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Elis [28]
3 years ago
9

A sample of gas exerts a pressure of 16 atm at a temperature of 340 K. What is the pressure of the gas in atm if the temperature

is raised to 408 K?
ANSWER FAST!!!!
Chemistry
1 answer:
gavmur [86]3 years ago
5 0

Answer:

P2 = 19.2atm

Explanation:

Initial pressure (P1) = 16atm

Initial temperature (T1) = 340K

Final temperature (T2) = 408K

Final pressure (P2) = ?

This question involves the use of pressure law

Pressure law states that the pressure of a fixed mass of gas is directly proportional to it's temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

Therefore,

P1 / T1 = P2 / T2 = P3 / T3 = ......=Pn / Tn

P1 / T1 = P2 / T2

We need to solve for P2

P2 = (P1 × T2) / T1

Now we can plug in the values and solve for P2

P2 = (16 × 408) / 340

P2 = 6528 / 340

P2 = 19.2atm

The final pressure (P2) of the gas is 19.2atm

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Explanation:

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Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used  otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

The attached diagrams portraying this notions is shown in the attached file below.

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
2 years ago
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