Cu⇒ 1 atom
N⇒2 atoms
O⇒6 atoms
Total 9 atoms
<h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
<em>(empirical formula) n = molecular formula </em>
Chemical formula : Cu(NO₃)₂
Number of Cu : 1
Number of N : 2
Number of O = 2 x 3 = 6
Total atoms in Cu(NO₃)₂ : 1 + 2 + 6 = 9
<span>Displaced volume :
final volume - initial volume
32.4 mL - 25.2 mL => 7.2 mL
Density = mass / volume
D = 22.6 g / 7.2 mL
D = 3.1388 g/mL
hope this helps!</span>
Answer:
Answers are in the explanation.
Explanation:
<em>Given concentrations are:</em>
- <em>SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60M</em>
- <em>SO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M </em>
- <em>And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10M</em>
<em />
In the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are equilbrium concentrations.</em>
<em />
Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are ACTUAL concentrations.</em>
<em />
If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
- Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reaction
- Q = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the left
- Q = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
<em />
Answer:
Explanation:
Hello!
In this case, according to the chemical reaction:
We can evidence the 2:1 mole ratio between hydrogen and tin, thus, we perform the following stoichiometric setup to obtain the mass of produced tin:
Best regards!
Answer:
Assuming pressure is held constant,the question reduces to a ratio and proportion type of question where;
At constant pressure,
21.0ft³-55.0°F
11.0fy³=11.0ft³/21.0ft³×55°F
The temperature is 28.809°F≈29°F
