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Kryger [21]
3 years ago
6

State the number of electrons that must be lost by atoms of

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
4 0

<u>The number of electrons that must be lost by atoms of each given to achieve a stable electron configuration are as follows: </u>

  • Li has to lose 1 electron
  • Ca has to lose 2 electrons
  • Ga has to lose 13 electrons
  • Cs has to lose 1 electron
  • Ba has to lose 2 electrons.

<u>Explanation: </u>

It is known that the elements in the last group or we can stay the noble gases have the most stable electron configuration. So if any other element has to attain stable electron configuration, either they have to loose electrons to attain the electronic configuration similar to its closest noble gas element or they have to attain extra electron to reach the electronic configuration of its nearest noble gas element.

In this case, the question is stated to determine the number of electrons lost by the atoms to reach its stable electron configuration. Thus, the outermost shell electrons or the valence state electrons can be lost by the element to attain the electronic configuration of its nearest noble gas element.

So, the elements given in the question need to lose the electrons such that the electronic configuration of the element after losing of electrons will be equal to the electronic configuration of noble elements. This attainment of noble gas configuration by the elements will be achieving the stable electron configuration.

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Answer:

3.74 M

Explanation:

We know that molarity is moles divided by liters. The first thing to do here is convert your 1500 mL of solution to L. There's 1,000 mL in 1 L, so you need to divide 1500 by 1000:

1500 ÷ 1000 = 1.50

Now you can plug your values into the equation for molarity:

5.60 mol ÷ 1.50 L = 3.74 M

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12.23 g of ammonia (NH3) are dissolved in enough water to make 560.0 mL of solution. How many moles of NH3 are added to the wate
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Answer:

0.719 moles of NH₃

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4 0
3 years ago
A student prepares an iron standard solution in a 50.00 mL volumetric flask. The iron stock concentration is 0.0001974 M. A stud
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Answer:

0.00011765 M

Explanation:

When a solution is prepared by dilution, the volumes and concentrations are related by:

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Where C1 is the concentration of the solution 1, V1 is the volume of the solution 1, C2 is the concentration of solution 2, and V2 is the volume of solution 2.

The stock solution is the solution 1, and the standard solution, the solution 2, so:

0.0001974*29.80 = C2*50.00

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