 gold atom has a mass of 3.27X10^-22 grams

x gold atoms have a mass of 57.8 * 10^-3 g

x = 57.8 * 10^-3 g/3.27X10^-22 g

x = 18 * 10^19 atoms

7 0
3 years ago
If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced
ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • P₄: 1 mole
  • H₂: 6 moles
  • PH₃:4 moles

Being the molar mass of the compounds:

  • P₄: 124 g/mole
  • H₂: 2 g/mole
  • PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

  • P₄: 1 mole* 124 g/mole= 124 g
  • H₂: 6 mole* 2 g/mole= 12 g
  • PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0
3 years ago
CAN YALL HELP A MANS OUT FOR A QUESTION?
Degger [83]
The answer is
b. mantle
3 0
3 years ago
WILLL GIVE BRAINLIEST IF YOU ANSWERRR PLEASEEEE IM LITERALLY BEGGING YOU I PUT THIS QUESTION IN SOO MANY TIMES I HAVEN'T GOTTEN
tino4ka555 [31]

Answer:

The answer is 375.54 g of AgBr

Explanation:

Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)

Mass = 2M x 1L x 187.77 g/mol

Mass = 375.54g

6 0
2 years ago
Which solution is more concentrated?
deff fn [24]

2 moles of NaOH dissolved in 1 litre of solution is the solution with more concentration.

Answer: Option A

<u>Explanation:</u>

Concentration of solution is the measure of the amount of solute dissolved in the solvent of the solution. So this is measured using the molarity of the solution. Molarity is determined as the number of moles of the solute present in the given amount of solvent.

\text {Molarity}=\frac{\text {Moles of solute}}{\text {Amount of solvent }}

In this present case, the option A gives the molarity of 2 M as

\text {Molarity}=\frac{2}{1}=2 M

But the second option, mass of NaOH is given. So we have to determine the molarity. First we have to find the molar mass of NaOH. We know that 1 mole of NaOH will contain 40 g/mole.

1 g of NaOH = 40 g of NaOH

1 g of NaOH = 1/40 moles

So 2 g of NaOH will contain \frac{2 g}{40 g/mole} which is equal to 0.05 moles of NaOH.

Thus, the molarity of 2 g of NaOH will be

Molarity = \frac{0.05 moles }{1 L}=0.05 M

Thus, the option A is having higher concentration as the molarity is more for 2 moles of NaOH dissolved in 1 l of solution.

7 0
2 years ago
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