<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:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
If the diatomic molecule consists of atoms from two different elements, then it is aheteronuclear diatomic molecule. There are seven elements that naturally occur as homonucleardiatomic molecules in their gaseous states: hydrogen,nitrogen, oxygen, fluorine,chlorine, bromine, and iodine
Ionic: transfer of electrons
Covalent: sharing of electrons
Metallic: sharing of free electrons in a structure of cations
Answer:
Concentration of nitrate in the new solution = 0.007 M
Explanation:
Given:
Concentration nitrate solution = 0.070 m
Volume of aliquote of the nitrate solution is add = 10.0 ml
Total volume = 100 ml
Find:
Concentration of nitrate in the new solution
Computation:
Number of M. mole = 0.070 m x 10.0 ml
Number of M. mole = 0.7 m-moles
Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml
Concentration of nitrate in the new solution = 0.007 M
