<span>Answer is: Van't Hoff factor
(i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT
=i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.
i = 1.63°C ÷ (0.512°C/m · 1.26 m).
i = 1.051.
C6H6 is the
most insoluble in water as it is non polar in nature. HCL is highly soluble in
water as it is an electrolyte. Kbr is also highly soluble in water. NaCl
dissolves in water very quickly. CH3CH3OH is also totally soluble in water
because of the Hydrogen bonding formation. It is a well known fact that the
hydrocarbons do not easily mix with water. C6H6 being a strong hydrocarbon does
not mix with water. So among NaCl, KBr, CH3CH3OH, HCl and C6H6, the hydrocarbon
is the least soluble in water.
8 moles of H 2O are produced.
First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.
<h3>How many moles of H2 are required to make one mole of NH3?</h3>
Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.
<h3>
How many moles of h2 are needed to produce a solution?</h3>
An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).
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Boyle's law gives the relationship between pressure and volume of gas.
for a fixed amount of gas, at constant temperature the pressure of gas is inversely proportional to its volume.
PV = k
where P - pressure, V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side.
substituting the values in the equation
1.2 atm x 1.0 L = P x 4.0 L
P = 0.3 atm
new pressure is 0.3 atm
