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zhenek [66]
2 years ago
9

PLS HELP ASAP, WILL GIVE 100 POINTS.

Chemistry
1 answer:
Ivan2 years ago
5 0

Answer: values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

location of the electron: In the 7th energy level away from the nucleus.

Explanation:

From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.

           The orbital designation of the describe electron is 7d

Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus.

Got it from quasarJose

Hope it helps

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8 0
2 years ago
When 0.40 mol Al is mixed with 0.40 mol Br2, the following reaction occurs: 2Al(s) + 3Br2(l) → 2AlBr3(s) Identify the limiting r
Lera25 [3.4K]

Answer:

d. The limiting reactant is Br2, and 0.13 mol Al should remain unreacted.

Explanation:

From the reaction, 2Al(s) + 3Br2(l) → 2AlBr3(s)

2 moles of Al combines with 3 moles of Br to form 2 moles of AlBr3

Hence 1 mole of Br combines with 2/3 moles of Al

or 0.4 moles of Br combines with 2/3×0.4 moles of Al or 0.267 Moles of Al leaving

0.4 - 0.267 = 0.133 moles of Al remaining unreacted

7 0
2 years ago
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
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Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

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x = 0.189 / 2 = 0.0945 moles

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A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
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Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

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For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

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for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

= 32.94 g of KrCl_4

So, 32.94 g of the product will form.

5 0
2 years ago
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