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3 years ago

A green laser pointer emits light with a wavelength of 542 nm. What is the frequency of this light?

1 answer:

6 0

As we know that wavelength and frequency is inversely proportional to each other. Greater the wavelength smaller the frequency and vice versa.

Solution:

The relation between wavelength and frequency is as follow,

υ = c / λ

where
  υ = frequency = ?
c = velocity of light = 3.0 × 10⁸ ms⁻¹
  λ = wavenumber = 542 nm = 542 × 10⁻⁹ m
Putting the given values,

  υ = 3.0 × 10⁸ ms⁻¹ / 542 × 10⁻⁹ m
Result:
  υ = 5.53 × 10¹⁴ s⁻¹

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The term “periodic” helps identify _______ that occur in the table, like electronegativity or atomic radii.

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Answer: ITS C OR B I THINK BUT I HAVE A FEELING ITS ONE OF THOSE

Explanation:

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3 years ago

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How many moles are in 135g of Teflon?

Oliga [24]

1.34 moles are in 135g of Teflon.

Explanation:

Molecular formula of teflon is (C2H4)n

mass of teflon = 135 grams

atomic mass of teflon = 100.01 gram/mole

number of moles of teflon = ?

Formula used to calculate number of moles in a substance is given as:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the above equation:

number of moles = $\frac{135}{100.01}$

number of moles = 1.34 moles

Teflon is polymer which is used for making non-stick coating, in 135 grams of teflon there are 1.34 moles in it.

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3 years ago

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C

Kipish [7]

Answer: 72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

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

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

Putting in the values we get:

$\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles$

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

According to stoichiometry:

1 mole of $C_3H_8$ requires 5 moles of oxygen

0.45 moles of $C_3H_8$ require= $\frac{5}{1}\times 0.45=2.25$ moles of oxygen

Mass of $O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g$

72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

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3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

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3 years ago

Draw the product(s) that are expected when tert-butyl bromide undergoes solvolysis in isopropanol, (CH3)2CHOH.

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Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo SN¹ and E¹ elimination reaction due to the formation of stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as nucleophile it will produce ether and when it acts as a base it will produce unsaturated compound. The reaction along with both products is shown below,

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3 years ago

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