Answer:
b. It should be dumped in a beaker labeled "waste copper" on one's bench during the experiment.
d. It should be disposed of in the bottle for waste copper ion when work is completed.
Explanation:
Solutions containing copper ion should never be disposed of by dumping them in a sink or in common trash cans, because this will cause pollution in rivers, lakes and seas, being a contaminating agent to both human beings and animals. They should be placed in appropriate compatible containers that can be hermetically sealed. The sealed containers must be labeled with the name and class of hazardous substance they contain and the date they were generated.
It never should be returned to the bottle containing the solution, since it can contaminate the solution of the bottle.
In the Solutions and Spectroscopy experiments there is always wastes.
Answer:
4.823 x 10^-19 J
Explanation:
Energy is calculated by E = hv where h - Planck's constant in joule.s
v - frequency.
in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s
-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec
now we can simply use Planck's constant in E=hv =
(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.
Explanation:
3 little bones in ear takes that vibration and empowers them. Then they transfer the vibrations to next part of ear.
Answer:
HI.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
- <em>(Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).</em>
- An unknown gas effuses at one half the speed of that of oxygen.
∵ Rate of effusion of unknown gas = 1/2 (Rate of effusion of O₂)
∴ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = 2.
Molar mass of O₂ = 32.0 g/mol.
∵ (Rate of effusion of O₂) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of O₂).
∴ 2.0 = (√molar mass of unknown gas) / √32.0.
(
√molar mass of unknown gas) = 2.0 x √32.0
By squaring the both sides:
∴ molar mass of unknown gas = (2.0 x √32.0)² = 128 g/mol.
∴ The molar mass of sulfur dioxide = 80.91 g/mol and the molar mass of HI = 127.911 g/mol.
<em>So, the unknown gas is HI.</em>
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