Answer:
[NH₄NO₃] at D → 0.279 M
Explanation:
This exercise involves a series of dilutions one after the other.
First of all, we calcualte ammonium nitrate's concentration at A.
15.71 g . 1 mol/ 80 g = 0.196 mol / 0.150 mL = 1.31 M
At B → 1.31 M . 20 mL/ 75 mL = 0.349 M
At C → 0.349 M . 15 mL / 25 mL = 0.209 M
[NH₄NO₃] at B = 0.349 M
[NH₄NO₃] at C = 0.209 M
So let's calculate the new moles
In 1 mL of B we have 0.349 mmoles
In 10 mL of B we have 3.49 mmoles
In 1 mL of C we have 0.209 mmoles
In 10 mL of C we have 2.09 mmoles
Volume of D = 10 ml + 10ml = 20 mL
Total mmmoles = 3.49 mmoles + 2.09 mmoles = 5.58 mmoles
[NH₄NO₃] at D = 5.58 mmoles / 20mL → 0.279 M
Because light travels faster than sound
Answer:
a)Anode-Zinc
Cathode - Iron
b) At the anode;
Zn(s) -----> Zn^+(aq) + 2e
At the cathode;
Zn^+(aq) + 2e -----> Zn(s)
c) ZnSO4
d) The electrode which is connected to the positive post of the battery is the anode.
The electrode which is connected to the negative post of the battery is the cathode.
Explanation:
Electroplating is an electrolytic process in which a metal is used to cover the surface of another metal. The plating substance is the anode while the plated substance is the cathode. The electrolyte is a salt of the plating metal.
The anode of an electrolytic cell must be maintained at a positive potential with respect to the cathode hence it is connected to the positive terminal of the battery. The cathode must be maintained at a negative potential hence it is connected to the negative terminal of the battery.
Answer:
The heat of reaction is 4938 kJ/mole
Explanation:
It is possible to calculate the heat of a reaction using:
For the reaction:
C₁₂H₂₂O₁₁ (s) + 12 O₂ (g) → 12 CO₂ (g) + 11 H₂O (l)
Specific heat: 7,50kJ/°C
dT = 20,2°C
Thus, q = 151,5 kJ
The moles of sucrose are: 10,5g /342,2965 g/mol = 0,030675 moles
The heat of reacton is:
I hope it helps!
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