Part 1 : Answer is only B substance is soluble in water.
In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.
Part 2 : Substance A is not soluble in water and substance B is soluble in water.
According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.
But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.
Answer:
2450 cm3
Explanation:
Volume of cylinder = V=πr2h
2.45L = 2450mL
1mL = 1 cm cubed
2450mL = 2450 cm cubed
Answer:
False. It should read that both plant and animal species are in danger of extinction, and climate change can destroy habitats.
Explanation:
Answer:
The NMR spectrum that corresponds best to p-bromoaniline is the one that is attached in the image below.
Explanation:
For the p-bromoaniline 3 types of hydrogen are observed. The first signal that appears at 3.7 ppm would be from the hydrogens of the NH2 group, the hydrogens in ortho position with respect to the NH2 group give a double at approximately 6.54 ppm, and finally the characteristic 7.21 ppm signal is observed for the hydrogens in meta position with with respect to the NH2 group.
Answer:
a) 38.2 % mass
b) 61.8 g solute/100 g solvent
c) 1.65 g/mL
Explanation:
Given the data:
mass of solute = 17.5 g
mass of solvent= 28.3 g
total solution volume= 27.8 mL
a)- mass percent= mass of solute/mass of solution x 100
mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g
mass % = 17.5 g/45.8 g x 100 = 38.2 % mass
b)- solubility = grams of solute/ 100 g solvent
= 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent
c)- density = massof solution/total volumesolution = 45.8 g/27.8 mL = 1.65 g/mL
