A covalent bond is formed between two non-metals that have similar electronegativities.
An <em>i</em><em>o</em><em>n</em><em>i</em><em>c</em><em> </em><em>b</em><em>o</em><em>n</em><em>d</em> is formed between a metal and a non-metal. Non-metals(-ve ion) are "stronger" than the metal(+ve ion) and can get electrons very easily from the metal. These two opposite ions attract each other and form the ionic bond.
The question is incomplete, the complete question is
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V
Answer:
only IV and V
Explanation:
If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.
On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.
This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Answer: Option D) covalent bonds between water molecules
In water, hydrogen bonds are best described as covalent bonds between water molecules
Explanation:
The hydrogen bonds between water molecules are covalent bonds because they are formed when oxygen attract the lone electron in hydrogen, thus resulting in the formation of a partially negative charge on the oxygen atom and a partially positive charge on two hydrogen atoms
Thus, the sharing of electrons between oxygen and hydrogen atoms is responsible for the covalent bonds between water molecules
Answer:
7.71x10^23 molecules
Explanation:
Avogadro's # = 6.022x10^23
1.28 mol SiO2 x 6.022x10^23/ 1 mol SiO2 = 7.71x10^23 molecules