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Luda [366]
3 years ago
7

What is the wavelength in nanometers of a photon of light that has a frequency on 1.5 x 10^15 hz

Chemistry
2 answers:
vladimir2022 [97]3 years ago
7 0
Equation:  

Ф = v / f where Ф is wavelength, v is velocity of wave (which is usually the speed of light 300,000 km per second, and f is frequency.

Ф = (300,000) / 1.5x10^15

Vladimir79 [104]3 years ago
5 0
We have<span> been looking at the wave properties of electromagnetic radiation. ...  the</span>wavelength<span> and </span>frequency<span> of </span>photons<span> with the following energies: E</span>photon<span>=1200 eV (x-ray) 1.0 </span>nm3.0x10-17 Hz; Ephoton=4.2x10-19<span> J (blue ... MeV</span>/c2<span> ) 8.2</span>x10-14 J 0.51 MeV; Proton (1.67x10-27<span> kg)</span>1.5x10<span>-10 J 0.94</span>
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You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
bonufazy [111]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

5 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
3 years ago
What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Aleksandr-060686 [28]

Answer:

The balanced chemical reaction is given as:

Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

3 0
3 years ago
What is the interval used on the y-axis of this graph?
bezimeni [28]

Answer:

5

Explanation:

the numbers go up by 5

7 0
3 years ago
58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2,
erica [24]

<em><u>Answer and Explanation:</u></em>

Greetings!

Let's~answer~your~question!

Partial ~pressure ~of ~gas ~can ~be ~directly~ calculated ~by ~multiplying ~the~ percentage\\ of~ pressure~ of~ gases~ to ~the ~total ~pressure.

\boxed{Pgas~ = ~P~total~ * \% ~P ~of ~gas}

<em><u>For % of N2 gas: </u></em>

<em><u /></em>100\% - (5\% + 12\%) = 83\% ~N2<em><u /></em>

<em><u /></em>

<em><u /></em>PN_2 ~= ~146~ atm~ *~ 0.83 ~ = 121.18 ~atm<em><u /></em>

<em><u /></em>PO_2 ~= ~146~ atm~ *~ 0.12~ = 17.52 ~atm<em><u /></em>

<em><u /></em>~PCO_2~ = ~146~ atm ~* 0.05 = 7.3 ~atm<em><u /></em>

3 0
3 years ago
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