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SSSSS [86.1K]
1 year ago
6

A balloon with a volume of 2.0 l at 25°c is placed in a hot room at 35°c. the pressure on the balloon is constant at 1.0 atm. fo

rmula to use: v1 t1 = v2 t2 how does the volume of the balloon change after moving it to the hot room?
Chemistry
2 answers:
shusha [124]1 year ago
5 0

The volume increases and the final volume is 2.1 L

<h3>Why volume increases?</h3>

The volume of a gas is directly related to the heat and pressure. If temperature increases, volume increases.

If pressure increases, volume decreases

Based on Boyle's law,

The absolute temperature of a gas is directly proportional to the volume of the gas.

As the balloon change increases its temperature, the volume increases

The Boyle's equation is:

V₁T₂ = V₂T₁

Where,

  • V is volume
  • T is absolute temperature of 1, initial state and 2, final state of the gas.

Replacing :

V₁ = 2.0L

T₂ = 35°C + 273.15 = 308.15K

V₂ = ?

T₁ = 25°C + 273.15 = 298.15 K

2.0L x 308.15 K = V₂ x 298.15 K

2.1 L = V₂

Final volume is 2.1L

Hence, The volume increases and the final volume is 2.1 L

Learn more about gas laws here ;

brainly.com/question/12669509

#SPJ1

Archy [21]1 year ago
4 0

Answer:

increases, 2.1

Explanation:

I got it right.

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Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because
olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = H_{picric}

So,  H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}

0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}    ;     ( where +/-  represent ± )

= \frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}

= \frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}

= \frac{-0.42+\sqrt {1.0496} }{2}     OR   \frac{-0.42-\sqrt {1.0496} }{2}

= \frac{-0.42+1.0245}{2}       OR    \frac{-0.42-1.0245}{2}

= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

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