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lisabon 2012 [21]
2 years ago
9

A photon of blue light has a frequency of 7.50 × 10¹⁴ Hz. Calculate the wavelength of this photon (c = 3.00 × 10⁸ m/s) in nanome

ters. please show work
Chemistry
1 answer:
Ivan2 years ago
3 0

Answer:  <u>Wavelength of photon </u><u> = 400 nm</u>

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<em>If it served you, give me 5 stars please, thank you!</em>

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Answer:

B is correct answer ...

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3 years ago
Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.
Softa [21]

Explanation:

PbBr_{2} will dissociate into ions as follows.

         PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)

Hence, K_{sp} for this reaction will be as follows.

                   K_{sp} = [Pb^{2+}][Br^{-}]^{2}

We take x as the molar solubility of PbBr_{2} when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

               [Pb^{2+}] = [Pb^{2+}]_{o} + x

               [Br^{-}]^{2} = [Br^{-}]_{o} + 2x

So, equilibrium solubility expression will be as follows.

            K_{sp} = ([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains [Br^{-}]_{o} = 0.10 and there will be no lead ions. So, [Pb^{2+}]_{o} = 0

So, [Br^{-}]_{o} + 2x will approximately equals to [Br^{-}]_{o}.

Hence, K_{sp} = x[Br^{-}]^{2}_{o}

            4.67 \times 10^{-6} = x \times (0.10)^{2}

                        x = 4.67 \times 10^{-4} M

Thus, we can conclude that molar solubility of PbBr_{2} is 4.67 \times 10^{-4} M.

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Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac
olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

n=0.089 M\times 0.200 L=0.0178 mol of HCL

HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

\frac{1}{1}\times 0.0178 mol=0.0178 mol of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

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