See the sketch attached.
<h3>Explanation</h3>
The Lewis structure of a molecule describes
- the number of bonds it has,
- the source of electrons in each bond, and
- the position of any lone pairs of electrons.
Atoms are most stable when they have eight or no electrons in their valence shell (or two, in case of hydrogen.)
- Each oxygen atom contains six valence electrons. It demands <em>two</em> extra electrons to be chemically stable.
- Each sulfur atom contains six valence electrons. It demands <em>two </em> extra electrons to be chemically stable.
- Each hydrogen atom demands <em>one</em> extra electron to be stable.
H₂O contains two hydrogen atoms and one oxygen atom. It would take an extra 2 + 2 × 1 = 4 electrons for all its three atoms are stable. Atoms in an H₂O would achieve that need by sharing electrons. It would form a total of 4 / 2 = 2 O-H bonds.
Each O-H bond contains one electron from oxygen and one from hydrogen. Hydrogen has no electron left. Oxygen has six electrons. Two of them have went to the two O-H bonds. The remaining four become 4 / 2 = 2 lone pairs. The lone pairs repel the O-H bonds. By convention, they are placed on top of the two H atoms.
Similarly, atoms in a SO₂ molecule demands an extra 2 × 2 + 2 = 6 electrons for its three atoms to become chemically stable. It would form 6 / 2 = 3 chemical bonds. Loops are unlikely in molecules without carbon. As a result, one of the two O atoms would form two bonds with the S atom while the other form only one.
Atoms are unstable with an odd number of valence electrons. The S atom in SO₂ would have become unstable if it contribute one electron to each of the three bond. It would end up with 3 × 2 + 3 = 9 valence electrons. One possible solution is that it contributes two electrons in one particular bond. One of the three bonds would be a coordinate covalent bond, with both electrons in that bond from the S atom. In some textbooks this type of bonds are also known as dative bonds.
Dots and crosses denotes the origin of electrons in a bond. Use the same symbol for electrons from the same atom. Electrons from the oxygen atoms O are shown in blue in the sketch. They don't have to be colored.
Answer) D. ketone. Im not sure if its right tho
<span>Correct answer is:
But how to get there?
Let's start with simple explanation of what exactly is cellular respiration.
Cellular respiration is a multistage biochemical oxidation process of organic substances when prime product is energy (ATP - adenosine triphosphate) and other are released waste products. Cellular respiration takes place even if other metabolic processes are stopped, but cellular respiration may differ in particular organism groups.Some reactions during whole process of cellular respiration are similar in all types of living organisms.
Cellular respiration is prime indication of declining living processes.Only viruses which are on the edge of living organism and chemical particle are not performing cellular respiration.But to the point :P
In cellular respiration all substrates which are in the cell might be organic, but mostly we are using sugar oxidation - glucose in the presence of oxygen. Chemical formula of sugar looks like this:
Oxygen is just
so for now we have just part of the equation:
But what would be on the right hand side?
It's quite simple, remember equation of full combustion? If we want to burn something we need oxygen like in the equation, so the product of this equation would be carbon dioxide, water and of course energy (ATP).Carbon dioxide formula looks like this:
As a reminder water formula:
Full formula would look like that:
But still as you see this equation is unbalanced, after balancing it would like that:
At the end I would like to explain one more thing. Energy which has been released during this process is part of high-energy connection which might be used to perform chemical reactions in the cell or to move organism for example in muscles. We need to remember that production of ATP is not happening with 100% efficiency and part of this energy is released as heat.</span>
Answer:
1.) 3
2.) 60 CM
Explanation:
1. Density=
= 
2. Length*Width*Height=3*10*2
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
