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.
Frequency, f = v / λ
f = 2.998 * 10⁸ / 3.55*10⁻⁸
f = 8.445 * 10¹⁵ Hz.
Answer:
228 mL
Explanation:
M1*V1 = M2*V2
M1 = 6.58 M
V1 = ?
M2 = 3.00 M
V2 = 500 mL
V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL
N=N₀*2^(-t/T)
N₀=200 g
T=10 d
t=30 d
N=200*2^(-30/10)=25 g
25 g will remain
Substances that cannot be separated and found on a periodic table are elements.
