Answer:
66.2 % of O
Explanation:
Our compound is the lithium nitrite.
LiNO₂
This salt is ionic and can be dissociated: LiNO₂ → Li⁺ + NO₂⁻
We determine the molar mass:
molar mass of Li + 3 . molar mass of N + 6 . molar mass of O
6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol
The mass of oxygen contained in 1 mol of lithium nitrite is:
6 . 16 g/mol = 96 g
So the percentage of oxygen present is:
(96 g / 144.94 g) . 100 = 66.2 %
Answer:
0.071 moles of Na₃PO₄ .
Explanation:
Given data:
Number of molecules of Na₃PO₄ = 4.3× 10²² molecules
Number of moles = ?
Solution:
1 mole contain 6.022 × 10²³ molecules
4.3× 10²² molecules × 1 mol / 6.022 × 10²³ molecules
0.71× 10⁻¹ mol
0.071 mol
The number 6.022 × 10²³ is called Avogadro number.
"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
Answer:
(c) P and Sb
Explanation:
We can determine the number of valence electrons of an element:
- If it belongs to Groups 1 and 2, the number of valence electrons is equal to the number of group and the differential electron occupies the s subshell.
- If it belongs to the groups 13-18, the number of valence electrons is equal to: "Number of group - 10" and the differential electron occupies the p subshell.
Which pair of elements have the same valence electronic configuration of np³?
(a) O and Se. NO. They belong to the group 16 and the valence electron configuration is ns² np⁴.
(b) Ge and Pb. NO. They belong to the group 14 and the valence electron configuration is ns² np².
(c) P and Sb. YES. They belong to the group 15 and the valence electron configuration is ns² np³.
(d) K and Mg. NO. They belong to the groups 1 and 2 and the valence electron configuration is ns¹ and ns².
(e) Al and Ga. NO. They belong to the group 13 and the valence electron configuration is ns² np¹.
For part 1, just copy them off of the periodic table. For example, element 1 is Hydrogen, and its symbol is H
Elements on the left usually lose electrons and elements on the right tend to gain them. Noble gases have no charge.
