• Take a look at the steps below to see how to balance this equation. Let's start by writing the unbalanced equation given the information.
Unbalanced Equation : C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O (g)
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Start by Balancing the Carbons : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + H₂O (g)
Now let's balance the Hydrogen : C₃H₈ (g) + O₂ (g) → 3CO₂ (g) + 4H₂O (g)
Balancing the Oxygen : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
Balanced Equation : C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)
• Let's apply dimensional analysis here,
0.7 L propane × (5 liters Oxygen / 1 liter Propane) = 3.5 Liters of Oxygen
• Similarly we can identify the liters of carbon dioxide produced in the reaction,
0.7 L propane × (3 liters Carbon Dioxide / 1 liter Propane) = 2.1 Liters of Carbon Dioxide
• 0.7 L propane × (4 liter water vapor / 1 liter propane ) = 2.8 Liters of Water Vapor
If it has a positive charge it is a cation if negative it is an anion .
I attached a chart that will help you know the charges of the elements
I recently did this assignment!
Instructions: Read each myth (untruth). Reword it to make a factual statement. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words.
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Answer:
Myth: A dead organism is the same as a nonliving thing in science.
o Fact: In science, dead is the same as nonliving.
o Evidence: Things that are nonliving never had the characteristics of life, and never will. Things that are dead once did have the characteristics of life, but when they die, they lose some of the characteristics. That is why dead and non-living are NOT the same thing.
Hope this helped!
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~Lola
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
Metal pots are good for cooking because they have heat conductivity.
