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Brilliant_brown [7]
2 years ago
7

Which electron transition results in the emission of energy

Chemistry
1 answer:
Yuki888 [10]2 years ago
3 0
Energy is emitted when an electron falls from a higher energy level to a lower one.

The transition from 3p to 3s would emit energy because the 3s sublevel is lower than the 3p.

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Which elements make up 95 percent (by weight) of the human body?
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Hope this helps.
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2 years ago
A flask with a volume of 125.0mL contains air with a density of 1.269 g/L. What is the mass of the air contained in the flask?
Aliun [14]
<span>In order to solve this problem you must first make sure all your numbers are in like terms. From the density value you can see that it is grams per liter. The first conversion you must do in convert the 125.0 mL value to Liters which you would do by dividing by 1000 because 1 liter is equal to 1000 mL. 125.0 divided by 1000 is 0.125 Liter. Now you will use the density equation to solve. The density equation is density is equal to mass divided by volume. Plug in your known numbers for density and volume. Then solve for mass. So Density (1.269 g/l is equal to mass divided by volume (.125 Liter) You must rearrange the equation to multiple density by volume which is 1.269 times 0.125 which will give you 0.1586. Because the Liters cancel each other out, the answer's unit will be grams. Your final answer is 0.1586 grams.</span>
4 0
3 years ago
2. When the vapor pressure of water is 80 kpa the temperature of the
timofeeve [1]

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3 0
2 years ago
Consider the reaction of Mg3N2 with H2O to form Mg(OH)2 and NH3. If 4.33 g H2O is reacted with excess Mg3N2 and 6.26 g of Mg(OH)
Len [333]

Answer:

89.34%

Explanation:

First, write a balanced reaction.

Mg3N2 + <u>6</u>H2O --> <u>3</u>Mg (OH)2 + <u>2</u>NH3

Next determine the moles of the known substance, or limiting reagent ( H2O)

n= m/MM

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n(H2O)= 0.2403

Use the mole ratio to find the moles of Mg(OH)2

0.2403 ÷2

n (Mg (OH)2) = 0.1202

Next, find the theoretical mass of Mg (OH)2 that should have been produced

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=7.007g

To find percentage yield, divide the experimental amount by the theoretical amount and multiply by 100.

6.26/ 7.007 × 100

=89.34%

7 0
2 years ago
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