Answer:
c) There are sharp emission lines demonstrating discrete energy levels.
Explanation:
When an element emits energy in the form of radiation, it produces a spectrum of colors on a photographic plate. This spectrum can either be continuous or discrete. In continuous spectrum the spectrum continues without any discrimination between two regions. This represents the continuous emission of radiation, and thus the continuous emission of energy without any break.
On the other hand, the line spectrum consists of discrete and sharp lines, which shows the emission of radiation in a certain amount in a certain time, with a break between emission. Hence, the line spectra supports the quantization of energy.
The correct option is:
<u>c) There are sharp emission lines demonstrating discrete energy levels.</u>
Some elements can have to electrons and be stable in their shell they are stable in a duplet state instead of a octet state
<span>85% ethanol | 25% ethanol | 50% ethanol
x | y | 20 gal
use x and y because you don;t know how much she needs.
0.85x | 0.25y | 20(0.5)
85% is 85/100 or 0.85, and you need that much of x, same goes for the 25% and 50% mixtures so now you can make up 2 equations
1) x + y = 20 2) 0.85x + 0.25y= 10 (you get 10 when you multiply 20 by 0.5) now you can solve for x or y using substitution.
first rewrite 1) in terms of x or y: x+ y= 20 ----> y= 20 - x now you can substitute 20- x for y in the second equation.. 0.85x + 0.25y= 10 0.85x + 0.25(20-x)= 10 distribute here..(0.25 * 20 and 0.25 * (-x) ) 0.85x + 5 - 0.25x = 10 combine like terms 0.6x +5 = 10 move the 5 over to the other side 0.6x= 10 -5 0.6x = 5 divide both sides by 0.6 x= 25/3 or 8.3 now you know the amount of x so you can substitue this back into the first equation to find y. 0.85x + 0.25y= 10 0.85(25/3) +0.25y= 10 85/12 + 0.25y= 10 0.25y = 10- 85/12 0.25y= 35/12 y= 35/3 or 11.6 you can check by putting these values into the euations: 1) x+ y= 20 25/3 + 35/3 =20 20= 20 good so far 2) 0.85x + 0.25y= 10 0.85(25/3) + 0.25(35/3)=10 10 = 10
so our values for x and y work
x= 25/3 and y= 35/3</span>
Answer:
The concentration would be 0.76 mol/L.
The most common way to solve this problem is to use the formula
c1V1=c2V2
In your problem,
c1 = 4.2 mol/L; V1 = 45.0 mL
c2 = ?; V2 = 250 mL
c2=c1×V1V2 = 4.2 mol/L × 45.0mL250mL = 0.76 mol/L
This makes sense. You are increasing the volume by a factor of about 6, so the concentration should be about ¹/₆ of the original (¹/₆ × 4.2 = 0.7).
Explanation:
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