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Roman55 [17]
4 years ago
5

Consider the following reaction:

Chemistry
1 answer:
rusak2 [61]4 years ago
8 0

Answer:

1. a) 0,18

2. e) The reaction proceeds to the left, forming more BrCl(g).

Explanation:

1. In a gas reaction as:

2SO₂(g) +O₂(g) ⇄ 2SO₃(g)

it is possible to convert kp to kc using:

kp = kc (RT)^Δn

Where kp is gas equilibrium constant, kc is equilibrium constant (13), R is gas constant (0,082atmL/molK), T is temperature (900K), and Δn is number of moles of gas products - number of moles of gas reactants (That is 2 - (2+1) = -1). Replacing:

kp = 13×(0,082atmL/molK×900K)^-1

<em>kp = 0,18</em>

<em></em>

2. Based on Le Chatelier's principle, the change in temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system. For the reaction:

2 BrCl(g) ⇄ Br₂(g) + Cl₂(g).

The addition of 0,050M of each compound cause <em>the reaction proceeds to the left, forming more BrCl(g)</em> because based on the reaction, you need two moles of BrCl per mole of Br₂(g) and Cl₂(g) to keep the system in the same. But you are adding the same proportion of moles of each compound.

I hope it helps!

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In this lab, you are to carry out the formation of a Grignard reagent from 1-bromo-benzene andits subsequent reaction with solid
yan [13]

Answer:

(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.

(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.

(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).

PRODUCTS=> C6H5-COOH and Mg^+Br(OH).

Explanation:

A Grignard reagent is a reagent that/which is an organometallic compound that is R -Mg- X. The R = alkyl, vinyl or allyl and the X = halogens.

It must be noted that an important reaction of Grignard reagent is its reaction with compounds containing the Carbonyl that is -CO functional group and this kind of Reaction is known as a Grignard Reaction.

So, in this question we are told that;

=> "1-bromo-benzene andits subsequent reaction with solid carbon dioxide (CO2) followed by acidic workup (using HCl asthe acid). "

Thus;

(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.

(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.

(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).

5 0
4 years ago
A closed, adjustable cylinder containing 6.2 moles of oxygen gas has a volume of 4.3 liters and a pressure of 3.8 atmospheres. I
il63 [147K]

Answer:

The new pressure is 1.05 atm

Explanation:

Step 1: Data given

Number of moles oxygen gas = 6.2 moles

Volume of the cylinder = 4.3 L

Pressure of the gas = 3.8 atm

The volume changed to 15.5 L

Step 2: Calculate the new pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure of the gas = 3.8 atm

⇒with V1 = the initial volume = 4.3 L

⇒with V2 = the increased volume = 15.5 L

⇒with P2 = the new pressure = TO BE DETERMINED

3.8 atm * 4.3 L = 15.5 L * P2

P2 = (3.8 atm *4.3 L) / 15.5 L

P2 = 1.05 atm

The new pressure is 1.05 atm

5 0
3 years ago
The half-life for the second-order decomposition of HI is 15.4 s when the initial concentration of HI is 0.67 M. What is the rat
larisa [96]

Answer:

The correct answer is option C.

Explanation:

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times A_0}

t_{\frac{1}{2} = half life = 15.4 s

k = rate constant =?

A_0 = initial concentration = 0.67 M

15.4 s=\frac{1}{k\times 0.67 M}

k=\frac{1}{15.4 s\times 0.67 M}=0.09692 M^{-1} s^{-1}=9.69\times 10^{-2}M^{-1} s^{-1}\approx 9.7\times 10^{-2} M^{-1} s^{-1}

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4 0
3 years ago
How many hydrogen atoms are present in 0.70 moles of ptcl2(nh3)2?
Rina8888 [55]
In this case, the hydrogen atoms are present in the (NH3)2 part of the molecule. The sign (NH3)2 mean that there are two NH3 molecules in their compound. That means the total atom of hydrogen would be: 3*2 = 6 atom for every compound molecule
Since there is 0.7 mole of compound, then the hydrogen atom would be:
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7 0
3 years ago
Fluid ounces are a metric unit for volume.<br><br> True<br> False
Stels [109]

Answer:

True

Explanation:

Hope this helps:) Have a good day!

5 0
3 years ago
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