Answer:
0.312 moles of H2O
Explanation:
no. of moles of ch4= mass ÷ molar mass
=2.5 ÷ 16.04
=0.156 moles of ch4
According to balanced chemical equation
CH4 : H2O
1 mole : 2 moles
0.156 moles : x moles
by cross multiplication
x= (0.156x2) ÷ 1
= 0.312 moles of H2O
Answer:
The correct answer is 146 g/mol
Explanation:
<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
<em>Therefore, the molar mass of the compound is 146 g/mol </em>
Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
Atomic weight is actually calculated by the sum of protons and neutrons of that atom. It is not equal to the number. You must add them for the result
Answer:
A. The reaction will proceed forward forming more CH4
B. The reaction will proceed forward forming more CH4
C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.
D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4
E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.
F. The following will favour CH4 at equilibrium
i. Catalyst to the reaction mixture,
ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture
iii. Adding more C to the reaction mixture.
