This is hard to show but here is how you would determine these. NOTE each dot is an electron.
<span>Question 1) </span>
<span>F-H </span>
<span>1) determine the valance electrons for each. F has 7 and H has 1 </span>
<span>2) one electron from both F and H form the bond "-" which means that you still have 6 electrons to place around F and none to place around H. Place the 6 in sets of 2 around the F </span>
<span>.. </span>
<span>F-H </span>
<span>¨ </span>
<span>Question 2) </span>
<span>2) H-O-H </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place </span>
<span>H-O-H: place two dots above and below the oxygen </span>
<span>Question 3) </span>
<span>3) O=N----H : NOTE: a double bond requires O and N to share two of their electrons each </span>
<span>O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place </span>
<span>N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place </span>
<span>place the 2 dots on top and bottom of oxygen. </span>
<span>place 2 above the N </span>
Answer:
1-46
2-18
Explanation:
c=12 H=1 O=16
ethanol (12×2)+(6×1)+(16)=46
water (2×1)+(16)=18
Answer: The correct answer is option E
Explanation:
Sodium/potassium pump is a mechanism that involves the movement of sodium ions (Na+) out of a cell and potassium ions (K+) into a cell, thereby regulating concentration of ions on both sides of a typical cell membrane.
In this situation, the sodium-potassium pump is usually helps in the establishment of the resting potential. The potassium voltage channels normally closes before the membrane potential is brought to a resting level.
In summary, sodium/potassium pump helps to maintain a balance in the system.
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m
We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
1 mm³ = (10⁻³)³ m³
Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³
When we multiply by this conversion, we will get:
32 m³ = 32 x 10⁹ mm³
This would be c as for the amswer
