Archive for October 9th, 2008
Chemistry Tutor Help - Chemistry Problem
If the mass of electrons = 9.1093897.10-31 kg, protons = 1.6726430.10-27 kg, and neutrons = 1.674954.10-27 kg. Calculate the mass of a single atom Carbon C-12?
Chemistry Tutor Problem Solving
From the periodic table of atoms we know that carbon-12 has 12 electrons, 12 protons, and 6 neutrons thus to find the mass of a single carbon atom we just calculate the masses of total species that build carbon atom. thus
Mass of 1 atom C
= (12x mass of e-) + (12x mass of protons) + (6x mass of n)
= (12 x 9.1093897.10-31 kg ) + (12 x 1.6726430.10-27 kg ) + (6×1.674954.10-27 kg)
= 3.0.10-26 kg
we can compare from the other way, 1 mole of Carbon has the mass of 12.01 g, since 1 mole of atom has 6.022.1023 molecules thus:
mass of 1 atom C
= 12.01 /6.022.1023
= 2.10-23 g
= 2.10-26 Kg
October 9th, 2008
Chemistry Tutor Help - Chemistry Problem
A sample of air contains 74.25% nitrogen, 24.90% oxygen, 0.05% carbon dioxide, and 0.80% argon, by volume. How many molecules of each gas are present in 1.00 L of the sample at 25°C and 1.00 atm?
The composition of each gas in the volume of 1 L are obtain by multiply the percentage and the volume
Chemistry Tutor Help - Chemistry Problem Solving
This is a simple question, why because only need a simple formula that involved gases. Use the “ideal gas law” to solve this problem, from it you would find the moles and from the moles you can find the number of molecules in each gas.
Do you know that we only need 3 variable of the state of the gas to know the fourth variable, according of this problem we have had, temperature, pressure, and the volume thus we can calculate the moles.
Volume of each gas can be calculate by multiply the percentage with volume of the sample that is:
Volume gas = % x Volume of sample
from the formula above we get each gas volumes
N2 = 0.7425 L
O2 = 0.2490 L
CO2 = 5.10exp-4 L
Ar = 8.10exp-3 L
Next use the formula of ideal gas to find the mole of each gas p is the pressure in atm, V is the volume in liter, T is the temperature in Kelvin, and R is the gas constant 0.08206 and the ideal gas formula is:
PV = nRT solve for the “n”
n = PV/RT
We get:
moles of N2 = 0.030 mol
O2 = 0.010 mol
CO2 = 2.05.10exp-5 mol
Ar = 3.271.10exp-4 mol
The number of molecules in each gas are obtain by:
Number of molecules = N x mole
N is the avogadro number that is 6.022.10exp23
thus
N2 = 1.81.10exp23 molecules
O2 =6.022.10exp21molecules
CO2 =1.235.10exp19 molecules
Ar = 1.98.10exp20 molecules
October 9th, 2008
Chemistry Help - Chemistry Problem
A solid White sample of Zn(OH)2 is added to 0.300 L of 0.420 M HBr. solution. The solution that remains is still acidic and It is then titrated with 0.420 M NaOH solution, and it takes 94.5 mL of the NaOH solution to reach the equivalence point.
What mass of Zn(OH)2 was added to the HBr solution?
Chemistry Tutor Answer
To solve this problem you only need to find how many moles of HBr that reacted with Zinc hydroxide Zn(OH)2 How you will find it? First you must find the initial moles of HBr and than subtracts with the moles of HBr that reacted with natrium hydroxide.
After that change the mol of HBr into moles of Zn(OH)2 using the coefficient of the reaction and multiply with the molecular mass to obtain the mass of zinc hydroxide. here are the problem solving of that chemistry problem.
moles of HBr initial
= 0.300 L x 0.420
= 0.126 mol
mol of NaOH
= 0.420 M x 94.5 mL
= 39.69 mmol
= 0.039 mol
NaOH reacted with HBr according this reaction, remember to find the mol of HBr you only need to compare the reaction coefficient since the coefficient of HBr and NaOH are equal thus their moles would be the same.
NaOH + HBr -> NaBr + H2O
from the reaction the mol of HBr are
= 1/1 x 0.039 mol
= 0.039 mol
thus the mol of HBr that reacted with Zn(OH)2 are
= Mol HBr initial - mol HBr that reacted with NaOH
= 0.126 mol - 0.039 mol
= 0.087 mol
Zn(OH)2 reacted with HBr according this reaction to give ZnBr2 and H2O, remember to find the mol of Zn(OH)2 you only need to compare the reaction coefficient since the coefficient of HBr and Zn(OH)2 are 2:1 thus the moles of Zn(OH)2.
Zn(OH)2 + 2 HBr -> ZnBr2 + 2 H2O
from the reaction we can find moles of Zn(OH)2
= 1/2 x moles of HBr
= 1/2 x 0.087 mol
= 0.0435 mol
and the mass
= mol x mr
= 0.0435 mol x 99.396
= 4.324 g
October 9th, 2008