Answer:
[C₆H₁₂O₆] = 0.139 M
Explanation:
Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.
We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.
For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)
(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL
We determine moles, we convert them to mmoles, we divide by mL
M = 0.139 M
Moles = 3.95 g . 1mol / 180g → 0.0219 mol
We convert mL to L → 158 mL . 1L/1000mL = 0.158L
M = 0.0219 mol / 0.158L = 0.139 M
Answer:
a.) 22.4 L Ne.
Explanation:
It is known that every 1.0 mol of any gas occupies 22.4 L.
For the options:
<em>It represents </em><em>1.0 mol of Ne.</em>
<em />
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 20 L.
The no. of moles of (20 L) Ar = (1.0 mol)(20 L)/(22.4 L) = 0.8929 mol.
using cross multiplication:
1.0 mol occupies → 22.4 L.
??? mol occupies → 2.24 L.
<em>The no. of moles of (2.24 L) Xe </em>= (1.0 mol)(2.24 L)/(22.4 L) = <em>0.1 mol.</em>
- So, the gas that has the largest number of moles at STP is: a.) 22.4 L Ne.
<span>Hi, friend.
Steepest - Being steep to the greatest degree.
Steep - S</span>harply angled.
Example: When hiking trails lead straight up mountainsides, they've got a steep incline.
Hope this helps!
A compound can be treated like an element when balancing chemical equations when : A. when there is no change to the atoms in the compound
If they remain together during the reaction, you can treat a compound as a singe atom.
hope this helps
Answer: 0.18 V
Explanation:-
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
=-0.40V[/tex]
=-0.24V[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCd%5E%7B2%2B%7D%2FCd%5D%7D)
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCd%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential = 0.16 V
![E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.16-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5B0.10%5D%7D%7B%5B0.5%5D%7D)
Thus the potential of the following electrochemical cell is 0.18 V.
