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The copper wire was sanded before burning in order to make sure that copper metal was exposed on the surface of the wire.
Answer: B
Explanation
The copper wire when placed in atmosphere without coating leads to oxidation of copper metal with respect to the impurities present in the atmosphere.
As copper is electropositive in nature, so electronegative ions present in the universe will try to react with copper and the copper will react easily with other elements.
So generally copper wire is coated with color or polymer coating.
In this case, the copper wire without any coating is sanded, so that the eddy sheets or polishing materials on friction with copper wire will remove the impurities by the electrostatic law of conservation of charges and charge transfer.
As the impurities are removed when copper wire is sanded, the copper atoms will be exposed on the surface of the wire leading to burning of copper in the copper wire.
PH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.
or basically 2
Answer:
18.7887 g of NaCl
Explanation:
<em>The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?</em>
Let us start by writing out the balanced equation of the reaction:
Na + Cl2 ---> NaCl2
1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.
mole = mass/molar mass
Therefore
18 g of Na = 18/23 = 0.7826 mole
23 g of Cl2 = 23/71 = 0.3239 mole
In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence
0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.
mass of 0.3239 mole NaCl = 0.3239 x 58 = 18.7887 g
<u>Hence, 18.7887 grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.</u>
The isoelectric point<span> (</span>pI, pH(I),IEP<span>), is the pH at which a particular molecule carries no net electrical charge in the statistical mean.</span>
