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BabaBlast [244]
4 years ago
7

35.5 moles of lithium fluoride is dissolved in 65 L of solution

Chemistry
1 answer:
kkurt [141]4 years ago
3 0

Answer:

Explanation:

L

=

1.10

L

of solution

Explanation:

The Molarity

M

is calculated by the equation comparing moles of solute to liters of solution

M

=

m

o

l

L

For this question we are given the Molarity 0.88M

We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride

We can convert the mass of LiF to moles by dividing by the molar mass of LiF

Li = 6.94

F = 19.0

LiF = 25.94 g/mole

25.2

g

r

a

m

s

x

1

m

o

l

25.94

g

r

a

m

s

=

0.97

moles

Now we can take the the molarity and the moles and calculate the Liters of solution

M

=

m

o

l

L

M

L

=

m

o

l

L

=

m

o

l

M

L

=

0.97

m

o

l

0.88

M

L

=

1.10

L

of solution i just did look at my papaer

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8 0
3 years ago
Part 1. A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
galben [10]

The mass of NaCl needed for the reaction is 91.61 g

We'll begin by calculating the number of mole of F₂ that reacted.

  • Volume (V) = 12 L
  • Temperature (T) = 280 K
  • Pressure (P) = 1.5 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Number of mole (n) =?

PV = nRT

1.5 × 12 = n × 0.0821 × 280

18 = n × 22.988

Divide both side by 22.988

n = 18 / 22.988

n = 0.783 mole

Next, we shall determine the mole of NaCl needed for the reaction.

F₂ + 2NaCl —> Cl₂ + 2NaF

From the balanced equation above,

1 mole of F₂ reacted with 2 moles of NaCl.

Therefore,

0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.

Finally, we shall determine the mass of 1.566 moles of NaCl.

  • Mole = 1.566 moles
  • Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
  • Mass of NaCl =?

Mass = mole × molar mass

Mass of NaCl = 1.566 × 58.5

Mass of NaCl = 91.61 g

Therefore, the mass of NaCl needed for the reaction is 91.61 g

Learn more about stiochoimetry: brainly.com/question/25830314

6 0
2 years ago
The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38
adell [148]

Answer:

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Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x10^{-3}

H = \frac{0.21}{1.38x10^{-3} }

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

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7 0
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butalik [34]

Answer:

2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO

7 0
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