Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.
We make a proportion out of the word problem
(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose
Answer:
Sample A is a mixture
Sample B is a mixture
Explanation:
For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.
For sample B, we are told that boiling started at 66.2°C and continued until 76.0°C. The implication of this is that B must be a mixture since it boils over a range of temperatures. Pure substances have a sharp boiling point.
You could write about a fear you may have had as a child, eg. the dark, or underneath the bed, etc...
