PLS HELP ASAP!!!!! IM TIMED!!!!!

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

You wish to prepare an HC2H3O2 buffer with a pH of 4.24. If the pKa of is 4.74, what ratio of C2H3O2 /HC2H3O2 must you use?

LenaWriter [7]

To solve this problem, we can use the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:<span>

<span>pH = pKa + log ([base]/[acid]) ---> 1</span></span>

Where,

[base] = concentration of C2H3O2 in molarity or moles

<span>[acid] = concentration of HC2H3O2 in molarity or moles</span>

For the sake of easy calculation, let us assume that:

[base] = 1

[acid] = x

<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x)

<span>log (1 / x) = - 0.5

1 / x = 0.6065 </span></span>

x = 1.65<span>

The required ratio of C2H3O2 /HC2H3O2 <span>is 1:1.65 or 3:5. </span></span>

3 0

3 years ago

Heating up a reaction increases the speed of a reaction until the ____

Vesnalui [34]

Heating up a reaction increases the speed of a reaction until the enzyme denatures.

<h3>What is enzyme denaturation?</h3>

Enzyme denaturation occurs when a biological protein catalyst does not work anymore due to a high temperature that alters its tridimensional conformation.

This cellular process (denaturation) is well known to be one of the main causes of enzymatic failure.

In conclusion, heating up a reaction increases the speed of a reaction until the enzyme denatures.

Learn more about enzymes here:

brainly.com/question/1596855

#SPJ12

5 0

2 years ago

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember the reaction among aqueous acetic acid ( $CH_3COOH$ ) and aqueous sodium hydroxide (NaOH)

$CH_3COOH + NaOH -> CH_3COONa + H_2O$

First step. Need to know how much moles of the substances are present

$0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH$ = 0.0025 mol NaOH

0.003 mol NaOH * $1 mol CH_3COOH$ / 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> $Na^{+} + OH^{-}$

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

$pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95$

6 0

3 years ago

How many moles of chlorine gas and 120°C and 33.3 ATM would occupy a vessel of 12 L?

Mashutka [201]

<h3>Answer:</h3>

12.387 moles

<h3>Explanation:</h3>

We are given;

Temperature of chlorine, T = 120°C

But, K = °C + 273.15

Therefore, T = 393.15 K

Pressure, P = 33.3 Atm

Volume, V = 12 L

We are required to calculate the number of moles of chlorine gas,

To find the number of moles we are going to use the ideal gas equation;

PV = nRT

R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore, rearranging the formula;

n = PV÷RT

Hence;

n = (33.3 atm × 12 L) ÷ (0.082057 × 393.15 K)

= 12.387 moles

Therefore, the number of moles of chlorine are 12.387 moles

8 0

3 years ago