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BlackZzzverrR [31]
2 years ago
14

The SI unit of time is the second, which is defined as 9,192,631,770 cycles of radiation associated with a certain emission proc

ess in the cesium atom. Calculate the wavelength of this radiation (to three significant figures).
Chemistry
2 answers:
Vanyuwa [196]2 years ago
8 0
To calculate the wavelength of this radiation if the SI unit of time is the second, which is defined as 9,192,631,770 cycles of radiation associated with a certain emission process in the cesium atom, the wavelength is one cycle of radiation, and therefore the wavelength is 1/9192631770
Cerrena [4.2K]2 years ago
8 0

<u>Given:</u>

Radiation emission in Cs atom = 9,192,632,770 cycles

<u>To determine:</u>

The wavelength of the above radiation

<u>Explanation:</u>

It is given that :-

1 sec equivalent to 9,192, 631, 770

Now, frequency (ν) = cycles /sec = 9,192, 631, 770/sec

Wavelength of a radiation is given as:

λ = c/ν

where c = speed of light = 3*10⁸ m/s

λ = 3*10⁸ ms⁻¹/9,192, 631, 770 s⁻¹ = 0.0326 m

Ans: Thus the wavelength of this radiation is 0.033 m


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if an element [x] contains 8 protons, 8 neutrons and 8 electrons, the atomic number of the element is ?
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Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1g of potassium nitrate?
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4 0
3 years ago
Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

3 0
3 years ago
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