Answer:
3.4x10⁻¹⁹J is the energy of the photon
Explanation:
To find the energy of a photon from its frequency we must use the equation:
E = hf
<em>Where E is the energy of the photon in J, our incognite</em>
<em>h is Planck's constant = 6.6262x10⁻³⁴Js</em>
<em>And f is the frequency = 5.2x10¹⁴Hz = 5.2x10¹⁴s⁻¹</em>
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Replacing:
E = 6.6262x10⁻³⁴Js*5.2x10¹⁴s⁻¹
E = 3.4x10⁻¹⁹J is the energy of the photon
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Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter
Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.
The algebraic expression would be:
<em>ppb [=] micrograms of compound/liter of solution</em>
We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.
For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.
The respective procedure is in a attached file.
Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)
According to the equation, an acid constant has the following form:
Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]
From pH, we can calculate the concentration of H⁺ and NO₂⁻:
[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]
Now, the acid constant can be calculated:
Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴
And finally,
pKa = -log Ka = 3.44
Answer:
Explanation:"Watch the video and identify which of the following statements are correct." Excuse no video
Answer:
16.02 g
Explanation:
the balanced equation for the decomposition of CuCO₃ is as follows
CuCO₃ --> CuO + CO₂
molar ratio of CuCO₃ to CO₂ is 1:1
number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol
according to the molar ratio
1 mol of CuCO₃ decomposes to form 1 mol of CO₂
therefore 0.364 mol of CuCO₃ decomposes to form 0.364 mol of CO₂
number of CO₂ moles produced - 0.364 mol
therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g
16.02 g of CO₂ produced