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Maurinko [17]
3 years ago
7

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o

f the work used to solve this problem.
2 K + F2 ---> 2 KF
Chemistry
1 answer:
Degger [83]3 years ago
8 0
1) Molar mass:

K = <span> 39.0983 g/mol
F</span>₂ = <span> 37.99 g/mol

moles ratio:

2 K + F</span>₂<span>   =   2 KF

2 x 39.0983 g K ----------------- 37.99 g F</span>₂
23.5 g K -------------------------- ( mass of F₂)
<span>
mass of F</span>₂ = 23.5 x 37.99 / 2 x 39.0983
<span>
mass of F</span>₂ = 892.765 / 78.1966
<span>
mass of F</span>₂ = 11.4169 g
<span>
Therefore:

</span>37.99 g F₂------------------ 22.4 L ( at STP)
11.4169 g F₂ --------------- ( volume F₂ )

Volume F₂ = 11.4169 x 22.4 / 37.99

Volume F₂ = 255.73856 / 37.99

Volume F₂ = 6.731 L
<span>


</span>
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<h2>Given :-</h2>

  • Moles = 2.10

  • Gas = Nitrogen

  • Pressure = 1.25 atm

  • Volume = 25L

<h2>To Find :-</h2>

  • Temperature

<h2>Solution :-</h2>

As we know that

PV = nRT

Here,

  • P = Pressure
  • V = Volume
  • n = Moles
  • T = Temperature

<u>According</u><u> </u><u>to</u><u> </u><u>the question</u><u>, </u>

  • 1.25 × 25 = 2.1 × 0.082 × T
  • 31.25 = 2.1 × 0.082 × T
  • 31.25 = 1.722T
  • 31.25/0.1722 = T
  • 181 ≈ T

Temperature is 181 K

  • \large\fbox{181 K}
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Does rapid burial increase exposure to oxygen?
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Explanation:

a gentle burial also increases an organisms chances of becoming more fossil like

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What determines the amount of chemical energy a substance has
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3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of wa
Ostrovityanka [42]

Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

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