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nlexa [21]
3 years ago
13

1.31 times 10^-22 / 6.6262 times 10^-34

Chemistry
1 answer:
victus00 [196]3 years ago
4 0
The first answer is -.595454 the second answer is -1.9488
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Formula equation is more informative than a word equation why​
Taya2010 [7]

Answer:

yes

Explanation:

this is because the formula equation shows the details on how they solved the equation

6 0
2 years ago
If two protons and two neutrons are added to the nucleus of a carbon atom, what nucleus does it become?
Shkiper50 [21]
If two protons are added to carbon, the chemical identity changes to oxygen, which is two spaces over horizontally on the periodic table. The number of protons is what we call the atomic number (Z), and this is what defines the identity of an element. Since we also added two neutrons, this is simply oxygen, or oxygen-16, the most abundant isotope of oxygen.
5 0
3 years ago
Which is not a layer found in the earth's mantle? A. Upper mantle B. asthenosphere C. lithosphere D.crust
Debora [2.8K]

Answer:

LITHOSPHERE

Explanation:

5 0
3 years ago
How many grams of ammonium chloride (gram formula mass= 53.5 g) are contained in .500 L of a 2.00 M solution?
Elza [17]

Answer:

53.5g of NH4Cl

Explanation:

First, we need to obtain the number of mole of NH4Cl. This is illustrated below:

Volume = 0.5L

Molarity = 2M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 2 x 0.5

Mole = 1mole

Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:

Molar Mass of NH4Cl = 53.5g/mol

Number of mole = 1

Mass =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass = 1 x 53.5

Mass = 53.5g

Therefore, 53.5g of NH4Cl is contained in the solution.

8 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
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