Answer:
Atomic mass of Ag-107 = 106.94 amu
Explanation:
Let the mass of Ag-107 be y
Since the mass ratio of Ag-109/Ag-107 is 1.0187, the mass of Ag-109 is 1.0187 times heavier than the mass of Ag-107
Mass of Ag-109 = 1.0187y
Relative atomic mass of Ag = sum of (mass of each isotope * abundance)
Relative atomic mass of Ag = 107.87
Abundance of Ag-107 = 51.839% = 0.51839
Abundance of Ag-109 = 41.161% = 0.48161
107.87 = (y * 0.51839) + (1.018y * 0.48161)
107.87 = 0.51838y + 0.49027898y
107.87 = 1.00865898y
y = 107.87/1.00865898
y = 106.94 amu
Therefore, atomic mass of Ag-107 = 106.94 amu
Answer:
11.842g
Explanation:
To find mass the formula is volume times density.
Whether water is frozen, in a gaseous state, or is a liquid, it is still H2O. So the chemical composition does stay the same.
Expanding in powers 4FCsixteen.
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<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>
