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sineoko [7]
3 years ago
9

2) Calculate the mass(g) of the following substances: b) 0.119 mole MgCl2

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

MgCl2 = 24 + 2(35.5)

= 95

mass of substance = mol × molar mass

= 0.119 × 95

= 11.305 g

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Give the symbol for the element with the following orbital diagram:
wel

Answer: 0%

Explanation:

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5 0
3 years ago
How many moles of sodium cyanide (NaCN) would be needed to produce 4.2 moles of sodium sulfate (Na2SO4)? H2SO4 + 2NaCN → 2HCN +
leonid [27]
The purpose of a chemical equation is to relate the amounts of reactants to the amounts of products based on the rate each is consumed. In this problem, one mole of sulfuric acid is consumed along with two moles of sodium cyanide to produce two moles of hydrocyanic acid and one mole of sodium sulfate. The relationship between sodium cyanide and sodium sulfate is 2:1, meaning that two moles of NaCN is required to produce one mole of sodium sulfate.

To produce 4.2 moles of sodium sulfate, two times this amount of NaCN is required. This means that you would need 8.4 moles of sodium cyanide.

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4 0
3 years ago
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List one reason as to why historical growth has slowed in recent years.
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Lower fertility and longer lifespans steadily increased the potential labor force relative to the total population
4 0
3 years ago
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
4 years ago
How many atoms are there in 3.559*10^-6 mol of krypton?
algol [13]

Answer:

The answer is

<h2>2.143 \times  {10}^{18}  \:  \: atoms</h2>

Explanation:

To find the number of atoms given the number of moles we use the formula

N = n × L

where

N is the number of entities

n is the number of moles

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

n = 3.559 \times  {10}^{ - 6}  \: mol

Substitute the values into the above formula and solve

That's

<h3>N  = 3.559 \times  {10}^{ - 6}  \times 6.02 \times  {10}^{23}  \\  = 2.1425 \times  {10}^{18}</h3>

We have the final answer as

<h3>N  = 2.143 \times  {10}^{18}  \:  \: atoms</h3>

Hope this helps you

3 0
4 years ago
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