Answer:
2.6 ×10^-42
Explanation:
From
∆G= -RTlnK
∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1
R= 8.314 Jmol-1K-1
T= 25°C + 273= 298K
-237.2×10^3= 8.314 × 298 × ln K
ln K= -237.2×10^3/2477.572
K = 2.6 ×10^-42
Answer:
In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.
Answer:
Shown below
Explanation:
a) for BrN3
80+3(14)=122amu
b) forC2H6
2(12) + 6(1) = 30amu
C) for NF2
14+2(19) = 52amu
D) Al2S3
2(27) + 3(32)= 150amu
E) for Fe(NO3)3
56 + 3 [14+3(16)] =242amu
F) Mg3N2
3(24) + 2(14)= 100amu
G) for (NH4)2CO3
2[14 +4(1)] +12 +3(16)=96amu
Answer:
1.7 * 10^-5
Explanation:
1- get the number of moles of PbCl2:
number of moles = mass / molar mass
number of moles = 0.45 / 278.1 = 1.618 * 10^-3 moles
2- get the concentration of Pb2+:
molarity = number of moles of solute / volume of solution in liters
molarity = (1.618 * 10^-3) / (0.1) = 0.0162 M
3- getting concentration of Cl-:
<span>PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)
</span>We can note that:
For a certain amount of Pb2+ formed, twice this amount of Cl- is formed.
This means that:
for 0.0162 M of Pb2+, 2*0.0168 = 0.0324 M of Cl- is formed
4- getting Ksp:
Ksp = [Pb2+][Cl-]²
Ksp = (0.0162)*(0.0324)²
Ksp = 1.7 * 10^-5
Hope this helps :)
Scene B depicts chemical change in matter at atomic change.
Composition distinguishes a chemical reaction from a physical reaction. In a chemical process, the makeup of the components changes; in a physical change, the appearance, smell, or straightforward exhibition of a sample of matter changes without changing its composition. Despite the fact that we refer to them as physical "reactions," nothing is actually changing. A change in the substance in question's elemental composition is necessary for a reaction to occur. Therefore, from now on, we will simply refer to bodily "reactions" as physical changes.
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