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Maru [420]
2 years ago
5

I NEED TOUR HELP, PLEASE.

Chemistry
1 answer:
kvasek [131]2 years ago
8 0
It would be 5.0 more in i did this
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Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according
Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

6 0
3 years ago
How are lead and plastic foam different?
kozerog [31]
First and foremost, they are completely different substances with each exhibiting unique properties. Both have different atoms involved on their structures which is the cause of the differing properties.
3 0
3 years ago
Read 2 more answers
A measured sample of argon gas has a volume of 20.0 L at a pressure of 660 mm Hg. What is the final volume
valkas [14]

Answer:

C . 24 L

Explanation:

Given data:

Initial volume of gas = 20.0 L

Initial pressure of gas = 660 mmHg

Final volume = ?

Final pressure = 550 mmHg

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

660 mmHg × 20.0 L = 550 mmHg × V₂

V₂ = 13200 mmHg. L/ 550 mmHg

V₂ = 24 L

3 0
3 years ago
An enclosed vessel contains 2.5g of 9b nitrogen and 13.3g of chlorine at s.T.P. Of What will be the partial pressure of the Il n
kow [346]

Answer:

0.535 atm

Explanation:

Since the volume of the tank is constant, we use Gay- Lussac's law to find the pressure at 180°C.

So, P₁/T₁ = P₂/T₂ where P₁ = pressure at S.T.P = 1 atm, T₁ = temperature at S.T.P = 273.15 K, P₂ = pressure of gas at 180 °C and T₂ = 180 °C = 273.15 + 180 K = 453.15 K

So, P₁/T₁ = P₂/T₂

P₂ = P₁T₂/T₁

Substituting the values of the variables into the equation, we have

P₂ = P₁T₂/T₁

P₂ = 1 atm × 453.15 K/273.15 K

P₂ = 1 atm × 1.66

P₂ = 1.66 atm

We now need to find the total number of moles of each gas present

number of moles of nitrogen = mass of nitrogen, m/molar mass of nitrogen molecule M

n = m/M

m = 2.5 g and M = 2 × atomic mass of nitrogen (since it is diatomic) = 2 × 14 g/mol = 28 g/mol

So, n = 2.5 g/28 g/mol

n = 0.089 mol

number of moles of chlorine, n' = mass of chlorine, m'/molar mass of chlorine molecule M'

n' = m'/M'

m' = 13.3 g and M = 2 × atomic mass of chlorine (since it is diatomic) = 2 × 35.5 g/mol = 71 g/mol

So, n' = 13.3 g/71 g/mol

n' = 0.187 mol

So, the total number of moles of gas present is n" = n + n' = 0.089 mol + 0.187 mol = 0.276 mol

So, the partial pressure due to nitrogen gas, P = mole fraction of nitrogen × pressure of gas at 180 °C

P = n/n" × P₂

P = 0.089 mol/0.276 mol × 1.66 atm

P = 0.322 × 1.66 atm

P = 0.535 atm

8 0
3 years ago
For the 52nd electron must pair with one of the<br> electrons in the 5p orbital.
Umnica [9.8K]

Answer:

This question is incomplete

Explanation:

This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as

[Kr] 4d¹⁰ 5s² 5p⁴ where [Kr] is the electronic configuration of krypton. Based on this, we can deduce that the 52nd electron will be in the first orbital of the P subshell (as attached in the picture). This is because when indicating the electrons in the subshell, one electron will be spread across each orbital and if any electron is still remaining, it will be added starting from to the first orbital of the subshell, however no two electrons in an orbital in a subshell can have the same spin and hence must face opposite direction based on pauli's exclusion principle (as seen in attached); thus for the 5p-orbital of elements with 52 or more electrons, when one electron each is represented in each box (3 boxes in total) in the 5p-orbital, the remaining electron is paired with the the first electron in the first box of the 5p-orbital

5 0
3 years ago
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