Answer:
2 ways in which a weather satellite can orbit the earth
Explanation:
Answer:
Then pressure will double
Explanation:
PV = nRT
P = nRT / V
If all other variables are kept constant, P will be proportional to n i.e
P1 / n1 = P2 / n2
n = 1
P1 = P
n2 = 2
P2 =?
P1 / n1 = P2 / n2
P /1 = P2/2
Cross multiply
P2 = 2P
Therefore, if an additional 1mole of the gas is added, the pressure will double
When Ksp = [A2+] [S2-]
when A is the metal: Fe, Ni, Pb, and Cu
When we have [S2-] = 0.1 m and we have Ksp for each metal So by substitution in Ksp formula we can get [A2+] for each metal and compare its value with solution concentration 0.01 M, when we have a concentration more than 0.01 M So there are no sulfides precipitates
- [Fe2+] = Ksp/[S2-]
by substitution with Fe2+ Ksp value:
= 6x10^2 / 0.1
= 6x10^3 M
when [Fe2+] > 0.01 M
∴ no precipitate- [Ni2+] = Ksp /[S2-]
by sustitution with Ni Ksp value :
= 8x10^-1 / 0.1
= 8 M
When [Ni2+] > 0.01 M
∴ no precipitate-[Pb2+] = Ksp / [S2-]
by substitution with Pb Ksp value:
= 6x10^-7 / 0.1
= 6 x 10^-6 M
when [Pb2+] < 0.01 M
∴PbS will be precipited-[Cu2+] = Ksp / [S2-]
by substitution with Cu2+ Ksp value:
= 6x10^-16 / 0.1
= 6x10^-15 M
when [Cu2+] < 0.01 M
∴ CuS will be precipited∴The sulfides precipitates are CuS & PbS
Answer:
940 J
Explanation:
This is a calorimtetry excersie where you apply the formula
Q = C . m . ΔT
We only have 2 data from here, C and ΔT (from the solution)
C → 3.97 J/g°C
ΔT → 2.3°C
We determine the solution's mass by density:
1.03 g/mL = Solution mass / Solution volume
We assume the HCl as the solution volume → 100 mL
1.03 g/mL . 100 mL = 103 g → solution's mass
We replace data: Q = 103 g . 3.97 J/g°C . 2.3°C → 940.4 J