Answer:
Iron and Oxygen
Explanation:
Fe is Iron and O2 is Oxygen
Answer:
0.171 M
Explanation:
Step 1: Given data
- Mass of H₃PO₄ (solute): 3.35 g
- Volume of solution (V): 200 mL
Step 2: Calculate the moles of solute
The molar mass of H₃PO₄ is 97.99 g/mol.
3.35 g × 1 mol/97.99 g = 0.0342 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
200 mL × 1 L/1000 mL = 0.200 L
Step 4: Calculate the molarity of the solution
We will use the definition of molarity.
M = moles of solute / liters of solution
M = 0.0342 mol/0.200 L = 0.171 M
Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
#1. An element or ion that has lost two electrons must have a net charge of 2+, because it has two more protons than electrons, therefore the answer is Mg2+
#2. aluminum ions have an oxidation state of 3+ and fluoride has an oxidation state of 1-, therefore I’m order for the charges to cancel you need 3 fluoride ions.
Therefore, the answer is AlF3
