Irony can be seen in which of the following lines? (Mark all that apply.)

Calculate the oxidation state of the underlined elements of K2 Cr2 O7

Semenov [28]

Answer:

+4

Explanation:

In PbO2, oxygen exhibits an oxidation number of -2 (since it's not a peroxide or superoxide):

Let the oxidation number of Pb be x. Then, for the compound to be neutral, the oxidation numbers of all atoms should add up to zero.

⇒ x + (−2) + (−2) = 0

x = +4

So the oxidation no. of Pb is +4.

I hope this helps.

3 0

2 years ago

Suppose the flask is already at equilibrium but then the volume of the reaction flask is then reduced what will happen

sveta [45]

1) Answer is: c) The reaction will proceed right.

Balanced chemical reaction: N₂(g) + 3H₂(g) ⇄ 2NH₃(g) ΔH = +92 kJ.

Reducing the volume of the system increase the partial pressures of the products and reactants.

With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable, there are 4 moles at the left side (three moles of hydrogen and one mole of nitrogen) and 2 moles (ammonia) at the right side of the reaction.

2) Answer is: d) The partial pressure of ammonia will increase.

This reaction is endothermic (enthalpy is higher than zero), which means that heat is added.

According to Le Chatelier's principle when the reaction is endothermic heat is included as a reactant and when the temperature increased, the heat of the system increase, so the system consume some of that heat by shifting the equilibrium to the right, producing more ammonia.

8 0

3 years ago

Convert 625 mg to kg.

svetlana [45]

Answer: 0.000625

Explanation:

If you don't know, one Milligram is equivalent to 0.000001 Kilograms. Hence, we'll multiply 625 by 0.000001.

With that being said, 625 times 0.000001 will equal to 0.000625.

Don't forget to add the units.

Milogram unit ⇒ mg

Kilogram unit ⇒ kg

Finally, your grand answer is 0.000625.

<em>Please comment down below for any questions about my answer.</em>

4 0

2 years ago

What substance would you add to KF(aq) to form a buffer solution? View Available Hint(s) What substance would you add to to form

AnnyKZ [126]

<h3>Answer</h3>

HF

<h3>Explanation</h3>

A buffer solution contains <em>a weak acid</em> and<em> its conjugate base</em>. The two species shall have a similar concentration in the solution. It's also possible for <em>a weak base</em> and <em>its conjugate acid</em> to form a buffer solution.

The KF solution already contains large number of $\text{F}^{-}$ ions. The objective is to thus find its conjugate acid or base.

$\text{F}^{-}$ contains no proton $\text{H}^{+}$ and is unlikely to be a conjugate acid. Assuming that $\text{F}^{-}$ is a conjugate base. Adding one proton to $\text{F}^{-}$ would produce its conjugate acid.

$\text{H}^{+} \; (aq) + \text{F}^{-} \; (aq) \rightleftharpoons \text{HF}\; (aq)$

Therefore $\text{HF}$ is the conjugate acid of $\text{F}^{-}$ . $\text{HF}$ happens to be a weak acid. As a result, combining $\text{F}^{-}$ with $\text{HF}$ would produce a solution with large number of both the weak acid and its conjugate base, which is a buffer solution by definition.