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amm1812
3 years ago
11

Describe the difference between red light and blue light. Describe the difference between red light and blue light. 700 nm corre

sponded to light in the red range and 450 nm corresponded to light in the blue range. 450 nm corresponded to light in the red range and 700 nm corresponded to light in the blue range. Red light can be produced by only neon and blue light can be produced by various elements. 700 nm corresponded to light in both red and blue ranges but different light speed. 450 nm corresponded to light in both red and blue ranges but different light speed.
Chemistry
1 answer:
fiasKO [112]3 years ago
6 0

Answer:700 nm corresponded to light in the red range and 450 nm corresponded to light in the blue range.

Explanation:The wavelengths of the visible spectrum are usually between 400 nm and 700 nmn with the the rays of red light having a longer wavelength, and blue light rays having shorter wavelengths. The red light has a low frequency of about 405–480 THz and , blue light has a higher frequency in the range of 620–680 THz.  This is due to the inverse relationship between wavelength and frequency which also occurs betwwn  wavelength and energy of light rays with the  red light having less photon energy of about  1.65–2.00 eV and the  blue light with high energy photons of about 2.64–2.75 eV.

Red light can be employed in the photosynthesis in plants as it helps in formation of flowers and the production of fruits. Blue light also is important in the production of chlorophyll and can cause a healthy and strong stem and leaves in excess exposure.

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Arrange the molecules in order from strongest intermolecular force to weakest.
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Answer:- C_4H_1_0>C_3H_8>C_2H_6>CH_4

Explanations:- Alkanes are non polar molecules as these only have carbons and hydrogens. Electron negativity difference of C and H is very low and it makes them non polar. These have weaker London dispersion forces.

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3 0
3 years ago
If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma
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Answer:-

29.07 gram of Cu(NO3)2 will be formed.

4.756 grams of AgNO3 will be left over when the reaction is complete.

Explanation:-

Atomic weight of Cu = 63.546 g mol -1

Molecular weight of AgNO3 = 107.87 x 1 + 14 x 1 + 16 x 3

= 169.87 g mol-1

Number of moles of Copper = 9.85 gram / (63.546 g mol-1)

= 0.155 mol

Number of moles of AgNO3 = 31 gram / ( 169.87 g mol-1)

= 0.183 mol

The balanced chemical equation for this reaction is

Cu + AgNO3 --> Cu(NO3)2 + Ag

According to the equation,

1 mole of Cu reacts with 1 mole of AgNO3.

∴0.155 mol of Cu react with 0.155 mol of AgNO3.

Number of moles of AgNO3 left over = 0.183-0.155=0.028 mol

Number of grams of AgNO3 left over = 0.028 mol x 169.87 grams mol-1

= 4.756 gram

Molecular weight of Cu(NO3)2 = 63.546 x 1 + (14 x 1 +16 x 3 ) x 2

=187.546 gram

Now from the balanced chemical equation,

1 Cu gives 1 Cu(NO3)2

∴ 63.546 g of Cu gives 187.546 gram of Cu(NO3)2

9.85 grams pf Cu gives 187.546 x 9.85 / 64.546 gram of Cu(NO3)2

= 29.07 gram of Cu(NO3)2

4 0
3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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