Explanation:
b. What useful functions do oxidation numbers serve?
It is used to show oxidation and reduction (loss and gain of electrons)
b. How many molecules are in 1 mole of molecules?
1 mole = 6.022 * 10^23 molecules
c. What is the name given to the number of molecules in 1 mole?
Avogadro's Number of molecules
21. a. What is the molar mass of an element?
This is the mass of an element divided by the number of moles.
Molar mass = Mass / Number of moles
b. Write the molar mass rounded to two decimal places of carbon, neon, iron and uranium.
amu = Atomic Mass Unit
Carbon = 12.01 amu
Neon = 20.18 amu
Iron = 55.85 amu
Uranium = 238.03 amu
The variable that is measured as data in an experiment is 1. the dependent variable.
The variables that are held constant in an experiment are 2. the controlled variables.
The variable that is changed by the experimenter is 5. the independent variable.
A count or measurement recorded during an experiment is 4. quantitative data.
Descriptions or observations during an experiment are 3. qualitative data.
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
Answer:
0.0258 mol <em>Answer</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is 
.
Explanation:
The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.
Now, in terms of chemical formulae this reaction equation will be as follows.
Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply HCl by 4 on reactant side and multiply 
by 2 on product side. Therefore, the equation can be rewritten as follows.
Hence, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.
Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is .
