Answer:
115.891
Explanation:
Molarity is equal to moles/liters so to get moles we multiply molarity by liters, after this we multiply the moles by the total atomic weight of the compound to get grams
Answer:
NiCO3 (s) + 2H+ (aq) → H2O (l) + CO2 (g) + Ni2+ (aq)
Explanation:
To write the complete ionic equation:
1. Start with a balanced molecular equation.
2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
3. indicate the correct formula and charge of each ion
4. indicate the correct number of each ion
5. write (aq) after each ion
6. Bring down all compounds with (s), (l), or (g) unchanged.
Answer:
1.) 0.1 M
2.) 0.2 M
3.) 1 M
4.) Solution #3 is the most concentrated because it has the highest molarity. This solution has the largest solute to solvent ratio. The more solvent there is, the lower the concentration and molarity.
Explanation:
To find the molarity, you need to (1) convert grams NaOH to moles (via molar mass from periodic table) and then (2) calculate the molarity (via the molarity equation). All of the answers should have 1 sig fig to match the given values.
Molar Mass (NaOH): 22.99 g/mol + 16.00 g/mol + 1.008 g/mol
Molar Mass (NaOH): 39.998 g/mol
4 grams NaOH 1 mole
---------------------- x ------------------ = 0.1 moles NaOH
39.998 g
1.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (1 L)
Molarity = 0.1 M
2.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.5 L)
Molarity = 0.2 M
3.)
Molarity = moles / volume (L)
Molarity = (0.1 moles) / (0.1 L)
Molarity = 1 M
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
The answer is D I hope this helps you !
