<h2>
</h2>
Explanation:
1. Water decomposition
- Decomposition reactions are represented by-
The general equation: AB → A + B.
- Various methods used in the decomposition of water are -
- Electrolysis
- Photoelectrochemical water splitting
- Thermal decomposition of water
- Photocatalytic water splitting
- Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
- The chemical equation will be -
Hence, balancing the equation we need to add a coefficient of 2 in front of 
on right-hand-side of the equation and 2 in front of 
on left-hand-side of the equation.
∴The balanced equation is -
→ 
2. Formation of ammonia
- The formation of ammonia is by reacting nitrogen gas and hydrogen gas.
→ 
Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.
∴The balanced chemical equation for the formation of ammonia gas is as follows -
→ 
.
- When 6 moles of
react with 6 moles of 4 moles of ammonia are produced.
0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)
Explanation:
First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume (L)
number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles
We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.
Now to find the mass of hydrated sodium borate we use the following formula:
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g
Learn more about:
molar concentration
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Answer:
z≅3
Atomic number is 3, So ion is Lithium ion (
)
Explanation:
First of all
v=f*λ
In our case v=c
c=f*λ
λ=c/f
where:
c is the speed of light
f is the frequency
Using Rydberg's Formula:
Where:
R is Rydberg constant=
z is atomic Number
For highest Energy:
n_1=1
n_2=∞
z≅3
Atomic number is 3, So ion is Lithium ion ()
Answer:
Explanation:
<u>1) Data:</u>
a) M = ?
b) mass of solue = 17 g
c) solute: NH₃
d) V = 0.5o liter
<u>2) Formulae:</u>
a) number of moles, n = mass in grams / molar mass
b) M = n / V (in liters)
<u>3) Solution</u>
a) Molar mass of NH₃ = 17.03 g/mol
b) n = mass in grams / molar mass = 17 g / 17.03 g/mol = 0.998 mol NH₃
c) M = n / V (in liters) = 0.998 mol / 0.50 liter = 1.996 M
d) Round to the appropiate number of significant figures, 2: 2.0 M.
Answer: 2.0 M
Answer: a) Anode: 
Cathode: 
b) Anode : Cr
Cathode : Au
c) 
d) 
Explanation: -
a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
At cathode:
At anode:
b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.
At anode which is a negative terminal, oxidation occurs which is loss of electrons.
Gold acts as cathode ad Chromium acts as anode.
c) Overall balanced equation:
At cathode: (1)
At anode: (2)
Adding (1) and (2)
d)
= -0.74 V
= 1.40 V
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAu%5E%7B3%2B%7D%5D%7D%7B%5BCr%5E%7B3%2B%7D%5D%5E%7D)
where,
n = number of electrons in oxidation-reduction reaction = 3
= standard electrode potential = 2.14 V
![E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.14-%5Cfrac%7B0.0592%7D%7B3%7D%5Clog%20%5Cfrac%7B%5B1.0%7D%7B%5B1.0%5D%7D)
Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.
