Another student measured the volume of an unmarked water container, obtaining a 35.0% (6pts) Given the actual volume of the cont
1 answer:
Answer:
Relative error = 0.35
observed value = 8.775
Explanation:
given data:
Actual value = 13.5 L
Percentage of error=35%
Relative error can be determined as follow:
Absolute error = actual value - observed value
Percentage of error
Then relative error
Relative error
Then absolute error =
Therefore experimental measurement value(observed value)
= actual value - absolute error
=13.5 - 4.725 = 8.775
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Answer:
Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over
Explanation:
For the reaction:
Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)
10,0g of Al₂O₃ are:
10,0g ₓ = <em>0,0980 moles</em>
And 10,0g of HCl are:
10,0 gₓ = <em>0,274 moles</em>
<em />
For a total reaction of 0,274 moles of HCl you need:
0,274× = <em>0,0457 moles of Al₂O₃</em>
Thus, limiting reactant is HCl
The grams produced of AlCl₃ are:
0,274 moles HCl × × 133
= <em>12,1 g of AlCl₃</em>
<em />
The moles of Al₂O₃ that don't react are:
0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles
And its mass is:
0,0523 molesₓ = <em>5,33 g of Al₂O₃ </em>
<em />
I hope it helps!
The coefficients in the order CH₄, O₂, CO₂, and H₂O, are 1,2,1,2
The equation becomes:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)
<h3><em>Further explanation</em></h3>
Complete combustion of hydrocarbon with oxygen will be obtained by CO₂ and H₂O compounds.
If O is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (specifically alkanes)
Equalization of chemical reaction equations can be done using variables.
Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and last atoms O atoms
Methana combustion reaction
CH₄ (g) + O₂ (g) -> CO₂ (g) + H₂0 (g)
We give the most complex compounds, namely Methane, we give the number 1
So the reaction becomes
CH₄ (g) + aO₂ (g) -> bCO₂ (g) + cH₂0 (g)
C atom on the left 1, right b, so b = 1
H atom on the left 4, right 2c, so 2c = 4 ---> c = 2
Atom O on the left 2a, right 2b + c, so 2a = 2b + c
2a = 2.1 + 2
2a = 4
a = 2
The equation becomes:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)
<h3><em>Learn more</em></h3>
the combustion of octane in gasoline
hydrogen and excess oxygen
Keywords: methane, combustion, hydrocarbons, equalization of reaction equations
Answer:
Explanation:
We are asked to find the new volume of a gas after a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:
The gas was heated to 150 degrees Celsius and had a volume of 1587.4 liters.
The temperature was 100 degrees Celsius, but the volume is unknown.
We are solving for the volume at 100 degrees Celsius, so we must isolate the variable V₂. It is being divided by 100°C and the inverse of division is multiplication. Multiply both sides of the equation by 100°C.
The units of degrees Celsius cancel.
The original measurement of volume has 5 significant figures, so our answer must have the same. For the number we calculated, that is the tenth place. The 6 in the hundredth place to the right tells us to round to 2 up to a 3.
The volume of the gas at 100 degrees Celsius is approximately <u>1058.3 liters.</u>
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