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1 year ago

LIMITED TIME I WILL MARK BRAINLYEST

Chemistry

1 answer:

adelina 88 [10]1 year ago

5 0

Explanation:

potential energy is the correct answer

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Select the correct answer. Two charged objects, A and B, exert an electric force on each other. What happens if the distance bet

VikaD [51]

Answer:

The electric force between them decreases

Explanation:

The force between two charged particle is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

r is the distance between charges

If the distance between the charges is increased, the electric force gets decreased as there is an inverse relation between force and distance.

Hence, the correct option is (c) "The electric force between them decreases"

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2 years ago

To name the compound written as cucl2 you would write\

MA_775_DIABLO [31]

Copper chloride ...............

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3 years ago

Calculate the wavelength of a photon of light having a frequency of 3.50 x 1015 Hz.

Aliun [14]

Answer:

Wavelength, $\lambda=8.57\times 10^{-8}\ m$

Explanation:

Given that,

Frequency, $f=3.5\times 10^{15}\ Hz$

We need to find the wavelength of a photon of light. The relation between frequency and wavelength is as follows :

$c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{3.5\times 10^{15}}\\\\\lambda=8.57\times 10^{-8}\ m$

So, the wavelength of the light is $8.57\times 10^{-8}\ m$ .

4 0

2 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago

Convert 3,400 into a scientific notation

kirza4 [7]

3.4 x10^3 best answer to the question

8 0

2 years ago