Answer:
0.250L of solution. 0.250 moles of solute.
Explanation:
As you can see in the image, there is a beaker with an amount of solution. 1/2L are 500mL and each line of the beaker represents 100mL. That means the volume of the solution is approximately 250mL = 0.250L
Molarity is an unit of concentration defined as the moles of solute per liter of solution. A solution that is 1.000M contains 1.000 moles of solute per liter of solution.
As the volume of the solution is 0.250L, the moles are:
0.250L * (1.000mol/L) = 0.250 moles of solute
percent by mass of carbon = 27,27 %
percent by mass of oxygen = 72.72 %
Explanation:
To calculate the percent composition of carbon dioxide (CO₂) we use the following algorithm.
Molecular mass of CO₂ = atomic weight of carbon × number of carbon atoms + atomic weight of oxygen + number of oxigen atoms
Molecular mass of CO₂ = 12 × 1 + 16 × 2 = 44 g / mole
Considering 1 mole of CO₂ we devise the following reasoning:
If in 44 g of CO₂ we have 12 g of carbon and 32 g of oxygen
Then in 100 g of CO₂ we have X g of carbon and Y g of oxygen
X = (100 × 12) / 44 = 27,27 % carbon
Y = (100 × 32) / 44 = 72.72 % oxygen
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percent by mass
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Answer:
The hypothesis becomes a theory if the results support it is answer.
Explanation:
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Mass of water required = 64.42 g
<h3>Further explanation</h3>
Given
Reaction
6CO2 + 6H2O → C6H12O6 + 6O2
157.35 g CO2
Required
mass of water
Solution
mol CO₂ = mass : MW CO₂
mol CO₂ = 157.35 g : 44.01 g/mol
mol CO₂ = 3.575
From the equation, mol ratio CO₂ : H₂O = 6 : 6, so mol H₂O = mol CO₂=3.575
Mass H₂O = mol x MW H₂O
mass H₂O = 3.575 x 18.02
mass H₂O = 64.42 g
Answer:
There are three significant figures in the number 20300