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sergij07 [2.7K]
3 years ago
15

If electrons were used in the two-slit experiment instead of light, what change would need to be made to the slit spacing in ord

er to see a diffraction pattern?
Chemistry
1 answer:
Andrei [34K]3 years ago
5 0
In order to conduct the two-slit experiment with a beam of electrons, the slit opening has to be smaller than the opening for the light waves. This is because the wavelength of the electron beam is lower, which means it will pass through the large slit without diffracting. Diffraction will occur if the spacing is reduced. The reason that electrons have a smaller wavelength is because they have a higher energy than light waves.
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Well the answer to number 1 is definitely B
 
number 2 is B

number 3 is D

Number 4 is C

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the correct answer is

A. Neutral atoms coming together to share electrons

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What is the unabbreviated electron configuration for Neptunium?
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PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!
Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)

This is acid - base type reaction where

CH_{3}COOH(aq) act as Acid

NaHCO_{3}(aq) act as weak base

Reactant :CH_{3}COOH(aq) ,NaHCO_{3}

Number of atoms of :

C = 2 (CH_{3}COOH(aq)) + 1 (NaHCO_{3})

   = 2 + 1

   = 3

H  = 4(CH_{3}COOH(aq)) + 1 (NaHCO_{3})

     = 4 + 1

       5

O = 2(CH_{3}COOH(aq)) + 3 (NaHCO_{3})

    = 5

Na = 1 (NaHCO_{3})

     = 1

Product :  CO_{2}(g),H_{2}O(l) , CH_{3}COONa(aq)

Number of atoms :

C = 1(CO_{2}(g)) + 2(CH_{3}COONa(aq))

   = 1 + 2

   = 3

H = 2(H_{2}O(l)) + 3(CH_{3}COONa(aq))

   = 2 + 3

   = 5

O = 1(H_{2}O(l)) + 2(CH_{3}COONa(aq))

        +2(CO_{2}(g)

    = 1 + 2 + 2

    = 5

Na = 1(CH_{3}COONa(aq)

     = 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2.CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

7 0
3 years ago
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