<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
0.447 is the mole fraction of Nitrogen in this mixture.
mole fraction of nitrogen= moles of nitrogen/total moles
mole fraction of nitrogen=0.85/1.90
mole fraction of nitrogen=0.447
The product of the moles of a component and the total moles of the solution yields a mole fraction, which is a unit of concentration measurement. Because it is a ratio, mole fraction is a unitless statement. The sum of the components of the mole fraction of a solution is one. In a mixture of 1 mol benzene, 2 mol carbon tetrachloride, and 7 mol acetone, the mole fraction of the acetone is 0.7. This is computed by dividing the sum of the moles of acetone in the solution by the total number of moles of the solution's constituents:
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Answer:
The answer is
<h2>0.89 atm </h2>
Explanation:
To convert from kPa to atm we use the conversion
101.325 kPa = 1 atm
If
101.325 kPa = 1 atm
Then
90.23 kPa will be
We have the final answer as
<h3>0.89 atm</h3>
Hope this helps you
This element is found in group 3A, period 3
<h3>Further explanation
</h3>
The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)
-
K shell (n = 1) maximum 2 x 1² = 2 electrons
- L shell (n = 2) maximum 2 x 2² = 8 electrons
- M shell (n = 3) maximum 2 x 3² = 18 electrons
- N shell (n = 4) maximum 2 x 4² = 32 electrons
Electron configuration of element X : 2.8.3 , so :
K shell = 2 ⇒1s²
L shell = 8⇒2s²2p⁶
M shell = 3⇒ 3s²3p¹
Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids
The outer shell 3s²3p¹ : located in group 3A and period 3
group⇒valence electron ⇒3
period⇒the greatest value of the quantum number n⇒3
Answer:
D. The rate decreases as reactants are used up.
Explanation:
Initially, the rate increases until the reaction is at equilibrium. At equilibrium, the rate is constant.
As the reaction progresses, the rate decreases to zero when reactants are used up ( for irriversible reactions only )
