A scientist hypothesizes that fertilizer containing 10 percent nitrogen will produce tomato plants with more leaves and blossoms
2 answers:
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Answer: a)
b)
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
a) Mass of Ba= 66.06 g
Mass of Cl = 34.0 g
Step 1 : convert given masses into moles.
Moles of Ba =
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ba =
For O =
The ratio of Ba: Cl= 1:2
Hence the empirical formula is
b) Mass of Bi= 80.38 g
Mass of O= 18.46 g
Mass of H = 1.16 g
Step 1 : convert given masses into moles.
Moles of Bi =
Moles of O=
Moles of H=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Bi=
For O =
For H=
The ratio of Bi: O: H= 1:3: 3
Hence the empirical formula is
Answer:- Mole fraction of methane in the original gas mixture is 0.854.
Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:
PV = nRT
V = 2.50 L
P = 1.45 atm
T = 20 + 273 = 293 K
Let's plug in the values in the equation:
n = 0.151
Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)
The combustion equations of methane and propane are:
From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.
From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.
So, total moles of carbon dioxide that we would get from methane and propane combustion are:
X + 3(0.151 - X)
From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.
moles of Carbon dioxide =
moles of carbon dioxide = 0.195 mol
Hence, 0.195 = X + 3(0.151 - X)
Let's solve this for X as:
0.195 = X + 0.453 - 3X
0.195 = 0.453 - 2X
2X = 0.453 - 0.195
2X = 0.258
X = 0.129
So, there are 0.129 moles of methane in the mixture.
moles of propane = 0.151 - 0.129 = 0.022
mole fraction of methane =
mole fraction of methane =
mole fraction of methane = 0.854
Hence, the mole fraction of methane gas in the original gas mixture is 0.854.