Answer:
0.00559 mol
Explanation:
The following data were obtained from the question:
V = 140mL = 0.14L
T = 296K
P = 98.2kPa = 98200Pa
Recall: 101325Pa = 1atm
98200Pa = 98200/101325 = 0.97atm
R = 0.082atm.L/Kmol
n =?
The number of mole (n) can be obtained as follows:
PV = nRT
n = PV /RT
n = (0.97x0.14)/(0.082x296)
n = 0.00559 mol
Therefore, the number of mole of the gas present is 0.00559 mol
The molar solubility is 7.4×
M and the solubility is 7.4×
g/L .
Calculation ,
The dissociation of silver bromide is given as ,
→
+ 
S
- S S
Ksp = [
] [
] = [S] [ S ] = 
S = √ Ksp = √ 5. 5×
= 7.4×
The solubility =7.4×
g/L
The molar solubility is the solubility of one mole of the substance.
Since , one mole of
is dissociates and form one mole of each
and
ion . So, solubility is equal to molar solubility but unit is different.
Molar solubility = 7.4×
mol/L = 7.4×
M
To learn more about molar solubility ,
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Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
= 61.25 g
= 88.75 g
Explanation:
=
= 50 g
⇒
=
= 1.25 (moles)
2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O
2 : 1 : 1 : 2
1.25 (moles)
⇒
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒
= 0.625 × 98 = 61.25 g
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒
= 0.625 × 142 = 88.75 g
Considerando la definición de molaridad, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5
.
La molaridad es una medida de la concentración de un soluto en una disolución que se define como el número de moles de soluto que están disueltos en un determinado volumen.
La molaridad de una solución se calcula dividiendo los moles del soluto por el volumen de la solución:

La Molaridad se expresa en las unidades
.
En este caso, sabes que una solución acuosa se preparó al mezclar 4 moles del ácido con suficiente agua hasta completar 8 litros de solución. Entonces, sabes que:
- número de moles de soluto= 4 moles
- volumen= 8 litros
Reemplazando en la definición de molaridad:

Resolviendo:
Molaridad= 0.5 
Finalmente, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5
.
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