Hey there!:
8) ΔTb = i*Kb*m
m is molality
Since same number of mol is added to same amount of water in both cases
m will be same for both
is 1 for glucose since it is covalent compound
is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻
So, ΔTb will be 4 times in aluminum nitrate case
So, boiling point will change by 4ºC
9) use Q = m* L
L = heat of vaporization so:
T1=T2=100ºC
5.40 * 1000 => 5400 cal/g
Q = 5400 / 540
Q = 10 grams
Hope that thlps!
Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:
Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL
First we have to calculate the heat gained by the solution in coffee-cup calorimeter.
where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 2.242 kJ
n = number of moles fructose = 
Therefore, the enthalpy change during the reaction is -199. kJ/mol.
Answer:
6.32 moles of Fe
Explanation:
The given chemical equation is presented as follows;
2Fe + 3Cl₂ → 2FeCl₃
The mass of Cl₂ in the reaction = 336 grams
The molar mass of chlorine gas Cl₂ = 35.435 g/mol
The number of moles, n = Mass/(Molar mass)
The number of moles of Cl₂ in the reaction, n = 336 g/(35.435 g/mol) ≈ 9.842 moles
From the given reaction, 3 moles of Cl₂ react with 2 moles of Fe to produce 2 moles of FeCl₃
By the law of definite proportions, we have that 9.482 moles of Cl₂ will react with approximately 9.482 × 2/3 = 6.32 moles of Fe to produce approximately 6.32 moles of FeCl₃
Therefore, approximately 6.32 moles of Fe will be required to react with 336 grams of Cl₂.
Answer:
Group A elements or the main group
Another name for the representative elements are the Group A elements or the main group elements
Explanation:
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
