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Tatiana [17]
4 years ago
5

What pressure would be exerted by 9.3 Newtons over a 48.9 m2 area?

Chemistry
1 answer:
VashaNatasha [74]4 years ago
5 0
Pressure = 9.3 / 48.9 = .... Pascals
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Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). (3 points)
AleksandrR [38]

Answer:

1. Adding hydrogen gas, b. shift to the right
2. Adding a catalyst, c. No effect
3. Decreasing the pressure, a. shift to the left

Explanation:

Hydrogen gas can be rewritten as H2. Whenever you add something to an equilibrium expression, it will shift to whichever side does not have this. So, since the reactant side has 3 moles of H2, adding more H2 to the reaction will shift to the products side, since there is no H2 there.


Adding a catalyst has no effect on equilibrium reactions.

When decreasing the pressure, equilibrium will shift to the side with the greater number of moles of gas. In this case, there are 4 moles of gas on the left, and 2 on the right, so it would shift to the left.

5 0
2 years ago
The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57
In-s [12.5K]

Answer:

1.135 M.

Explanation:

  • For the reaction: <em>2HI → H₂ + I₂,</em>

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

  • To solve this problem, we can use the integral law of second-order reactions:

<em>1/[A] = kt + 1/[A₀],</em>

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>

8 0
3 years ago
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