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frosja888 [35]
3 years ago
9

An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not

ation.) How many atoms will there be in 5.00 g of gold.
Chemistry
1 answer:
Kobotan [32]3 years ago
5 0

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

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In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0
3 years ago
The prefix ‘kilo-‘ means a thousand.
Blababa [14]

Answer:

0.97 kg, 1.0kg

Explanation:

As you have mentioned in the question, to convert kilo-gram to grams, this conversion is used, 1 kilogram = 1,000 grams

Initially, when one has to convert from one unit to another, there is a lot of confusion. To avoid any such, multiply and divide the number with units to which it has to be converted.

Here,

970 g = 970 g \times\frac{1 kg}{1 kg}

This makes no difference to the number.

970 g =  970 g \times\frac{1 kg}{1000 g}

As, 1 kg= 1,000 g

970 g =  \frac{970}{1000} kg

970 g = 0.97 kg

Rounding this off to the nearest tenth of a kilogram.

The tenths place is immediately to the right of the decimal point.

Here, 9 is at the tenth's place. Round the tenth's place up if the digit at hundredth's place is greater than or equal to 5. Since, 7 > 5, up the number at tenth's place.

So, answer is 1.0 kg

8 0
3 years ago
The oxidation number of all the element in free state is 0​
Art [367]

Answer:

Zero-Nine

Explanation:

this is becasue these numbers are rather small and if you plug these numbers into an equation you will most likely get 0.

8 0
3 years ago
Which tool was most likely used in a procedure if the lab report shows that approximately 300 mL of water was used ?
Travka [436]

Graduated Cylinder is the answer to this.


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3 years ago
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Which can be used to determine the H?
mamaluj [8]
F,g,a can be used to get ''h''
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