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Klio2033 [76]
3 years ago
8

Complete the equation and indicate if a precipitate forms.

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + NO₃⁻(aq) → PbBr₂(s) + NO₃⁻(aq) + Li⁺(aq)

A precipitate forms

Explanation:

To solve this question we must know the general solubility rules:

All group 1A (Li, Na, K...) and NH₄⁺ ions are always soluble.

The nitrates (NO₃⁻) and acetates are always solubles.

Cl⁻, Br⁻ and I⁻ are soluble except in the presence of Ag⁺, Pb²⁺, Cu⁺ and Hg₂²⁺

That means in the mixture of ions that we have, the Br⁻ will react with Pb²⁺ to produce PbBr₂(s):

<h3>Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + NO₃⁻(aq) → PbBr₂(s) + NO₃⁻(aq) + Li⁺(aq) </h3>

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8. If 0.795 moles of ammonia gas occupies 42.5 L at 670 torr, what is the Celsius temperature?
maksim [4K]

<u>Answer:</u> The temperature in degree Celsius is 301.31°C

<u>Explanation:</u>

To calculate the temperature, we use the equation given by ideal gas, which is:

PV=nRT

P = pressure of the gas = 670 torr = 0.882 atm   (Conversion factor: 1 atm = 760 torr)

V = volume of gas = 42.5 L

n = number of moles of gas = 0.795 mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of gas

Putting values in above equation, we get:

0.882atm\times 42.5L=0.795mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=574.31K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

574.31=T(^oC)+273\\T(^oC)=301.31^oC

Hence, the temperature in degree Celsius is 301.31°C

6 0
3 years ago
What does this do to the electrons outside the nucleus in the gaseous atoms
AleksandrR [38]

Answer:

Explanation:

As you know, ionization energy is the energy needed to remove one mole of electrons from one mole of atoms in the gaseous state

X

+

energy

→

X

+

+

e

−

Right from the start, you can tell that the harder it is to remove an electron from an atom, the higher the ionization energy will be.

Now, the periodic trends for ionization energy can be describe as follows

ionization energy increases as you move from left to right across a period

ionization energy decreases as you go down a group

As you mentioned, if you compare the first ionization energies for oxygen and chlorine using these two trends, you will get conflicting results.

If you follow the way ionization energy increases across period, chlorine would have a higher ionization energy, since it's closer to the noble gases.

On the other hand, if you go by how ionziation energy decreases from top to bottom in a group, oxygen would have higher ionization energy, since it's located in period 2, as compared with period 3 for chlorine.

As it turns out, the trend for groups overpowers the trend for periods. As aresult, oxygen will have a higher ionization energy than chlorine.

This happens because the smaller oxygen atom has its outermost electrons held tighter by the nucleus. By comparison, chlorine's outermost atoms are located further away from the nucleus.

Not only that, but they are screened from the charge of the nucleus better, since they're located on the third energy level.

Oxygen's outermost electrons are screened by

2

electrons, while chlorine's are screened by

8

electrons.

All these factors will make chlorine's outermost electrons a little easier to remove, which implies a smaller ionization energy than that of oxygen.v

6 0
3 years ago
The tundra is characterized by _____.
muminat
Extremely cold temperatures
5 0
3 years ago
If .758 moles of gas occupy a volume of 80.6L, how many moles will occupy a volume of 270.9L?
egoroff_w [7]

Answer:

n₂ = 2.55 mol

Explanation:

Given data:

Initial number of moles = 0.758 mol

Initial volume = 80.6 L

Final volume = 270.9 L

Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = initial number of moles

V₂ = Final volume

n₂ =  Final number of moles

now we will put the values in formula.

V₁/n₁ = V₂/n₂

80.6 L / 0.758 mol = 270.9 L/ n₂

n₂ = 270.9 L× 0.758 mol / 80.6 L

n₂ =  205.34 L.mol /80.6 L

n₂ = 2.55 mol

4 0
3 years ago
Draw all resonance structures for the nitryl fluoride molecule, NO2F.(a) Explicitly draw all H atoms.(b) Include all valence lon
Dima020 [189]

Answer:

Resonance structures are represented in the picture below.

Explanation:

When there is a double pair of electrons shared between atoms in a molecule, the position of these electrons can be changed, without changing the molecule conformation. This occurs to stabilization, the electrons are relocated. These structures are called resonance structures.

In the molecule of NO₂F, nitrogen has 5 electrons in its valence shell, so it needs 3 electrons to be stable. Oxygen has 6 electrons and needs 2 to be stable, and fluor has 7 electrons and needs one electron to be stable.

Nitrogen still has electrons after the sharing, so it can also share one pair and will have a partial positive charge. One of the oxygens will not complete the octet, so will share only one pair f electron and will have a partial negative charge, that will compensate the positive charge in nitrogen.

The two resonance structures are shown below:

5 0
3 years ago
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