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Elza [17]
3 years ago
12

What is the wavelength of an electron with a mass of 9.109×10−31 kg and a velocity of 3.43×107 ms? Use 6.626×10−34kg m2s for Pla

nck's constant. Your answer should include three significant figures.
Chemistry
1 answer:
anyanavicka [17]3 years ago
6 0

<u>Answer:</u> The wavelength of an electron is 2.121\times 10^{-11}m

<u>Explanation:</u>

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{mv}

where,

\lambda = De-Broglie's wavelength = ?

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

m = mass of electron = 9.109\times 10^{-31}kg

v = velocity of electron = 3.43\times 10^{7}m/s

Putting values in above equation, we get:

\lambda=\frac{6.626\times 10^{-34}Js}{(9.109\times 10^{-31}kg)\times (3.43\times 10^{7}m/s)}

\lambda=2.121\times 10^{-11}m

Hence, the wavelength of an electron is 2.121\times 10^{-11}m

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
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Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

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First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 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 CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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