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noname [10]
3 years ago
5

Calculate wavelength in metres and nanometers of green light whose frequency is 5.7 x 10^14 Hz.

Chemistry
1 answer:
Nookie1986 [14]3 years ago
7 0

Answer:

5.3 x 10⁻⁷ m

Explanation:

Data Given

frequency = 5.7 x 10¹⁴ Hz.

Convert Hz to per seconds (s⁻¹)

1 Hz = 1 s⁻¹

5.7 x 10¹⁴ Hz = 1 x 5.7 x 10¹⁴ = 5.7 x 10¹⁴ s⁻¹

wavelength =?

Solution:

Formula used to calculate wavelength

      f = c/λ

rearrange the above equation

               λ = c/f ........(1)

Where:

c = speed of light = 3 x 10⁸ ms⁻¹

λ= wavelength

put the values in above equation 1

   λ = 3 x 10⁸ ms⁻¹ / 5.7 x 10¹⁴ s⁻¹

   λ = 5.3 x 10⁻⁷ m

So the wavelength of light =  5.3 x 10⁻⁷ m

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What is the IUPAC name of this compund?​
mr_godi [17]

Answer:

3-ethyl-4-methyl Heptane

Explanation:

longest chain is straight .

we start assigning number from where we are close to the branch to this mean its from left hand side. and we saw two chain 1 is ethyl and other is methyl . In alphabetical order e comes first as compare to m so we write ethyl first after that methyl and than in last we write longest chain name with ane (bcz we only have single vonds).

4 0
3 years ago
Read 2 more answers
3. ¿Cuál de las siguientes operaciones es correcta para calcular el número de moles de hidrógeno necesarios para producir 6 mole
Aloiza [94]

Answer:

the answer is d that is 6×2/3

4 0
3 years ago
HI(aq)+NaOH(aq)→ <br> what the final balanced chemical equation with the phases included
julia-pushkina [17]

Answer:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Explanation:

Hello there!

In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:

HI(aq)+NaOH(aq)\rightarrow NaI+H_2O

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Which is also balanced as the number of atoms of all the elements is the same at both sides.

Best regards!

7 0
3 years ago
How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium
Sedbober [7]

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

  • HCl: 2 moles
  • Na: 1 mole
  • NaCl: 2 moles
  • H₂: 1 mole

You know the following masses of each element:

  • H: 1 g/mole
  • Cl: 35.45 g/mole
  • Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
  • Na: 23 g/mole
  • NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
  • H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • Na: 1 mole* 23 g/mole= 23 g
  • NaCl: 2 moles* 58.45 g/mole= 116.9 g
  • H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}

mass of H₂= 4.8 g

<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>

3 0
3 years ago
Read 2 more answers
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
3 years ago
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