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3 years ago

How many formula units are in 325 grams of iron (III) chloride?

Chemistry

1 answer:

OLEGan [10]3 years ago

7 0

Answer:

1.204 × 10^24 formula units

Explanation:

Using the following formula to convert mass of iron (III) chloride to moles;

mole = mass/molar mass

Molar mass of FeCl3 = 56 + 35.5(3)

= 56 + 106.5

= 162.5g/mol

Mole = 325/162.5

mole = 2mol

Next, to calculate the formula units of FeCl3, we multiply the number of moles of FeCl3 by Avagadro number (6.02 × 10^23)

Formula units of FeCl3 = 2 × 6.02 × 10^23

12.04 × 10^23

= 1.204 × 10^24 formula units.

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In which of these diatomic molecules would you NOT find an octet of electrons

Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³ → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so each atom has an octet of electrons.

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹ → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, we can not find an octet.

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴ → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, each oxygen atom has an octet.

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵ → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, each fluorine atom has an octet of electrons around.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!

8 0

3 years ago

Identify the effect of increasing acidity on the solubility of the given compounds.

Ilia_Sergeevich [38]

All the given compounds are basic on acidifying their solubility increases.

Explanation:

$C a_{3}\left(P O_{4}\right)_{2}$ solubility will increase, since it is basic in nature, on acidifying solubility increases.
$F e S$ being basic on acidification, the solubility increases.
$M g C O_{3}$ is alkaline in nature, it gets neutralized and there is no increase in acidity.
$M g B r_{2}$ is a salt and it shows no effect on acidification.
$A g I$ is a salt but insoluble in water and shows no effect on acidification.
$M g(O H)_{2}$ is strongly basic in nature, on acidifying the solubility increases.

4 0

3 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

7 0

3 years ago

What is the overall equation for the covalent bond in H2O . Is it H2 + O2=H2O OR. H2 + O = H2O

AlexFokin [52]

Answer:

The answer is in the photo

Explanation:

I hope that is useful for you :)

4 0

3 years ago

what would be the effect on the melting point of a sample if it were not dried completely after filtering the recrystallized sam

jok3333 [9.3K]

The melting point of the sample if it is not dried completely after filtering the recrystallized product will have a broad range and will occur at lower range than the actual value.

What is melting point?

Melting point is the temperature at which the solid form of a given substance changes to the liquid form at atmospheric pressure. It is important because, by using the physical property of a substance the substance can be identified.

The sharp range melting point of the substance indicates the purity of the substance. If the sample is not dried completely after recrystallisation, the melting point will have a broad range.

Therefore, if the sample given is not dried completely, it will be impure and the decreases the melting point of the substance. So the actual melting point of the substance cannot be determined.

To learn more about the melting point click on the given link brainly.com/question/40140

#SPJ4

7 0

1 year ago