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.
Answer:
The 3rd option is correct
Explanation:
The element in group 13 and period 3 is aluminium which has the electron configuration

or [Ne] 3s^2 3p^1
You should classify it as co2 solid as CO2(s) and that small left comes to the bottom right hand side and dry ice is a form of liquid carbon dioxide so u should write it as CO2(l) and that L comes to the bottom right hand sise
Answer:
<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>
Explanation:
[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>
Ionic equation:

first, we get moles of potassium hydroxide in 28.94 ml :

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.
Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

for the molar mass: