Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature
a. It is likely that more rock candy will be formed in batch A.
b. It is likely that less rock candy will be formed in batch A.
c. It is likely that no rock candy will be formed in either batch.
d. I need more information to predict which batch is more likely to form rock candy.
Answer: Option A
Explanation:
More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.
Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.
The sugar will be dissolved in water until the time all the space is filled sugar molecules.
Hence, the correct answer is Option A.
Weather can be predicted only as probable, not definite because the daily weather conditions depends on winds and storms.
It is common observation that weather forecasts are often given as a probability and never in definite terms.
This is because, the weather condition at anytime depends on the temperature of the earth's atmosphere which causes air masses to move leading to wind.
The temperature of the earth's atmosphere changes frequently hence weather conditions also change frequently.
Learn more: brainly.com/question/21209813
Explanation:
Soaps attach to both water and grease molecules.
The grease molecules are attracted more strongly towards each other as compared to water molecules. Also, water molecules are smaller in size hence, strong intermolecular force is required to break the hydrogen bonds of water molecule so that grease or oil molecules can enter the water molecule.
A soap molecule goes in between water and grease molecule and helps them to bind. The force for linkage between water and grease molecule through the soap molecule is weak london dispersion force.
The soap molecule has its salt end as ionic and water soluble. When grease or oil is added to the soap and water solution then the soap acts as an emulsifier. The soap forms miscelles of the non-polar tails and grease molecules are trapped between these miscelles. This miscelle is easily soluble in water hence, the grease is washed away.
Thus, it can be concluded that the nonpolar end of a soap molecule attaches itself to grease.
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution
Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine
Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine
Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine
Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine
Based on the above calculations, the correct answer is (d)</span>
Basis: 100 mL solution
From the given density, we calculate for the mass of the solution.
density = mass / volume
mass = density x volume
mass = (1.83 g/mL) x (100 mL) = 183 grams
Then, we calculate for the mass H2SO4 given the percentage.
mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams
Calculate for the number of moles of H2SO4,
moles H2SO4 = (179.523 grams) / (98.079 g/mol)
moles H2SO4 = 1.83 moles
Molarity:
M = moles H2SO4 / volume solution (in L)
= 1.83 moles / (0.1L ) = 18.3 M
Molality:
m = moles of H2SO4 / kg of solvent
= 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m
