Answer:
0.200 m K3PO3
Explanation:
Let us remember that the freezing point depression is obtained from the formula;
ΔTf = Kf m i
Where;
Kf = freezing point constant
m = molality
i = Van't Hoff factor
The Van't Hoff factor has to do with the number of particles in solution. Let us consider the Van't Hoff factor for each specie.
0.200 m HOCH2CH2OH - 1
0.200 m Ba(NO3)2 - 3
0.200 m K3PO3 - 4
0.200 m Ca(CIO4)2 - 3
Hence, 0.200 m K3PO3 has the greatest van't Hoff factor and consequently the greatest freezing point depression.
I would believe the answer is:
.9211 ATM
9.51 L
a. 2683.168 g
b. 2683.168 g
c. 663.468 g
<h3>Further explanation</h3>
Given
Compounds
Required
The mass
Solution
Mass = mol x MW
a) P4010
(MW=283.9 g mol−1)
mass = 6.15 x 283.9 g mol−1
mass = 1745.985 g
b) AgC2H302
(MW=436.2875 g/mol)
mass = 6.15 x 436.2875 g/mol
mass = 2683.168 g
c) Fe(CN)2 (MW=107.881 g/mol)
mass = 6.15 x 107.881 g/mol
mass = 663.468 g
The statement that is true among the given statements is ΔH1 = 2ΔH2. The correct option is d. ΔH1 = 2ΔH2
From the question,
We are to determine which of the given statements is true.
From the given reactions,
We have
1. N2 (g) + O2 (g) ----> 2NO (g) ΔH1
and
2. 1/2 N2 (g) + 1/2 O2 (g) ---> NO (g) ΔH2
ΔH1 is the enthalpy change for reaction 1
and
ΔH2 is the enthalpy change for reaction 2
Enthalpy is an extensive property. The enthalpy change for a given reaction depends on the stoichiometry of the reaction.
From the given equations,
We can observe that the stoichiometry of reaction 1 is twice that of reaction 2.
Therefore, the enthalpy of reaction 1 will also be twice the enthalpy of reaction 2
That is,
ΔH1 = 2ΔH2
Hence, the statement that is true among the given statements is ΔH1 = 2ΔH2. The correct option is d. ΔH1 = 2ΔH2
Learn more here: brainly.com/question/15516625
According to <span>Le Chatelier's Principle," when a system at equilibrium is subjected to any external stress by changing concentration, pressure or temperature, it will modify itself in such a way so as to minimize the effect of that change made in it".
Change in Concentration:
Suppose following reaction at equilibrium,
A + B </span>⇆ C
When the concentration of reactants (A or B) is increased the equilibrium will shift in forward direction and will produce more product in order to cancel out the effect of extra reactant added. Or, if product (C) is added then it will respond in opposite direction.
Change in Pressure:
Suppose following reaction at equilibrium,
A + B ⇆ C
If pressure is applied on this equilibrium, the equilibrium will shift in forward direction because there is a decrease in volume at product side (1 mole as compare to reactants which are 2 moles). And the opposite case will be observed if the pressure is decreases.
Change in Temperature:
Suppose following reaction at equilibrium,
A + B ⇆ C + Heat
This is an exothermic reaction (heat is evolved) and the heat produced is in the product side. When the temperature is increased the equilibrium will shift in the backward direction. And decrease in temperature will shift the equilibrium in forward direction.