How many sulfur atoms would be present in 41.8 moles of sulfur?

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

3 years ago

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0

3 years ago

If a student did not remove all of the bubbles from inside the buret before reading the initial volume and beginning the titrati

matrenka [14]

Hello!

If there's an air bubble inside the buret, and the bubble escapes the buret during the titration the initial volume lecture (Vi) would be lower (closer to 0) than the actual one, and the recorded consumed volume (ΔV=Vf-Vi) would be higher than the actual one and thus the calculated concentration of the hydrochloric acid would be higher than the real one.

Have a nice day!

5 0

3 years ago

How does chlorine react with ethene

PolarNik [594]

It creates

<span>THE HALOGENATION OF ALKENES</span>

5 0

2 years ago

Read 2 more answers

What is not a guideline required for a valid evidence test?

kirill115 [55]

I just took that quiz , it’s A

4 0

3 years ago