Answer:
the number of laps in the case when he run for 50 minutes is 18,333.33
Explanation:
The computation of the number of laps in the case when he run for 50 minutes is shown below:
Given that
He runs 440m lap in 1.2 minutes
So in 50 minutes he can have laps of
= 440 × 50 ÷ 1.2
= 18,333.33 laps
hence, the number of laps in the case when he run for 50 minutes is 18,333.33
Answer:
6.1×10^8
Explanation:
The reaction is;
Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)
E°cell = E°cathode - E°anode
E°cathode= -0.14 V
E°anode= -0.40 V
E°cell = -0.14-(-0.40)
E°cell= -0.14+0.40
E°cell= 0.26 V
But
E°cell= 0.0592/n log K
E°cell= 0.0592/2 log K
0.26= 0.0296log K
log K = 0.26/0.0296
log K= 8.7838
K= Antilog (8.7838)
K= 6.1×10^8
Answer:
B. products are found on the write side of the arrow in a chemical reaction.
Answer:
Mass of KNO3 in the original mix is 146.954 g
Explanation:
mass of 
in original 254.5 mixture.
moles of 
moles of
= 0.2926 mol of BaSO4
Therefore,
0.2926 mol of BaCl2,
mass of 
= 60.92 g
the AgCl moles 
= 1.3891 mol of AgCl
note that, the Cl- derive from both, 
so
mole of Cl- f NaCl 
mol of Cl-
mol of NaCl = 0.8039 moles
then
KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3
Mass of KNO3 in the original mix is 146.954 g
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