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ratelena [41]
3 years ago
10

PLEASE HELP ASAP!!!!!

Chemistry
1 answer:
swat323 years ago
4 0

Answer:

Periods indicate that a new period at a new principal energy level which is filled with electrons. Groups indicate that the elements share similar properties because their shells are filled with the same number of valence electons.

Explanation:

We see that starting at period 2 Li has a  +1 charge , He has +2 charge, B has +3 charge, C has the potential to have a +4 charge or -4 charge (because it can gain or loose electrons) and N has a -3 charge, etc. Groups however, have similar properties and every element in a 1A - 8A has the same number of valence electrons. All except the noble gases in which He only needs 2 electrons to be in a perfect state. This is its octet as it would be a completely different element if it had more than 2 electrons in its shell. An example of the groups sharing similar properties is if we were to look at group 1A. In group 1A we find H, Na, K, etc. These are all highly unstable elements when alone (not in a compound) and they are very reactive. If you were to put Na for instance in water, it would light up and you'd see a flame sitting on top of the water. A scientist could do that with the rest of the elements in the group and they would see this increasing in reactivity as you go down the periods. Francium isn't visible long enough to be able to see this happen though. It's more of that kind of element that scientists prove that it exists, but not a very common element to get to learn much about.

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Answer:

Answer is 'B' I think

Explanation:

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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
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<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

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