The two electrons in a helium atom occupy

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <---> N2(g) 3H2(g) At equilibrium, it was fou

Softa [21]

Answer:

Kc for this equilibrium is 2.30*10⁻⁶

Explanation:

Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.

Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is defined as:

$Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }$

In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.

In this case, being:

2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)

the equilibrium constant Kc is:

$Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }$

Being:

[N₂]= 0.0551 M
[H₂]= 0.0183 M
[NH₃]= 0.383 M

and replacing:

$Kc=\frac{0.0551*0.0183^{3} }{0.383^{2} }$

you get:

Kc= 2.30*10⁻⁶

<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>

8 0

4 years ago

A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat

Lana71 [14]

Answer:

Pressure = 1.14 atm

Explanation:

Hello,

This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

P = kT k = P / T

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

P1 / T1 = P2 / T2

P1 = 1.0atm

T1 = -15°C = (-15 + 273.15)K = 258.15K

P2 = ?

T2 = 21.5°C = (21.5 + 273.15)K = 294.65K

P1 / T1 = P2 / T2

P2 = (P1 × T2) / T1

P2 = (1.0 × 294.65) / 258.15

P2 = 1.14atm

The pressure of the gas after attaining equilibrium is 1.14atm

3 0

3 years ago

Please help me out with this

d1i1m1o1n [39]

Answer
I think it’s 5
Explanation

4 0

3 years ago

35. In the collision theory, a collision that leads to the formation of products is called an

FinnZ [79.3K]

Answer:

It's Effective Collision.

Explanation:

Hope my answer has helped you!

7 0

3 years ago

calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate

kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

#SPJ4

4 0

1 year ago