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3 years ago

Analyze the transition of a photon

1 answer:

5 0

Photons may be generated by the transition of an electron from one energy level in an atom or molecule to a lower energy level. Photons may be absorbed as they cause an electron to be raised from a lower energy level to a higher energy level (in an atom or molecule).
The photon itself does not undergo a transition of energy: it either exists (with an energy defined by its wavelength), or it doesn't exist (it was destroyed!). You could say that the emitting or absorbing atom/molecule/etc. undergoes a change, or transition, in energy. But "transition" is usually used as a name for the process of jumping in energy.
Hope it help

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What is the Representative Particles in 1.00 mol of atomic nitrogen

docker41 [41]

Answer:

a molecule.

Explanation:

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2 years ago

Low melting point and poor conductor of electricity

WINSTONCH [101]

i THINK THE ANSWER IS A

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3 years ago

What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.150 mol of HCl were

irinina [24]

Answer:

pH = 2.21

Explanation:

Hello there!

In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

$NaF+HCl\rightarrow NaCl+HF$

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

$n_{HF}^{new}=0.300mol+0.150mol=0.450mol\\\\n_{NaF}^{new}=0.200mol-0.150mol=0.050mol$

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:

$[HF]=\frac{0.450mol}{1.0L} =0.450M$

$[NaF]=\frac{0.050mol}{1.0L} =0.050M$

Now, calculated the pKa of HF:

$pKa=-log(6.8x10^{-4})=3.17$

We can proceed to the HH equation:

$pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21$

Best regards!

6 0

3 years ago

3 Mg + 1 Fe2O3 --> 2 Fe + 3 MgO

Gala2k [10]

Answer:

Fe₂O₃ is the limiting reactant.

7.57 g of MgO are formed.

Explanation:

3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO

First we convert the given masses of both reactants into moles, using their respective molar masses:

15.6 g Mg ÷ 24.305 g/mol = 0.642 mol Mg

10.0 g Fe₂O₃ ÷ 159.69 g/mol = 0.0626 mol Fe₂O₃

0.0626 moles of Fe₂O₃ would react completely with (3 * 0.0626 ) 0.188 moles of Mg. As there are more Mg moles than required, Mg is the reactant in excess; thus, Fe₂O₃ is the limiting reactant.

We now calculate how many MgO moles are produced, using the number of moles of the limiting reactant:

0.0626 mol Fe₂O₃ * $\frac{3molMgO}{1molFe_2O_3}$ = 0.188 mol MgO

Finally we convert moles of MgO into grams:

0.188 mol MgO * 40.3 g/mol = 7.57 g

8 0

3 years ago

will adding 525 g of ethylene glycol to 3.00 kg of water ensure that the water will not freeze at -25 degrees celsius?

san4es73 [151]

Yes it will ensure that it doesn’t freeze

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3 years ago