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katovenus [111]
3 years ago
5

A chemist is using radiation with a frequency of 6.0 x 10^13 Hz

Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Answer:

λ = 0.5×10⁻⁵ m

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

Explanation:

Given data:

Frequency of radiation = 6.0 ×10¹³ Hz

Wavelength of radiation = ?

Type of radiation = ?

Energy of radiation = ?

Solution:

Formula:

speed of light = wavelength × frequency

3 ×10⁸ m/s = λ × 6.0 ×10¹³ Hz

Hz = s⁻¹

λ = 3 ×10⁸ m/s /  6.0 ×10¹³  s⁻¹

λ = 0.5×10⁻⁵ m

Energy of radiation:

E = hf

h = planck's constant

f = frequency

E = 6.63×10⁻³⁴ j.s × 6.0 ×10¹³  s⁻¹

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

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soldi70 [24.7K]
<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

  • A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
  • Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
  • In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
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      Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

8 0
3 years ago
Given that 7.25 moles of carbon monoxide gas are present in a container of volume 11.90 L, what is the pressure of the gas (in a
d1i1m1o1n [39]

Answer:17.955atm

Explanation:Pv=nrt

P= nrt/v

P= 7.25*0.08205*360/11.90

P= 214.1505/11.90

P=17.995atm

6 0
3 years ago
A nuclear power plant operates at 40.0% efficiency with a continuous production of 1042 MW of usable power in 1.00 year and cons
Sergio039 [100]

Answer:

3.00 x 10^-11 joules / atom of U-235

Explanation:

We know that the formula for Power = Work done (w)/Time (t)

We need to get the joules from power , since Joules is the SI unit of work.

From the formula P = W/t

W = Power (P) * Time (t)

The SI unit for Time is seconds, hence we change 1 year in seconds

1yr * 365 days/yr * 24hrs/day * 60mins/hr * 60 secs/min = 31536000 secs

It was stated in the question that the plant operates at an efficiency of 40%,

Thus to get the true power we divide the power provided in the question by 0.4 or 40%

= X(0.4) = 1042MW

True Power X = 1042/0.4 = 2605MW

Thus true power = 2605 * 10^6 Watts

Now we have the time in seconds and true power in Watts, we then find the work done.

From our above formula P = W/t

W = P*t = (2605* 10^6) (31536000) =

Finally, we can solve for our energy (work):

P = W / T        PT = W = (2880x10^6) (31536000) = 8.22 x 10^16 joules

We then calculate the amount of energy released by only 1 single uranium-235 atom.

= 8.22 x 10^16 joules / 1.07x10^6 g U-235 (235 g / 1 mol)(1 mol/6.0210^23 atoms)

= 3.00 x 10^-11 joules / atom of U-235

5 0
3 years ago
What type of compound is CaBr2 ?
kirza4 [7]

Answer:

Calcium bromatum

Explanation:

3 0
3 years ago
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Describe what it would be like to be an atom
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Explanation:

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3 0
3 years ago
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