All categories

2 years ago

Consider the following reversible reaction:

1 answer:

8 0

The given equilibrium reaction indicates the reaction between carbon monoxide and the oxygen and forms carbon dioxide. In this reaction, by increasing the concentration of the carbon monoxide, the equilibrium shifts towards the right. Since, the reactant concentration increases, the equilibrium stress decreases the concentration of the reactants and therefore, the equilibrium shift towards the right side of the equation.

reaction indicates that carbon monoxide reacts with oxygen and forms carbon dioxide gas.

By increasing the concentration of carbon monoxide image, the equilibrium in the reaction shifts towards the right.

You might be interested in

What is the PH of a solution whose [H+] is 0.0007

cupoosta [38]

To determine the pH of the solution. All you would have to do is use the equation:

pH = -log(H3O^+)
pH = - log(0.0007)
pH = 3.15.

Note that H+ and H3O+ are the same, it is more accurate to say H3O+ because of how the H2O molecules become after receiving the additional proton.

7 0

3 years ago

Iron (ll) and Oxygen formula and correct name.

VikaD [51]

iron is called Fe amd oxygen is called O

8 0

3 years ago

Read 2 more answers

How many bromine atoms are present in 39.4 g of CH2Br2

strojnjashka [21]

37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2

4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms

7 0

4 years ago

What is the half-life of a first-order reaction if it takes 4.4 x 102 seconds for the concentration to decrease from 0.50 M to 0

elena55 [62]

Answer: The half-life of a first-order reaction is, $3.3\times 10^2s$

Explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken = 440 s

$[A_o]$ = initial amount of the reactant = 0.50 M

[A] = left amount = 0.20 M

Putting values in above equation, we get:

$k=\frac{2.303}{440s}\log\frac{0.50}{0.20}$

$k=2.083\times 10^{-3}s^{-1}$

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

Putting values in this equation, we get:

$t_{1/2}=\frac{0.693}{2.083\times 10^{-3}s^{-1}}=332.69s=3.3\times 10^2s$

Therefore, the half-life of a first-order reaction is, $3.3\times 10^2s$

4 0

3 years ago

A letter stating three things that need urgent fixing in your school

ioda

Answer and Explanation:

Computers ( The computers are very old and very slow, and we have to notice that almost everyone are using computers at the same time at the school, which makes it even harder for it to load up assignments. )

The Rick Rolling ( Everyone keeps sending links to teachers and students saying that it is part of some assignment but then you have to listen to Rick Astley, they should really block these links. )

The lockers ( The lockers are also very old and they are breaking down and rusting a lot from the moisture in the hallways. One of the lockers even broke down today!!! I hope they can fix this so no one else gets hit with a locker door. )

Hope this helps! ;)

3 0

3 years ago