Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
Answer:
It will be reported too low.
Explanation:
To measure the specific heat of the metal (s), the calorimeter may be used. In it, the metal will exchange heat with the water, and they will reach thermal equilibrium. Because it can be considered an isolated system (there're aren't dissipations) the total amount of heat (lost by metal + gained by water) must be 0.
Qmetal + Qwater = 0
Qmetal = -Qwater
The heat is the mass multiplied by the specific heat multiplied by the temperature change. If c is the specific heat of the water:
m_metal*s*ΔT_metal = - m_water *c*ΔT_water
s = -m_water *c*ΔT_water / m_metal*ΔT_metal
So, if m_water is now less than it was supposed to be, s will be reported too low, because they are directly proportional.
Answer:
Reaction 1 is balanced but 2 is not balanced , the balance equation are :
1. 
2.
Explanation:
Balanced Equations : These are the equation which follows the law of conservation of mass .
The total number of atoms present in reactant is equal to total number of atoms present in product.
1.
This is acid - base type reaction where
act as Acid
act as weak base
Reactant : ,
Number of atoms of :
C = 2 () + 1 ()
= 2 + 1
= 3
H = 4() + 1 ()
= 4 + 1
5
O = 2() + 3 ()
= 5
Na = 1 ()
= 1
Product : 
,
, 
Number of atoms :
C = 1() + 2()
= 1 + 2
= 3
H = 2() + 3()
= 2 + 3
= 5
O = 1() + 2()
+2(
= 1 + 2 + 2
= 5
Na = 1(
= 1
Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction
2.
This is double displacement reaction .
Check the balancing in both reactant and products should be :
Na = 2
H = 2
Ca = 1
C = 2
O = 6
Cl = 2
