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snow_lady [41]
2 years ago
5

A computer monitor has a width of 14.15 inches and a height of 10.61 inches. What is the area of the monitor display in square m

eters?
Chemistry
1 answer:
givi [52]2 years ago
3 0

This problem is providing us with the width, 14.15 inches and height, 10.61 inches of a computer monitor, so the area of this display in square meters is required. At the end, the result turns out to be 0.09659 m².

<h3>Dimensional analysis:</h3>

In chemistry, we use dimensional analysis in order to figure out results given some numerical information; however, despite not having a specific formula for every problem, one can use the required units as a reference to come up with a convenient mathematical setup.

In this case, one can firstly calculate the area in square inches:

A=14.15in*10.61in\\\\A=150.1in^2

Next, we use the equivalence statement relating inches with meters, 1 in = 0.0254 m, to obtain:

A=150.1in^2*(\frac{0.0254m}{1in} )^2\\\\A=0.09659m^2

Learn more about dimensional analysis: brainly.com/question/10874167

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Which of the following elements has the lowest ionization energy? Rb Sr Cs Ba Xe
Misha Larkins [42]
Its been almost a year since i've been in chemistry, so sorry if its not right. Cs, Ba, Rb, Sr, Xe. I think that the levels of ionozation increase from the bottom left corner of the periodic table and increase as you go up and to the right
5 0
3 years ago
The half-life of carbon-14 is 5,730 years. how old is a sample that is 75% daughter isotope and 25% parent isotope?
Hitman42 [59]

Half-life it tells you about the amount of time needed that half of the quantity of an isotope to disintegrate.

For carbon-14, assuming that the daughter isotope is a stable one and does not disintegrate further, you have:

<u>parent isotope</u>           <u>daughter isotope</u>             <u>years</u>

100%                            0%                                     0

50%                             50%                                  5,730

25%                             75%                                  11,460

5 0
3 years ago
Atoms form chemical bonds to satisfy the rule and to become .
soldi70 [24.7K]

Answer: Atoms form chemical bonds to satisfy the<u> Octet</u> rule and to become <u>stable.</u>

Explanation:

The tendency of atoms to attempt to get a noble gas configuration that is eight valence electrons is said to be octet rule.  This is done to attain noble gas configuration and stability.

In order to attain stability the atoms tends to have eight electrons in its valence shell which can be obtained by either by sharing of electrons or complete transfer of electrons.

For example : As we know that the sodium has one valence electron, so if giving it up then the result in the same electron configuration as the neon and chlorine has seven valence electrons, so if it takes one it will have eight and the result in the same electronic configuration as the argon which is stable.

4 0
4 years ago
Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3
NARA [144]

Answer: Option (4) is the correct answer.

Explanation:

Relation between potential energy and charge is as follows.

                 U = \frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]

As it is given that q_{1} = 1 \times 10^{-8} C, q_{2} = 2 \times 10^{-8} C, and q_{3} = 3 \times 10^{-8} C.

        Distance between the charges = 1 cm = 1 \times 10^{-2} m  (as 1 cm = 0.01 m)

Hence, putting these given values into the above formula as follows.

                 U = \frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]

            = 9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]    

            = 9 \times 10^{9} [2 + 6 + 1.5]

            = 85.5 \times 10^{-5} J

            = 0.00085 J

Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.              

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