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3 years ago

14) A wavelength of radiation has a frequency of 2.10 x 1014 Hz. What is the wavelength of this radiation in nm, and

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

3 0

Answer:

λ = 1.43 x 10³ meters (radio waves)

Explanation:

c = f·λ => λ = c/f

λ = wavelength = ?

f = frequency = 2.10 x 10¹⁴ Hz = 2.10 x 10¹⁴ cycles/sec

c = speed of light (vacuum) = 3.0 x 10⁸m/sec

λ = c/f = 3.0 x 10⁸m/sec / 2.10 x 10¹⁴sec⁻¹ = 1.43 x 10³ meters (radio waves)

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Determine the volume (L) of nitrogen monoxide gas that is created at STP when 32.2 g

brilliants [131]

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2 years ago

Polonium is a rare element with 33 radioisotopes. The most common one, 210Po, has 84 protons and 126 neutrons. When 210Po decays

Fofino [41]

Answer:

Lead Pb

Explanation:

Firstly, we need to know what occurs when a radioisotope emits an alpha particle. An alpha particle is an helium atom. When an isotope emits an alpha particle, it loses an helium atom corresponding to subtracting 4 from its mass number and 2 from its atomic number. This of course coupled with the release of radiation.

Now, we polonium has a proton number of 84 and a mass number of 210. Subtracting 2 and 4 respectively from its proton and mass numbers will yield 82 and 206 proton and mass numbers respectively.

Hence, the decomposition of the Po-210 isotope will yield an element with 82 proton number and 206 mass number. This corresponds to the element Lead.

210Po ——> 206Pb + alpha particle + radiation

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2 years ago

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

Archy [21]

Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) Alice heats the gas at constant volume until its pressure is doubled:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) Bob further heats the gas at constant pressure until its volume is doubled:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$