Natural: Land mounds, Rivers
Human made: Country borders, cities
Answer:
2VO + 3Fe2O3 —> V2O5 + 6FeO
Explanation:
The skeletal equation for the reaction is given below below:
VO + Fe2O3 —> V2O5 + FeO
We can balance the equation above by doing the following:
There are 2 atoms of V on the right side and 1 atom on the left side. It can be balance by putting 2 in front of VO as shown below:
2VO + Fe2O3 —> V2O5 + FeO
Now, we have a total of 5 atoms of O on the left and 6 atoms on the right side. We can balance it by putting 3 in front of Fe2O3 and 6 in front of FeO as shown below:
2VO + 3Fe2O3 —> V2O5 + 6FeO
Now, we can see that the equation is balanced
If you think of it endothermic is when there is energy needed for the reaction to occur and exothermic is when the reaction releases energy
1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
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The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
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