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Talc is not soluble in water, but it is slightly soluble in dilute mineral acids. In any case, solving it will keep intact its structure. This is naturally true for the case of silicates, which are characteristics by their insolubility.
A is milk in water, actually is does not have to be in water since milk already is a mixture of fats and water. ... A colloid is just a mixture where a substance of dispersed insoluble particles is suspended throughout another substance. Chalk powder is very little soluble in water therefore this will form an suspension.
Answer:
%age Yield = 85.36 %
Solution:
The Balance Chemical Reaction is as follow,
C₆H₁₂O + Acid Catalyst → C₆H₁₀ + Acid Catalyst + H₂O
According to Equation ,
100 g (1 mole) C₆H₁₂O produces = 82 g (1 moles) of C₆H₁₀
So,
4.0 g of C₆H₁₂O will produce = X g of C₆H₁₀
Solving for X,
X = (4.0 g × 82 g) ÷ 100 g
X = 3.28 g of C₆H₁₀ (Theoretical Yield)
As we know,
%age Yield = (Actual Yield ÷ Theoretical Yield) × 100
%age Yield = (2.8 g ÷ 3.28 g) × 100
%age Yield = 85.36 %
Answer:
Above 7
Explanation:
The equivalence point of any titration can be read off from the appropriate titration curve.
A titration curve is a plot of the pH of analyte against the volume of titrant added.
For a strong base and weak acid, the equivalence point lies above 7.
Answer:
a) Molarity KCl = 0.755 M
b) molality HNO = 5.09 m
Explanation:
- Formality (F) = moles sto / L sln
- Molarity (M) = # dissolved specie / L sln
- molality (m) = moles sto / Kg ste
- %p/p = ( g sto / g ste ) * 100
a) KCl ↔ K+ + Cl-
moles KCL:
⇒ 20 g KCl * ( mol / 74.6 g ) = 0.268 mol KCl
⇒ F = 0.268 mol KCl / 0.355 L = 0.755 F
⇒ M [ K+ ] = 1 * ( 0.755) = 0.755 M
b) 24% HNO:
calculation base: 1 g solution:
⇒24 = ( g sto / g sln) * 100
⇒ 0.24 = g sto / 1
⇒ g sto = 0.24g
⇒g sln = 1 - 0.24 = 076g sln
⇒Kg ste = 0.76 g * ( Kg / 1000g ) = 7.6 E-4 Kg ste
moles sto (HNO):
⇒ 0.24g * ( mol / 62.03g) = 3.869 E-3 moles HNO
⇒ m = 3.869 E-3 moles HNO / 7.6 E-4 Kg sln
⇒ m = 5.09 mol/Kg
