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BartSMP [9]
3 years ago
9

What is the main exception in the atomic radius? Explain.

Chemistry
1 answer:
yarga [219]3 years ago
7 0

Answer:

The atomic radius of atoms generally decreases from left to right across a period. There are some small exceptions, such as the oxygen radius being slightly greater than the nitrogen radius. Within a period, protons are added to the nucleus as electrons are being added to the same principal energy level.

hope this helps!

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How many moles are in 2.00g of H2O​
gavmur [86]

n = m/M = 2/18 = 1/9 ~0,1 mol

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What are the carbon (C) and oxygen (O) in carbon dioxide (CO2)?
Fynjy0 [20]
A. Atoms. Because I learned this in 6th grade.
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3 years ago
State two ways of measuring the quantity of reactant or product
Bezzdna [24]

Answer:

If a reaction produces a gas such as oxygen or carbon dioxide, there are two ways to measure the reaction rate: using a gas syringe to measure the gas produced, or calculating the reduction in the mass of the reaction solution.

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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
Determine the number of grams of C4H10 that are required to completely react to produce 8.70 mol of CO2 according to the followi
Nitella [24]

Answer:

126.4 g of C_{4}H_{10} are required

Explanation:

Balanced reaction: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

According to balanced reaction-

8 moles of CO_{2} are produced from 2 moles of C_{4}H_{10}

So, 8.70 moles of CO_{2} are produced from (\frac{2}{8}\times 8.70) moles of C_{4}H_{10} or 2.175 moles of C_{4}H_{10}

Molar mass of C_{4}H_{10} = 58.12 g/mol

So, mass of C_{4}H_{10} required = (2.175\times 58.12)g = 126.4 g

Hence 126.4 g of C_{4}H_{10} are required

5 0
3 years ago
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