The electronic configuration of Chlorine is 1s2 2s2 2p6 3s2 3p5.
There are three energy levels in chlorine
First energy level is n=1 has 1s2 so total 2 electrons
Second energy level is n=2, 2s2 2p6 so total 2+6= 8 electrons
Third has 3s2 3p5 electrons just 7 total... p can contain 6 electrons but only 5 are present. So the third level has lesser number of electrons than that can be filled
There are 2071.4662 grams of glucose in 11.5 moles.
Per 1 mole there are 180.15588 grams of glucose. 180.5588 x 11.5 =2076.4262
When we balance the given equation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We will get
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
Solution:
Balancing the given equaation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We have to balance the given number of O
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
We get balanced equation
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
The reaction quotient will be
Qc = [product] / [reactant]
Qc = [SO₂F₂(g)] / [SF₆(g) + SO₃(g)]
To learn more click the given link
brainly.com/question/16025432
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Some examples:
- the sun
- a flashlight turned on
- a light bulb
- stars
- fireflies
...hope i helped :)
The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)
Molarity of the urea solution ;
Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL
(1 g = 0.001 kg)
Molality of the urea solution ;
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
