The most common form of matter in the universe

The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2HI(9) =H2(g) +129) An equilibrium mixture of the

Mrrafil [7]

Explanation:

The equilibrium constant of the reaction = $K_c=1.80\times 10^{-2}$

Initial concentration of HI = 0.311 M

After addition 0.174 mol of HI(g)

Concentration of HI added = $\frac{0.174 mol}{1 L}=0.174 M$

New concentration of HI = 0.311 M + 0.174 M = 0.485 M

$2HI\rightleftharpoons H_2+I_2$

Initial concentration:

0.311 M $4.71\times 10^{-2} M$ $4.71\times 10^{-2} M$

At equilibrium:

(0.485 M - x) $(4.71\times 10^{-2} M+x)$ $(4.71\times 10^{-2} M+x)$

$K_{eq}=\frac{[H_2][I_2]}{[HI]^2}$

$1.80\times 10^{-2}=\frac{(4.71\times 10^{-2} M+x)\times (4.71\times 10^{-2} M+x)}{ (0.485 M-x)^2}$

$\sqrt{1.80\times 10^{-2}}=\frac{(4.71\times 10^{-2} M+x)}{ (0.485 M-x)}$

$0.1342=\frac{(4.71\times 10^{-2} M+x)}{(0.485 M-x)}$

$0.065087 M-0.1342x=4.71\times 10^{-2} M+x$

$0.065087 M-4.71\times 10^{-2} M=1.1342x$

$x=\frac{0.017987 M}{1.3142}=0.01586 M$

Equilibrium concentrations:

$[HI]=0.485 M-x = 0.485 M - 0.01586 M= 0.46914 M$

$[H_2]=[I_2]=4.71\times 10^{-2} M+x=4.71\times 10^{-2} M+0.01586 M$

$[H_2]=[I_2]=0.06296 M$

3 0

3 years ago

1. What is the difference between charging by conduction and charging by induction?

telo118 [61]

Answer:a

Explanation:

7 0

3 years ago

hi pls I need ASAP. 1 the tubes responsible for the moment of blood from the heart to d other parts of the body is called 2. the

IRISSAK [1]

Answer:

$\large \boxed{\text{1. Veins; 2. 600 W; 3. 45 N}}$

Explanation:

1. Arteries are the tubes responsible for the movement of blood from the heart to other parts of the body.

2. Power

(a) Work performed

w = mgh = 12 kg × 9.807 m·s⁻² × 15 m = 1800 J

(b) Power level

P = w/t = 1800 J/3 s = 600 W

3. Machine load

The formula for mechanical advantage (MA) is

$MA = \dfrac{F_{o}}{ F_{e}}$

where

Fₒ = the force of the object

Fₑ = the force exerted to overcome that of the object

$\begin{array}{rcl}MA & = &\dfrac{F_{o}}{ F_{e}}\\\\6 & = &\dfrac{F_{o}}{ \text{7.5 N}}\\\\F_{o} & = & \text{45 N}\\\end{array}\\\text{The machine can carry a maximum load of $\large \boxed{\textbf{45 N}}$}$

7 0

4 years ago

a sample of argon gas is cooled and its volume went from 380. mL to 250. mL. If its final temperature was -55.0C, what was its o

inn [45]

Answer:

The initial temperature was 58.4°C

Explanation:

Given the following data:

initial volume = V₁ = 380 mL = 0.38 L

final volume = V₂ = 250 mL = 0.25 L

final temperature = T₂ = -55°C = 218 K

According to Charles's law, the volume of a gas is <em>directly proportional to the temperature</em> (in Kelvin). The mathematical expression is:

V₁/T₁= V₂/T₂

So, we calculate the initial temperature (V₁) as follows:

T₁ = T₂/V₂ x V₁ = 218 K/(0.25 L) x 0.38 L = 331.36 K ≅ 331.4 K

Finally, we convert the initial temperature from K to °C:

T₁= 331.4 K - 273 = 58.4°C

6 0

3 years ago

Someone help me pls!

Naddika [18.5K]

Answer:Warm ocean currents originate near the equator and move towards the poles or higher latitudes while cold currents originate near the poles or higher latitudes and move towards the tropics or lower latitude.

Explanation:

8 0

3 years ago