Answer:
1 atm
Explanation:
The normal boiling point of a liquid is temperature at which the vapour pressure of the liquid equals 1 atm.
If the normal boiling point of cyclohexane is 81.0 °C, the vapour pressure of the liquid is 1 atm.
The graph below shows that the vapour pressure of cyclohexane reaches 1 atm at about 81 °C.
It would be: 1s2, 2s2, 2p6, 3s2
Answer:
1.15 g/mL
Explanation:
The following data were obtained from the question:
Mass (m) = 55.6 g
Volume (V) = 48.5 mL
Density (D) =?
Density is simply defined as the mass of the substance per unit volume of the substance. It is represented mathematically as:
Density (D) = mass (m) /volume (V)
D = m/V
With the above formula, we can obtain the density of the object as follow:
Mass (m) = 55.6 g
Volume (V) = 48.5 mL
Density (D) =?
D = m/V
D = 55.6/48.5
D = 1.15 g/mL
Therefore, the density of the object is 1.15 g/mL
Answer:
a) 7.94 x 10²³ molecules, b) 5.62 g SO2, c) 1.97 L
Explanation:
a) We need to first convert Cu2S to moles. Since molar mass of Cu2S is 159.14 g/mol, 14.0 g = 0.0880 mol. Using molar ratios (3 mol O2/2 mol Cu2S, 0.0880 mol of Cu2S = 0.132 mol O2. Since 1 mol contains 6.02 x 10²³ molecules, 0.132 mol O2 = 7.94 x 10²² molecules O2.
b) Since the molar ratios of Cu2S to SO2 is 1:1, 0.0880 mol of Cu2S produces 0.0880 mol SO2. To convert mol to grams, we use the molar mass of SO2 (64.06 g/mol) to figure out that 0.0880 mol SO2 = 5.63 g SO2.
c) At STP, 1 mol occupies 22.4 liters. 0.0880 mol SO2 x 22.4 L/1 mol = 1.97 L
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Answer:
Relative volume of ether to water that should be used for the extraction = 1.205
Explanation:
The extraction/distribution coefficient of an arbitrary solvent to water for a given substance is expressed as the mass concentration of the substance in the arbitrary solvent (C₁) divided by the mass concentration of the substance in water (C₂).
K = (C₁/C₂)
Let the initial mass of the organic substance X in water be 1 g (it could be any mass basically, it is just to select a right basis, since we are basically working with percentages here).
If 94% of the organic substance X is extracted by ether in a single extraction, 0.94 g ends up in ether and 0.06 g of the organic substance X that remains in water.
Let the volume of ether required be x mL.
Let the volume of water required be y mL.
Relative volume of ether to water that should be used for the extraction = (x/y)
Mass concentration of the organic substance X in ether = (0.94/x)
Mass concentration of organic substance X in water = (0.06/y)
The distribution coefficient , Ko (Cether / C water), for an organic substance X at room temperature is 13.
13 = (0.94/x) ÷ (0.06/y)
13 = (0.94/x) × (y/0.06)
13 = (15.667y/x)
(x/y) = (15.667/13) = 1.205
Hope this Helps!!!
