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1 year ago

What is the wavelength of light with a frequency of 5.92x10^14

Chemistry

1 answer:

Nikitich [7]1 year ago

6 0

The relationship between the frequency and the wavelength of light is given by this equation:

λ = C/f

Where λ is the wavelength, f is the frequency and C is the speed of light (a constant). We know these values:

f = 5.92 * 10^14 Hz = 5.92 * 10^14 1/s

C = 3.00 * 10^8 m/s

With those values we can solve our problem.

λ = C/f

λ = 3.00 * 10^8 m/s / (5.92 *10^14 1/s)

λ = 5.07 *10^(-7) m

Answer: the wavelength is 5.07 *10^(-7) m.

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How many neutrons and protons are there in the nuclei of the following atoms. a. Li-7 b. O-16 c. Th-232 d. Pu-239

ANEK [815]

Answer:

P = 3 , N = 4

P = 8 , N = 8

P = 90 , N = 142

P = 94 , N = 145

Explanation:

Mass number = Number of protons + Number of neutrons

Also,

Atomic number = Number of protons

Li - 7

Given, Mass number of Lithium = 7

For lithium, atomic number = 3

So,

Number of protons = 3

Number of neutrons = Mass number - Number of protons = 7 - 3 = 4

O - 16

Given, Mass number of oxygen = 16

For oxygen, atomic number = 8

So,

Number of protons = 8

Number of neutrons = Mass number - Number of protons = 16 - 8 = 8

Th - 232

Given, Mass number of Thorium = 232

For Thorium, atomic number = 90

So,

Number of protons = 90

Number of neutrons = Mass number - Number of protons = 232 - 90 = 142

Pu - 239

Given, Mass number of Plutonium = 239

For Plutonium, atomic number = 94

So,

Number of protons = 94

Number of neutrons = Mass number - Number of protons = 239 - 94 = 145

3 0

3 years ago

A student has a solution with a pH of 7. The student adds HF, hydrogen floride, to the solution, hoping to increases the pH of t

Allushta [10]

Low pH = high acidity.
HF has a very low pH, so when added to a solution, it will lower the pH of the solution and therefore make it more acidic. So the answer should be B.

5 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

How many miles of glucose are in 19.1 g of glucose

geniusboy [140]

Answer:

did you mean moles? If so, answer is down below.

Explanation:

there are 0.106 moles of glucose in 19.1 g of glucose.

7 0

3 years ago

If the solubility of a gas is 10.5 g/L at 525 kPa pressure, what is the solubility of the gas when the pressure is 225 kPa? Show

Talja [164]

Answer:

4.5 g/L.

Explanation:

To solve this problem, we must mention Henry's law.

Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

It can be expressed as: P = KS,

P is the partial pressure of the gas above the solution.

K is the Henry's law constant,

S is the solubility of the gas.

At two different pressures, we have two different solubilities of the gas.

∴ P₁S₂ = P₂S₁.

P₁ = 525.0 kPa & S₁ = 10.5 g/L.

P₂ = 225.0 kPa & S₂ = ??? g/L.

∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.

8 0

3 years ago