<u>First part:</u>
2 atoms of Na
6 atoms of N
<u>Second part:</u>
2 atoms of N
2 atoms of Na
Why?
We are given the chemical reaction, so, we can extract the number of atoms for both reactants and products.
We are given the chemical equation:

We can rewrite it as follow:

From the reactants, we have:


From the products we have:



Hence, we have that:
There are <u>6 atoms of N in the reactants</u> and <u>6 atoms of N in the products.</u>
and
There are <u>2 atoms of Na in the reactants</u> and<u> </u><u>2 atoms of Na in the products</u> since we have a balanced equation.
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Answer:
Yes
Explanation:
Because they are lighter than water
Answer:
Types of atomic orbitals present in the third principal energy are <u>s, p and d only .</u>
Explanation:
- <u>OPTION A-: s and p atomic orbitals -</u> these two orbitals are present in second principal energy level. Therefore , the option is incorrect.
- <u> OPTION B-: p and d only -</u> This option is wrong as there is no such principal level energy where , s atomic orbital is absent .
- <u>OPTION C-: s , p and d only -</u>these orbitals are present in<u> third principal energy level</u>. The third major level of energy has one orbital, three orbitals of p, and five orbitals of d, each of which can contain up to 10 electrons. The third stage thus holds a maximum of 18 electrons. This option is correct .
- <u>OPTION D-: s , p, d and f only -</u>There is also a f sublevel at the <u>fourth and higher stages,</u> containing seven f orbitals, which can accommodate up to 14 electrons at most. Therefore, up to 32 electrons will hold the fourth level: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals. This option is incorrect .
<u>Thus , the correct option is C (s , p and d only .)</u>
Answer:
Mass of water produced= 1.8 g
Explanation:
Given data:
Mass of water produced = ?
Mass of butane = 1.36 g
Mass of oxygen = excess
Solution:
Chemical equation:
2C₄H₁₀ +13 O₂ → 8CO₂ + 10H₂O
Number of moles of butane:
Number of moles = mass/molar mass
Number of moles = 1.36 g/ 58.12 g/mol
Number of moles = 0.02 mol
Now we will compare the moles of butane with water.
C₄H₁₀ : H₂O
2 : 10
0.02 : 10/2×0.02 = 0.1 mol
Mass of water produced:
Mass = molar mass × molar mass
Mass = 0.1 mol × 18 g/mol
Mass = 1.8 g
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