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vfiekz [6]
2 years ago
8

How many joules of energy are produced when 3.0 × 10-28 kilograms of mass are lost? [(1 j = 1 ) and the speed of light = 3.00 ×

108 ]
Chemistry
1 answer:
Aleksandr [31]2 years ago
8 0

The equation that we are going to use on this problem is the famous Einstein field<span> equation. E = mc^2 where E is the energy (to be computed), m is the mass (</span>3.0 × 10-28 kilograms) and c is the speed of light (3.00 × 10^8). If we plug in the given into the equation the answer will be 2.7*10^-11 KJ or <span>2.7*10^-8 J. </span>

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mafiozo [28]

The correct answer is B. on Apex!

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2 years ago
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Identify the hybridization of each carbon atom for the molecule above
ipn [44]

Carbons starting from the left end:

  1. sp²
  2. sp²
  3. sp²
  4. sp
  5. sp

Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

  • A carbon atom with four electron domains is sp³ hybridized;
  • A carbon atom with three electron domains is sp² hybridized;
  • A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

  • The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
  • The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
  • The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
  • The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
  • The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.

8 0
3 years ago
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Explanation:

8 0
2 years ago
Maya prepared 0.50 liters of a solution by dissolving 2.0 moles of an unknown compound in water. What is the molarity of the sol
Elena-2011 [213]

Answer:

4 M

Explanation:

Molarity can be represented by the following ratio:

Molarity = moles / volume (L)

Since you have been given both the mass and volume, you can plug the values into the equation and solve for molarity.

Molarity = moles / volumes

Molarity = 2.0 moles / 0.50 L

Molarity = 4 M

6 0
1 year ago
Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be
kakasveta [241]

Explanation:

Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).

The equation is show below as:

[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃  ⇒  [(CH₃)₂CH]₂N⁻Li⁺  + CH₃CH₂CH₂CH₃

N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence,  LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)

Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)

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3 years ago
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