In order to identify the reducing agent let us check the oxidation states
I2 + Ti+4 ----> Ti + IO3- reducing agent
I2 : oxidation state of I = 0
Ti^+4 = oxidation state of Ti = +4
Ti : oxidation state = 0
IO3^-1 , oxidation state of Iodine = +5
So here the Ti undergoes reduction hence it is oxidizing agent
Where iodine undergoes oxidation hence it is reducing agent
Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
Answer:
The amount of energy transferred to the diamond while being cut is thus Q = 852000 J
Explanation:
Since quantity of heat transferred Q = mcΔθ where m = mass of substance , c = specific heat capacity of substance and Δθ = temperature change.
Now, given that for diamond, m = mass of diamond = 600 g, c = specific heat capacity of diamond = 710 J/g°C and Δθ = temperature change = 2 °C.
So, the amount of energy transferred to the diamond while being cut is thus
Q = mcΔθ
Q = 600 g × 710 J/g°C × 2 °C
Q = 852000 J
So, the amount of energy transferred to the diamond while being cut is thus Q = 852000 J
The number of electrons in a completely filled second shell of an atom is 8.
Hope this helped.
Answer:
I don't know
Explanation:
I'm sorry I didn't learn this in my life,I have never reach in this level