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16 days ago

12

A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.Th

is emitted energy can be used to determine the
A) mass of the sample
B) volume of the sample
C) identity of the element
D) number of moles of the element

2 answers:

castortr0y [2.9K]16 days ago

5 0

The answer is C. The specific amount of energy released when excited electrons fall back to the ground state produces an emission spectrum. That energy is emitted as photons with precise wavelengths corresponding to the energy differences between levels. Because each element yields a characteristic set of wavelengths, the emission spectrum can be used to identify the element in the sample.

Alekssandra [2.8K]16 days ago

4 0

Answer:

C.

Explanation:

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According to the law of conservation of mass, if an element A has an atomic mass of 2 mass units and element B has an atomic mas

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The principle of conservation of mass asserts that mass cannot be created or eliminated. Given that element A has a mass of 2 g/mol and element B has a mass of 3 g/mol, the total mass of compound AB equals the combined molar masses: 2 g/mol + 3 g/mol results in 5 g/mol for AB. As for A2B3, the calculation is as follows: A2 has 2 multiplied by 2, yielding 4 g/mol, while B3 equals 3 multiplied by 3, which gives 9 g/mol. Consequently, A2B3 amounts to 13 g/mol.

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1 month ago

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

eduard [2645]

Details:

The equation to calculate work done is defined as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now we will insert the given values into the formula above to compute the work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is $3.68 \times 10^{-18} J$ .

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23 days ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

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Explanation:

The following data has been provided:

Energy of radiation absorbed by the electron in the hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed in the form of a photon, the frequency is calculated accordingly:

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

According to the De-Broglie equation $\lambda = \frac{h}{p}$

with p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$ = $3.6 \times 10^{-26} J/m$

Squaring both sides gives us:

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

$12.96 \times 10^{-52}$ = $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where m = mass of the electron

Therefore, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Our conclusion is that the kinetic energy gained by the electron in the hydrogen atom is $7.1 \times 10^{-22} J$ .

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28 days ago

A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

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Response:

702 mL

To elaborate:

Given the following:

Mass of sodium hydroxide = 13.20 g

Molarity of H₂SO₄ = 0.235 M

We're tasked with finding the volume of acid necessary to neutralize the sodium hydroxide solution

Step 1: Write the balanced reaction equation

The reaction between H₂SO₄ and NaOH can be summarized as follows:

2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

Step 2: Calculate the moles of NaOH

Moles are calculated by dividing mass by molar mass

The molar mass of NaOH is 40.0 g/mol

Hence;

Number of moles of NaOH = 13.20 g ÷ 40 g/mol

= 0.33 moles of NaOH

Step 3: Determine moles of H₂SO₄ that react

According to the balanced equation, 2 moles of NaOH react with 1 mole of H₂SO₄

Therefore, the ratio giving moles of H₂SO₄ = Moles of NaOH ÷ 2

= 0.33 moles ÷ 2

= 0.165 moles

Step 4: Find the volume of H₂SO₄

Molarity indicates the concentration of the solution in moles per liter

Molarity = Moles ÷ Volume

By rearranging the formula, we find volume = Moles ÷ Molarity

= 0.165 moles ÷ 0.235 M

= 0.702 L

= 702 mL

Thus, the volume of the 0.235 M H₂SO₄ acid solution required equals 702 mL

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14 days ago