Chemistry, 20.11.2020 17:40 kelseeygee

One student rubs a balloon against his hair several times. Another student passes a comb through her hair several times. The students place the comb near the balloon and observe that the comb repels the balloon. Which is the most likely reason the comb is able to repel the balloon?

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The most likely reason the comb is able to repel the balloon is that, the balloon and the comb have the same electrical charge.

The question is incomplete, the complete question is;

One student rubs a balloon against his hair several times. Another student passes a comb through her hair several times. The students place the comb near the balloon and observe that the comb repels the balloon. Which is the most likely reason the comb is able to repel the balloon?

A.The balloon has a positive charge and the comb is neutral.

B. The balloon and the comb have the same electrical charge.

C. The balloon has a negative charge and the comb is neutral.

D. The balloon and the comb have opposite electrical charges

When the student rubs a balloon against his hair several times or another student passes a comb through her hair several times, static charges are built up on the balloon and the comb.

One of the ways by which a neutral object can be charged is by friction. That is, rubbing it against another object repeatedly.

The fact that when the students place the comb near the balloon and observe that the comb repels the balloon means that both the comb and the balloon acquired the same kind of charge after friction.

Therefore, the balloon and the comb have the same electrical charge.

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Answer from: feyundre52

They have like charges

Explanation:

Like charges repel, opposite charges attract

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

It is sure ( overimitation)

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