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3 years ago

How many grams of NaOH are produced from 1.20 x 10^2 of Na2O

1 answer:

5 0

Write an balance the equation

Na2O + H2O -> 2 NaOH

Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.

Na = 23
O=16
H=1

Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40

For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g

120 g Na2O x 80g NaOH / 62g Na2O=

154.8 g NaOH

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An element that is likely to have this property is; A. bromine B. carbon C. sodium D. tellurium ?

vladimir2022 [97]

The answer is D. Most common semiconducting materials are crystalline solids. A<span>morphous and liquid semiconductors are also known to be.</span>

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3 years ago

Read 2 more answers

Two methods by which we can conserve water and water the plants.

Natalka [10]

Answer:

Two methods which help us to conserve water are:

Sprinkler irrigation system: this irrigation has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.

Drip irrigation system: this irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.

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2 years ago

The first electron shell of an atom is complete when it has ___ electrons and the second electron shell is complete when it has

elena-s [515]

Answer:

2, 8, 8

Explanation:

The first electron shell always has 2 electrons when it is complete. The second electron shell and 3rd shell can hold 8 when completed.

5 0

2 years ago

The reaction of 4.8g of sulfur and 5.4g aluminum yields 4.5g Al2S3. 3S+2AL-->Al2S3 Determine the percent yield of Al2S3.

Goshia [24]

Answer:

59.9% is the percent yield for the 4.5 g of produced Al₂S₃

Explanation:

Let's determine the reaction:

3S + 2Al → Al₂S₃

First of all, let's determine the limiting reactant. We need to convert the mass to moles:

4.8 g /32.06g/mol = 0.150 moles of S

5.4 g / 26.98 g/mol = 0.200 moles of Al

3 moles of S react to 2 moles of Al

Then, 0.150 moles of S may react to (0.150 . 2)/3 = 0.1 ,moles of Al

We have 0.200 moles and we only have 0.1. As we have excess of Al, this is the excess reactant. In conclussion, the limiting reagent is S.

2 moles of Al react to 3 moles of S

Then 0.2 moles of Al may react to (0.2 . 3) /2 = 0.3 moles of S. (We only have 0.150 moles)

Let's go to the product, 3 moles of S can produce 1 mol of Al₂S₃

Then 0.150 moles of S, may produce (0.150 . 1) /3 = 0.05 moles.

We convert moles to mass to determine the thoretical yield:

0.05 mol . 150.15g /mol = 7.50g

Percent yield = (Produced yield/Theoretical yield) . 100

% = (4.5g / 7.5g) . 100 = 59.9%

8 0

2 years ago

NEED HELP ASAP WITH THESE QUESTIONS GIVING FAIR AMOUNT OF POINTS IF HELPED WITH ALL QUESTIONS Violet light has a wavelength of 4

scoray [572]

Answer:

a) 7.14e19 Hz

b) 2.298e-27 J

c) 2.793e-19 J; 7.117e9 nm

d) 7.5e14 Hz; 4.96e-19 J

e) 6.2947e-18 J; 31.6 nm

f) 2.21e-22 J

g) 7.1e-19 J; 1.1e15 Hz

h) 3.422e-19 J; 581 nm

i) 4.2e14 Hz

j) 1.92e8 m

k) 7.14e16 Hz; Ultraviolet

Explanation:

Frequency: ν Wavelength: λ Energy: E Speed of light: C (3.00e8) Planck's Constant: h (6.626e-34)

ν -> λ λ = C/ν

λ -> ν ν = C/λ

For either of these equations, wavelength must be converted to meters or nanometers, depending on the equation.

For ν -> λ, after doing the equation, convert the wavelength into nanometers by dividing by 1e-9.

For converting λ -> ν, convert the wavelength into meters by multiplying by 1e-9.

For energy: E = hν = hc/λ

Now that the setup is out of the way:

a) Violet light has a wavelength of 4.20 x 10-12 m. What is the frequency?

λ -> ν ν = C/λ

$\frac{3.00e8}{4.20e-12}$ = 7.14e19 Hz

b) A photon has a frequency (n) of 3.468 x 106 Hz. Calculate its energy

E = hν = hc/λ

(6.626e-34) (3.468e6) = 2.298e-27 J

c) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 4.215 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (4.215e14) = 2.793e-19 J

ν -> λ λ = C/ν

$\frac{3.00e8}{4.215e14}$ = 7.117 m

$\frac{7.117m}{1}*\frac{1nm}{1e-9m}$ = 7.117e9 nm

d) Calculate the frequency and the energy of blue light that has a wavelength of 400 nm (h = 6.62 x 10-34 J-s).

λ -> ν ν = C/λ

$\frac{400 nm}{1} *\frac{1e-9m}{1nm}$ = 4e-7 m

$\frac{3.00e8}{4e-7}$ = 7.5e14 Hz

E = hν = hc/λ

(6.626e-34) (7.5e14) = 4.96e-19 J

e) Calculate the wavelength and energy of light that has a frequency of 9.5 x 1015 Hz.

ν -> λ λ = C/ν

$\frac{3.00e8}{9.5e15}$ = 3.16e-8 m

$\frac{3.16e-8m}{1}*\frac{1nm}{1e-9m}$ = 31.6 nm

E = hν = hc/λ

(6.626e-34) (9.5e15) = 6.2947e-18 J

f) A photon of light has a wavelength of 0.090 cm. Calculate its energy.

E = hν = hc/λ

Convert the wavelength from cm to meters:

$\frac{0.090cm}{1} *\frac{1m}{100cm}$ = 9e-4 m

$\frac{(6.626e-34)(3.00e8)}{9e-4}$ = 2.21e-22 J

g) Calculate the energy and frequency of red light having a wavelength of 2.80 x 10-5 cm.

E = hν = hc/λ

Convert the wavelength from cm to meters:

$\frac{2.80e-5cm}{1} *\frac{1m}{100cm}$ = 2.8e-7 m

$\frac{(6.626e-34)(3.00e8)}{2.8e-7}$ = 7.1e-19 J

λ -> ν ν = C/λ

Convert the wavelength from cm to meters:

$\frac{2.80e-5cm}{1} *\frac{1m}{100cm}$ = 2.8e-7 m

$\frac{3.00e8}{2.8e-7}$ = 1.1e15 Hz

h) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 5.165 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (5.165e14) = 3.422e-19 J

ν -> λ λ = C/ν

$\frac{3.00e8}{5.165e14}$ = 5.81e-7 m

$\frac{5.81e-7m}{1}*\frac{1nm}{1e-9m}$ = 581 nm

i) The wavelength of green light from a traffic signal is centered at 7.20 x 10-5 cm. Calculate the frequency.

λ -> ν ν = C/λ

Convert the wavelength from cm to meters:

$\frac{7.20e-5 cm}{1} *\frac{1m}{100cm}$ = 7.2e-7 m

$\frac{3.00e8}{7.2e-7}$ = 4.2e14 Hz

j) If it takes 1.56 seconds for radio waves (which travel at the speed of light) to reach the moon from Earth, how far away is the moon?

All we want to do here is to convert frequency (speed) to wavelength (distance). This problem requires a bit of thought, but it isn't bad once you realize that frquency is speed and wavelength is distance. It becomes just like the other problems after that. Also, I'll leave this distance in meters, but I think you can figure out how to convert it if it wants it in another unit.

One second is equal to 1 Hertz, so our frequency is 1.56 Hz.

ν -> λ λ = C/ν

$\frac{3.00e8}{1.56}$ = 1.92e8 m

The actual distance from the earth to the moon via google is 3.84e7, but sometimes problems like this will mess with the numbers to make sure that you didn't just look up the answer. I'm still pretty sure that this is right, however.

k) Calculate the frequency of light that has a wavelength of 4.20 x 10-9m. Identify the type of electromagnetic radiation.

First, we convert wavelength to frequency, as normal:

λ -> ν ν = C/λ

$\frac{3.00e8}{4.20e-9}$ = 7.14e16 Hz

Then we identify the electromagnetic wave type. You can look up a conversion chart for these on google, but since our frequency is in the e15 - e17 range, this light is considered ultraviolet.

5 0

3 years ago