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olga55 [171]
2 years ago
8

How many half-lives are required for the concentration of reactant to decrease to 12. 5% of its original value?.

Chemistry
1 answer:
Tcecarenko [31]2 years ago
5 0

Answer:

Three half-lives

Explanation:

12.5% is one-eighth. One eighth is the cube of one half, so it would take three half-lives to reduce a reactant's concentration to 12.5%

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For every meter, the equivalent measurements is 1000 millimeters. Hence in the problem where the number of millimeters is given, we divide the number by 1000 to get the number of meters. The answer here is 0.01123 m.
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Write in exponential form: 34⋅34⋅34⋅34⋅34⋅34⋅34⋅34.
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7 0
3 years ago
Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
nignag [31]

Answer:  1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

  • In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

  • Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the  molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

  • As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element  that would be completey consumed, and the maximum mass of ammonia will be produced from it.
  • We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

  • In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

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3 years ago
What has modern media shown you about atoms
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2 years ago
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How many neutrons does element X have if its atomic number is 30 and its mass number is 70?
Yanka [14]
The answer is 40.

We can solve this by finding out the number of protons, and neutrons. Atomic number of an element means the number of protons in that element. So, the atom has 30 protons if the atomic number is 30.

On the other hand, mass number is the total number of protons and neutrons, but not electrons, because they're too light comparing to the other 2. Therefore, we can simply solve the number of neutrons in the atom by subtracting the number of protons from the mass number. 70 - 30 = 40.

Therfore, the number of neutrons is 40.
3 0
3 years ago
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