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svlad2 [7]
3 years ago
15

The volume of a gas decreases to half of its original volume, but the gas maintains the same number of moles and temperature. Ac

cording to the ideal gas law, what will most likely happen to the pressure
Chemistry
1 answer:
Leya [2.2K]3 years ago
4 0
It will double

Key
P = pressure of the gas
n= number of moles of the gas
V= volume of the gas
R = the gas constant
T= absolute temperature of the gas
If V is halved when n and T remain the same
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A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4,
Sav [38]

Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

  • 2HA + Ba(OH)₂ → BaA₂ + 2H₂O

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We <u>convert mass of phthalate to moles</u>, using its molar mass:

  • 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol

Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:

  • 9.27 mmol HA * \frac{1mmolBa(OH)_{2}}{2mmolHA} = 6.64 mmol Ba(OH)₂

Finally we calculate the molarity of the Ba(OH)₂ solution:

  • 6.64 mmol / 35.8 mL = 0.129 M

(B) The reaction between Ba(OH)₂ and HCl is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

So<u> the moles of HCl that reacted </u>are:

  • 17.1 mL * 0.129 M * \frac{2mmolHCl}{1mmolBa(OH)_2} = 4.41 mmol HCl

And the <u>molarity of the HCl solution is</u>:

  • 4.41 mmol / 18.6 mL = 0.237 M

3 0
2 years ago
A sample of pure NO2 is heated to 335 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+
vodka [1.7K]

The equilibrium constant for the reaction is 0.00662

Explanation:

The balanced chemical equation is :

2NO2(g)⇌2NO(g)+O2(g

At t=t  1-2x ⇔ 2x + x moles

The ideal gas law equation will be used here

PV=nRT

here n= \frac{w}{W} = \frac{w}{V}= density

P = \frac{density RT}{M}       density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm

putting the values in reaction

0.75 = \frac{0.525 x 0.0821 x 608.15 }{M}

  M    = 34.61

         

to calculate the Kc

Kc=\frac{ [NO] [O2]}{NO2}

  \frac{1-2x}{1+x} x M NO2 + \frac{2x}{1+x} M NO+ \frac{x}{1+x} M O2

Putting the values as molecular weight of NO2, NO,O2

\frac{46(1-2x) +30(2x)+32x}{1+x}

34.61= \frac{46}{1+x}

x= 0.33

Kc= \frac{4x^2)x}{1-2x^2}

    putting the values in the above equation

Kc = 0.00662

     

5 0
2 years ago
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