Answer:
E = 3 × 10¹⁰ J
Explanation:
Mass, m = 100 kg
We need to find energy made by the loss of 100 kg of mass. The formula between the mass and energy is given by :
E = mc²
Where c is speed of light
Putting all the values, we get :
E = 100 kg × (3×10⁸ m/s)²
= 3 × 10¹⁰ J
So, the required energy is 3 × 10¹⁰ J.
To calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance, then, add the results together. If the natural abundance of 63Cu is assigned x, the natural abundance of 65Cu is 1-x (the two abundance always add up to 1). So the solution is: (63)(x)+(65)(1-x) = 63.55, 63x+65-65x=63.55, x=0.725=72.5%. The natural abundance of 63Cu is 72.5%, that of 65Cu is 1-72.5%=27.5%.
Answer: it’s b)
Explanation: that’s the only difference that is listed
The answer to your question is C. A solution is a homogeneous mixture composed of two or more substances, so it couldn't have been A and D. Since a solution can't have its substances separated by a chemical means because they are chemically bonded, thus they are able to be separated by physical means
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
brainly.com/question/23038117
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