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.
x + y = 100
0.10x + 0.45 y =25
eliminate
x + y = 100
x + 4.5 y = 250
-3.5 y = -150
y = 42.85
x = 57.15
hope this helps
Answer:
0.693kg of CuFeS2 is needed to produce 240 grams of Cu.
Explanation:
In 1 mole of CuFeS2 is associated 1 mole of Cu.
Molar mass of CuFeS2 = 183.5 g/mol
Molar mass of Cu = 63.5g/mol
No of moles of copper present in 240g of chalcopyrite = 240/63.5 = 3.779moles
we know
1 mole of Cu → 1 mole of CuFeS2
in 3.779 moles of Cu → 3.779 moles of CuFeS2
3.779 moles of CuFeS2 = 3.779 * 183.5 = 693.45grams
CuFeS2 in kg = 0.693kg
Answer:
CH₂
Explanation:
From the question given above, the following data were obtained:
Mass of compound = 1 g
Mass of CO₂ = 3.14 g
Mass of H₂O = 1.29 g
Empirical formula =?
Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:
For Carbon, C:
Mass of CO₂ = 3.14 g
Molar mass of CO₂ = 12 + (2×16)
= 12 + 32
= 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?
Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂
Mass of C = 12/44 × 3.14
Mass of C = 0.86 g
For hydrogen, H:
Mass of C = 0.86 g
Mass of compound = 1 g
Mass of H =?
Mass of H = (Mass of compound) – (mass of C)
Mass of H = 1 – 0.86
Mass of H = 0.14 g
Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:
Mass of C = 0.86 g
Mass of H = 0.14 g
Divide by their molar mass
C = 0.86 / 12 = 0.07
H = 0.14 / 1 = 0.14
Divide by the smallest
C = 0.07 / 0.07 = 1
H = 0.14 / 0.07 = 2
Thus, the empirical formula of cyclopropane is CH₂
