(i) 1 mole of Ar =6.022×10^23
atoms of Ar
52 mol of Ar=52×6.022×10 ^23
atoms of Ar =3.131×10 ^25 atoms of Ar
(ii) Atomic mass of He = 4amu
Or
4amu is the mass of He atoms = 1
Therefore 52 amu is the mass of He atoms=0.25×52=13 atoms of He
(iii) Gram atomic mass of He = 4g
Or
4g of He contains =6.022×10 ^23
atoms
Therefore 52 g of He contains=
6.022×10 ^23/4×52 =7.83×10 ^24 atoms
Answer:
0.0159m
Explanation:
9 M
Explanation:
Lead(II) chloride,
PbCl
2
, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.
Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,
K
sp
, will be established between the solid lead(II) chloride and the dissolved ions.
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
Now, the molar solubility of the compound,
s
, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.
Notice that every mole of lead(II) chloride will produce
1
mole of lead(II) cations and
2
moles of chloride anions. Use an ICE table to find the molar solubility of the solid
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
I
−
0
0
C
x
−
(+s)
(
+
2
s
)
E
x
−
s
2
s
By definition, the solubility product constant will be equal to
K
sp
=
[
Pb
2
+
]
⋅
[
Cl
−
]
2
K
sp
=
s
⋅
(
2
s
)
2
=
s
3
This means that the molar solubility of lead(II) chloride will be
4
s
3
=
1.6
⋅
10
−
5
⇒
s
= √
1.6
4
⋅
10
−
5 =
0.0159 M
Answer:
81.04°C
Explanation:
Heat loss by water = Heat gained by Aluminum
Heat loss by water;
H = MCΔT
ΔT = 100 - T2
M = 580g
c = 4.2
H = 580 * 4.2 (100 - T2)
H = 243600 - 2436T2
Heat ganed by Aluminium
H = MCΔT
ΔT = T2 - 24
M = 900g
c = 0.9
H = 900 * 0.9 (T2 - 24)
H = 810 T2 - 19440
243600 - 2436T2 = 810 T2 - 19440
243600 + 19440 = 810 T2 + 2436T2
263040 = 3246 T2
T2 = 81.04°C
Assumption;
Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.
Answer:
Explanation:
Hello,
In this case, we can apply the Boyle's law in order to understand the pressure-volume relationship as an inversely proportional relationship relating the initial and the final volume:
Next, we compute the final pressure P2:
Thus we validate, the higher the volume the lower the pressure.
Best regards.
