Box C will have the greatest density.
All boxes have the same volume.
Explanation:
We calculate the density using the following formula:
density = mass / volume
density of Box A = 10 g / 20 cm³ = 0.5 g/cm³
density of Box B = 30 g / 20 cm³ = 1.5 g/cm³
density of Box C = 170 g / 20 cm³ = 8.5 g/cm³
Box C will have the greatest density.
All boxes have the same volume.
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Answer:
63. 55 amu
Explanation:
Copper is known to exist in two different isotopes which are Cu-63 and Cu-65.
Cu-63 has an atomic mass of 62.93 amu and it has an abundance of 69.15%.
Similarly,
Cu-65 has an atomic mass of 64.93 amu and it has an abundance of 30.85%
Therefore, using the weighted average mass method, the atomic mass of copper is:
Atomic mass of copper = (0.6915*62.93) amu + (0.3085*64.93) amu = 43.52 amu + 20.03 amu = 63.55 amu
Thus, the atomic mass of copper (express in two decimal places) is 63.55 amu
The vapor pressure is obtained as 23.47 torr.
<h3>What is the vapor pressure?</h3>
Given that; p = x1p°
p = vapor pressure of the solution
x1 = mole fraction of the solvent
p° = vapor pressure of the pure solvent
Δp = p°(1 - x1)
Δp =x2p°
Δp = vapor pressure lowering
x2 = mole fraction of the of the solute
Number of moles of glycerol = 32.5 g/92 g/mol = 0.35 moles
Number of moles of water = 500.0 g/18 g/mol = 27.8 moles
Total number of moles = 0.35 moles + 27.8 moles = 28.15 moles
Mole fraction of glycerol = 0.35 moles/28.15 moles = 0.012
Mole fraction of water = 27.8 moles/28.15 moles =0.99
Δp = 0.012 * 23.76 torr
Δp = 0.285 torr
p1 = p° - Δp
p1 = 23.76 torr - 0.285 torr
p1 = 23.47 torr
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Answer:
336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.
Explanation:
In this case, the balanced reaction is:
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:
- C₃H₈: 1 mole
- O₂: 5 moles
- CO₂: 3 moles
- H₂O: 4 moles
Being the molar mass of each compound:
- C₃H₈: 44 g/mole
- O₂: 16 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by stoichiometry, the following quantities of mass participate in the reaction:
- C₃H₈: 1 mole* 44 g/mole= 44 grams
- O₂: 5 moles* 16 g/mole= 80 grams
- CO₂: 3 moles* 44 g/mole= 132 grams
- H₂O: 4 moles* 18 g/mole= 72 grams
So you can apply the following rules of three:
- If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?
mass of CO₂= 336.6 grams
- If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?
mass of H₂O= 183.6 grams
<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>
