Answer: The red blood cell will shrink due to water loss
Explanation:
Because 1% NaCl solution is slightly more concentrated than its isotonic 0.9% form, the red blood cell will lose its cell containing fluids such as water to the MORE concentrated environment.
This water loss after a prolonged period will result in shrinking of the red blood cell.
I looked at the question and there isn't a lot of information. im assuming the answer is carbon dioxide. its the only substance that would make sense I hope it helps.
Answer:
The equilibrium will be shifted to lift with the formation of a brown gelatinous precipitate of Fe(OH)₃.
Explanation:
- Le Chatelier's principle states that <em>"when any system at equilibrium for is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract the effect of the applied change and a new equilibrium is established that is different from the old equilibrium"</em>.
- The addition of NaOH will result in the formation of Fe(OH)₃ precipitate which has a brown gelatinous precipitate.
- The formation of this precipitate cause removal and decrease of Fe³⁺ ions.
- According to Le Chatelier's principle, the system will be shifted to lift to increase Fe³⁺ concentration and reduce the stress of Fe³⁺ removal and readjust the equilibrium again. So, the [Fe(SCN)²⁺] decreases.
- Increasing [Fe³⁺] will produce a yellow color solution that contains a brown gelatinous precipitate of Fe(OH)₃.
Answer:
CCl4- tetrahedral bond angle 109°
PF3 - trigonal pyramidal bond angles less than 109°
OF2- Bent with bond angle much less than 109°
I3 - linear with bond angles = 180°
A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°
Explanation:
Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.
Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.
Answer:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
