<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>
Answer: 3.69 × 10^27
Explanation:
Amount of energy required = 7.06 × 10^4 J
Frequency of microwave (f) = 2.88 × 10^10 s−1
Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum
Recall ;
Energy of photon = hf
Therefore, energy of photon :
(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1
= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J
Hence, number of quanta required :
(7.06 × 10^4)J / (19.0944 × 10^-24)J
= 0.369 × 10^(4 + 24) = 0.369×10^28
= 3.69 × 10^27
Answer:
Write down the phases and what happens during each phase.
Explanation:
Answer:
0.45 moles
Explanation:
The computation of the number of moles left in the cylinder is shown below:
As we know that
we can say that
where,
n1 = 1.80 moles of gas
V2 = 12.0 L
And, the V1 = 48.0 L
Now placing these values to the above formula
So, the moles of gas in n2 left is
= 0.45 moles
We simply applied the above formulas so that the n2 moles of gas could arrive
Answer:
Mass of water = 73.08 g
Explanation:
Given data:
Mass of hydrogen = 35 g
Mass of oxygen = 65 g
Mass of water = ?
Solution:
First of all we will write the balanced chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass/ molar mass
Number of moles of hydrogen = 35 g/ 2 g/mol
Number of moles of hydrogen = 17.5 mol
Number of moles of oxygen = 65 g / 32 g/mol
Number of moles of oxygen = 2.03 moles
Now we compare the moles of water with moles hydrogen and oxygen.
H₂ : H₂O
2 : 2
17.5 : 17.5
O₂ : H₂O
1 : 2
2.03 : 2× 2.03 =4.06 mol
Number of moles of water produced by oxygen are less so oxygen is limitting reactant.
Mass of water:
Mass of water = number of moles × molar mass
Mass of water = 4.06 mol × 18 g/mol
Mass of water = 73.08 g
