Answer:the initial composition of the reactants is
40cm^3 of CH4
40cm^3of H2
100cm^3 of H2O
Explanation:
Balanced reaction is
CH4 +H2+5/2O2______
CO2 +3H2O
Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases
So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.
Since CO2 in the equation is 1 mole
Means 1mole represent 40cm^3
So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.
The molality of a 0.677 m ethanol ( c 2 h 5 o h ) solution whose density is 0.588 g/ml.
What is molality and molarity?
The number of solvent moles per kilogram is known as molality. Because the mass of the solute and solvent in the solution remains constant, molality is the preferred concentration transfer method.
Molarity, also referred to as molar concentration, is the quantity of a material expressed as moles per litre of solution. Solutions marked with a molar concentration have a capital M. One mole of solvent per litre is present in a 1.0 M solution.
Molar mass (w) of C2H5OH = 46g
Molarity(M) = 0.677M
Density(D) = 0.588g/mol
<u>Molality(m)</u><u> = M/D * (1000 + w*M)</u>
<u>= 1187.2m</u>
To learn more about molality and molarity from the given link below,
brainly.com/question/14770448
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Answer:
See explanation below
Explanation:
The question is incomplete. However, here's the missing part of the question:
<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>
<em>C(s) + 2H2(g) <--> CH4(g). </em>
<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>
With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.
The expression of Kp for this reaction is:
Kp = PpCH4 / (PpH2)²
We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:
PpCH4 = Kp * PpH2²
*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.
Now solving for PpCH4:
PpCH4 = 0.455 * (1.78)²
<u><em>PpCH4 = 1.44 atm</em></u>
Given Data:
P₁ = 122 atm
P₂ = 112 atm
V₁ = 113 L
V₂ = ?
Solution:
Let suppose the gas is acting ideally. According to Ideal gas equation, keeping the temperature constant,
P₁ V₁ = P₂ V₂
Solving for V₂,
V₂ = P₁ V₁ / P₂
Putting Values,
V₂ = (122 atm × 113 L) ÷ 112 atm
V₂ = 113 L
Converting Volume to Moles,
As,
22.4 L is occupied by = 1 mole of He gas
So,
113 L will occupy = X moles of He gas
Solving for X,
X = (113 L × 1 mole) ÷ 22.4 L
X = 5.04 Moles of He
