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3 years ago

Explain how a neutral paper fragment can be attracted to a charged rod

Chemistry

1 answer:

Yuliya22 [10]3 years ago

3 0

Answer:

The particles in the neutral paper can shift, causing the paper to become polarized and attracted to the rod.

Explanation:

The neutral paper has an even distribution of its electrons throughout the paper. If a charged rod is brought near the neutral paper, this can cause the electrons in the paper to shift. If the rod is negative, the electrons will be repelled from the rod and cause the molecules in the paper to have a slight positive charge on the part of the paper closest to the rod. If the rod is positive, the electrons will be attracted to the rod and cause a slight negative charge on the side of the paper closest to the rod.

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Answer:

Boron has 3 valence electrons

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3 years ago

a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ladessa [460]

Answer:

The molarity of the solution is 1.1 $\frac{moles}{liter}$

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

$molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

In this case

number of moles of solute= 0.564 moles
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Replacing:

$molarity=\frac{0.564 moles}{0.510 L}$

Solving:

molarity= 1.1 $\frac{moles}{liter}$

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2 years ago

What do we call the type of chemical reaction that occurs in this step? Explain how the name of this type of reaction relates to

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3 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

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3 years ago