Explain how much risk you think your area has for water scarcity????

What is the wavenumber of the radiation emitted when a hydrogen

LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{hc}{\lambda}$

where,

E = energy of the radiation = $1.634\times 10^{-18}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of radiation = ?

Putting values in above equation, we get:

$1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m$

$\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}$

Thus wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

8 0

3 years ago

If a sample of sodium chloride with a mass of

alex41 [277]

Original molarity was 1.7 moles of NaCl

Final molarity was 0.36 moles of NaCl

Given Information:

Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity

Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity

1. Solve for the molarity of the original (concentrated) solution.

Molarity (M) = moles of solute (mol) / liters of solution (L)

Convert the given information to the appropriate units before plugging in and solving for molarity.

Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)

2. Solve for the molarity of the final (diluted) solution.

Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.

Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution

Molarity (M) of the final solution = 0.36 M NaCl

I hope this helped:))

6 0

2 years ago

Choose the correct option for the redox reaction below : 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) 4H2O(l) H+ (oxid

hodyreva [135]

Explanation:

8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

To determine if a substance was reduced or oxidized, the oxidation number at the reactant and product side is compared.

When oxidation occurs, an element loses electrons and its oxidation number increases (becomes more positive). When reduction occurs, an element gains electrons and its oxidation number decreases or is reduced (becomes more negative).

H+ (oxidized or reduced) Answer 1 Choose...

There is an increase in oxidation number from 0 to 1. This means H+ was oxidized.

MnO4- (oxidizing or reducing agent) Answer 2 Choose...

In chemistry, an oxidizing agent is a substance that has the ability to oxidize other substances — in other words to accept their electrons

MnO4- is a strong oxidizing agent.

Fe2+ (oxidizing or reducing agent) Answer 3 Choose...

A reducing agent means it will itself get oxidized. In this reaction, Fe2+ is the reducing agent.

Fe2+ (oxidized or reduced) Answer 4 Choose...

Oxidized

O in MnO4- (oxidized or reduced) Answer 5 Choose...

There is no change in oxidation number of , so it was neither reduced nor was it oxidized.

Mn in MnO4- (oxidized or reduced) Answer 6 Choose...

There is a decreaase in oxidation number from +7 to +2. This means reduction.

6 0

3 years ago

Four acids are analyzed and there KA values are determined which of the following values represents the strongest acid

Advocard [28]

x= the coefficients in front of the substance in the balanced chemical equation

[H+]= the concentration of hydrogen ions

[A-]= the concentration of the other ion that broke off from the H+

[HA]= the un-disassociated acid concentration

The higher the Ka value, the greater amount of disassociation of the reactants into products. As for acids, they will break down to form H+ ions. The more the H+ ions, the stronger acidity of the solution. Thus since A has the highest Ka value, that represents the strongest acid.

You can determine the Ka value from a number of ways. If equilibrium concentrations are given of a certain acid solution, you can find the proportion of the concentration of ions to the concentration of the remaining HA molecules, using the equation above. Also, pH and KpH can be used in a number of ways. This gets more complicated and depends on the situation, and requires more advanced equations.

Hope this helped a little, its obviously not my best work

6 0

3 years ago

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SKJUGCiudsgcksuch clksjgc iuszskh sjgf a kuysdgf akg

taurus [48]

<span>SKJUGCiudsgcksuch clksjgc iuszskh sjgf a kuysdgf akg</span>

8 0

3 years ago

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