Matter is every object that stands on this Earth
Answer:
115g/mol
Explanation:
To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.
We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m
Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml
The value of x is obtained as follows:
x = (750 * 0.29)/1000 = 0.2175 moles
Now since we have the number of moles, we can then obtain the molar mass.
Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol
Answer:
Complete ionic: 
.
Net ionic: 
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water: 
, 
, and 
. These three salts will exist as ions:
- Each
formula unit will exist as one 
ion and one 
ion. - Each formula unit will exist as one
ion and two 
ions (note the subscript in the formula 
.) - Each formula unit will exist as one and two ions.
On the other hand, 
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite , , and (three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each formula unit will exist as only one ion and one ion. However, because the coefficient of 
in the original equation is two, 
alone should correspond to two 
ions and two 
ions.
Do not rewrite the salt because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of and two units of . Doing so will give:
.
Simplify the coefficients:
.
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
