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3 years ago

Pls help tysm <3 :D:D

Chemistry

1 answer:

Marianna [84]3 years ago

5 0

Not sure about 2 or 3 but I believe that for number 1 is B or claim 2!!

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Which unit would be used in energy calculations in thermodynamics?

prisoha [69]

Answer:

kJ/mol

Explanation:

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2 years ago

2NH.

liq [111]

Answer:

N=2

H=6

Explanation:

1.Balance a chemical equation in terms of moles.

2.Use the balanced equation to construct conversion factors in terms of moles.

3.Calculate moles of one substance from moles of another substance using a balanced chemical equation.

The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products.

(P.s it could also be where you have to solve it in which you have to simplify it first then solve it.) like adding them all up.

Hope this is the answer. :)

6 0

3 years ago

What is the molecular formula of the compound

Ipatiy [6.2K]

Molecular formula: C10H15Cl5

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2 years ago

What physical change happens to atoms and molecules in a solution that is strongly heated in a flame? Is there a phase change? I

Charra [1.4K]

When atoms and molecules speed up or slow down, that is a physical change. When they change state from liquid to solid or from gas to liquid, that is a physical change. ... The ions or molecules can still come back together to form the original substance

6 0

2 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 orange line is 3.26 x 10⁻¹⁹ J.

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

the diagram of the emission spectrum has been added.

From the given chart;

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;

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

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

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

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

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 orange line is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

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1 year ago