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Naya [18.7K]
3 years ago
5

3. On your graph, the data points between the black squares are data for elements with atomic numbers 3 through 9. Locate these

elements on your periodic table. What term or description would you use to identify these elements with respect to the periodic table?
Chemistry
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

All the given elements belong to the second period of the periodic table

Explanation:

Let's locate the elements with the given atomic numbers:

  • Z = 3, this corresponds to lithium (Li);
  • Z = 4, this corresponds to beryllium (Be);
  • Z = 5, this corresponds to boron (B);
  • Z = 6, this corresponds to carbon (C);
  • Z = 7, this corresponds to nitrogen (N);
  • Z = 8, this corresponds to oxygen, (O);
  • Z = 9, this corresponds to fluorine, (F).

Now notice that each of these 9 elements can be found in the same row of the periodic table. This is period 2. We may then summarize that all of these elements belong to the same period, period number 2.

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This is for chemistry:/
Andreyy89

Answer:

Its the bottom one :) ......

5 0
3 years ago
Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of
harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

6 0
3 years ago
The formula weight of magnesium hydroxide is________ amu.<br> amu.
wlad13 [49]

Answer:

58.316  is the formula weight of magnesium hydroxide

5 0
2 years ago
Why is the law of conservation of mass law and not a theory?
Elodia [21]

Answer:

In physics and chemistry, the law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change, so quantity can neither be added nor be removed.

Explanation:

That is what I think on the subject

8 0
3 years ago
What mass of carbon dioxide (co2) can be produced from 15.6 g of c6h14 and excess oxygen?
madreJ [45]
C6H14+9.5O2=6CO2 +7H20
Number of moles of C6H14=15.6/86=0.1814 moles
so moles of CO2 = 6(0.1814)=1.088 
As the c6h14 has 1 is to 6 ratio with co2 
so
0.1814=mass/44
mass of co2 produced = 47.9 g

6 0
3 years ago
Read 2 more answers
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