Answer:
See below
Step-by-step explanation:
Matter is either a <em>pure substance</em> or a <em>mixture.
</em>
Pure substances
- Are composed of one type of atom or molecule.
- Have a constant chemical composition
- Have fixed chemical properties
- Have fixed physical properties
• For example, melting point, boiling point, density, solubility
Mixtures:
- Consist of two or more substances not chemically combined
- Have a variable composition
- Can be separated into two or more components by physical means
• For example, filtration, distillation, centrifugation
- Each component retains its own properties
Answer:
Formula: Na2S2O3
we get solubility.
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .
Solution given:
mass of sodium thiosulphate [m1]=25.5g
mass of water [m2]=40g
at temperature [t]=25°C
we have
<u>solubility in g/dm^3</u> :
- =
- =63.75g /litre=63.75g/dm³
<u>solubility in g/dm^3 :63.75g/dm³</u>
<u>n</u><u>o</u><u>w</u>
solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
