The freezing point of the sucrose solution is -0.435°C.
<h3>What is the freezing point of the solution?</h3>
The freezing point of the solution is determined from the freezing point depression formula below:
Kf(H₂O) = 1.86 Cm
m is molality of solution = moles of solute/mass of solvent
moles of sucrose = 8.0/342.3 = 0.0233 moles
m = 0.0233/0.1 = 0.233 molal
ΔT = 0.233 m * 1.86°C/m.
ΔT = 0.435 °C.
Freezing point of sucrose solution = 0°C - 0.435°C
Freezing point of sucrose solution = -0.435°C.
In conclusion, the freezing point of sucrose solution is determined from the freezing point depression.
Learn more about freezing point depression at: brainly.com/question/19340523
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<span>PV=nRT= a universal constant
For any condition
P1V1/n1T1=R
and
P2V2/n2T2=R
i.e
P1V1/n1T1=P2V2/n2T2
Becomes
V1/n1=V2/n2
rearranging and solving
V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>
Respuesta:
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)
Explicación:
Consideremos la ecuación no balanceada que ocurre cuando cloruro férrico acuoso reacciona con carbonato de sodio sólido para formar carbonato férrico sólido y cloruro de sodio acuoso. Esta es una reacción de doble desplazamiento.
FeCl₃(aq) + Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Vamos a usar el método de tanteo. Empezaremos balanceando los átomos de C, multiplicando Na₂CO₃ por 3.
FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Luego, balancearemos los átomos de Fe, multiplicando FeCl₃ por 2.
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)
Finalmente, obtendremos la ecuación balanceada, multiplicando NaCl por 6.
2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)
Answer : The concentration of 
and 
at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of 
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of 
at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M
First find the RFM (Relative Formula Mass) using a periodic table
the RFM of Ca(OH)2 is = 74
then use the equation Moles = Mass / RFM
Moles = 7.4 / 74 = 0.1mol
<span>hope that helps</span>
