A student has a balloon that looks shiny like a metal and wants to perform an experiment to determine if it contains any metal. Which of the following procedures could possibly determine if the balloon contains metal? Charge the balloon, hold it near an electroscope, and determine if the electroscope leaves move. Charge the balloon, hold it near an electroscope, and determine if the electroscope leaves move. A. Charge the balloon, hold it near small pieces of paper, and see if it picks up any of the paper. Charge the balloon, hold it near small pieces of paper, and see if it picks up any of the paper.
B. Hold a positively charged rod near the uncharged balloon, then hold a negatively charged rod near the uncharged balloon, and observe whether the balloon acquires a net charge. Hold a positively charged rod near the uncharged balloon, then hold a negatively charged rod near the uncharged balloon, and observe whether the balloon acquires a net charge.
C. Hold a positively charged rod near the uncharged balloon, then hold a negatively charged rod near the uncharged balloon, and observe whether the balloon is attracted or repelled by the rods.

Charge the balloon, hold it near an electroscope, and determine if the electroscope leaves move.

Explanation:

The gold leaf electroscope is an instrument used to detect if a body is charged. It has two gold leafs suspended from a brass stem in a vacuumed glass jar and connected to a metal cap(Toppr).

When the test body is allowed to touch the metal cap, a change in the size of the leaves shows whether the body is charged or not.

Since we are suspecting the balloon to be made up of a metal; metals can be charged. We can test if there is really a charge on the balloon by bringing it near an electroscope to see if the electroscope moves.

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M