All categories

2 years ago

How does the electrical force relate to the distance between two charged objects?

Chemistry

2 answers:

vivado [14]2 years ago

6 0

Answer

It is inversely proportional to the square of the distance

Explanation

This is the inverse square law , where in electrostatics, the electric force between two charged objects is inversely connected to the distance of separation between the two objects. This is to mean an increase in separation distance between objects will cause a decrease in the force of attraction between the objects.

77julia77 [94]2 years ago

5 0

Answer:

The correct answer option is 'It is inversely proportional to the square of the distance.'

Explanation:

The electrical force is inversely proportional to the square of the distance.

The relation of the electrical force to the distance between the two charged objects is given by the Coloumb's law which states that:

$ef=k\frac{q_1*q_2}{r^2}$

where $ef$ is the electrical force, $q_1$ and $q_2$ are the two charges which are separated by the distance $r$ .

You might be interested in

A chart that organizes elements by their chemical properties and increasing atomic number is the

asambeis [7]

Periodic table!

The periodic table organizes elements by their chemical properties (mainly through vertical columns known as groups) and increasing atomic number.

8 0

2 years ago

The maximum level of lead allowed in drinking water is 15 mg/kg. What is this concentration in units of parts per million?

Alecsey [184]

15 ppm
1.5 × 10-2 ppm
1.5 × 104 ppm
3.1 ppm
1.5 × 10-2 ppm

6 0

2 years ago

Given the reaction at equilibrium: 2H2(g) + O2(g) ↔ 2H2O(g) + energy Which concentration changes occur when the temperature of t

ivanzaharov [21]

Both the H2 and O2 increases!

4 0

2 years ago

In doing electrolysis, will the copper sulphate solution decolourise?

Elan Coil [88]

Answer:

yea ......... ..........

8 0

2 years ago

The concentration of Pb²⁺ in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of t

svetoff [14.1K]

Answer:

0.005404 M

Explanation:

$Pb^{2+}(aq) + Na_{2}CO_{3}(aq) ---> PbCO_{3}(s) + 2Na^{+}(aq)$

Since you added an excess of sodium carbonate you warrantied that all the $Pb^{+2}$ in the sample reacted with it. So we can say that the insoluble lead (II) carbonate $PbCO_{3}$ contains all the $Pb^{+2}$ ions in the original sample.

The moles of $PbCO_{3}$ are:

$moles-of-PbCO_{3}=\frac{mass-of-PbCO_{3}}{Molecular-weight-of- PbCO_{3}}=\frac{0.1443g}{267\frac{g}{mol}}=0.00054 mol$

One mol of $Pb^{+2}$ is required to form one mol of $PbCO_{3}$ . So, the stoichiometric relationship between them is 1:1.

$Pb^{+2} +CO_{3}^{-2} - - - > PbCO_{3}$

Knowing this, 0.00054 is also the number of moles of $Pb^{+2}$ in the original sample.

So, the concentration of $Pb^{+2}$ in the original sample is:

$M = \frac{mol-of-Pb^{+2}}{volume-wastewater-(liters)}=\frac{0.00054}{0.1L}=0.005404 M$

8 0

3 years ago