I believe it's B: series circuit
good luck
Chlorine has a smaller atomic size.
Explanation: As you move towards right of the periodic table, the atomic size decreases. This is because the number of protons is increasing towards the right of the periodic table, which applies a greater inward force on the electrons. that is why the elements on the right of the periodic table have a smaller atomic size when compared to the elements on the left. Since chlorine is on the right side of aluminium, it has a smaller atomic size.
<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
