The chemical formula for the compound can be written as,
CxHyOz
where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
CxHyOz + O2 --> CO2 + H2O
number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C
number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O
This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H
Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g
Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O
The formula of the compound is,
C0.0163H0.01635O0.0109
Dividing the numbers by the least number,
C3/2H3/2O
The empirical formula of the compound is therefore,
<em> C₃H₃O₂</em>
Answer:
The correct answer is 0.300 * 10^23 ions.
Explanation:
Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.
The moles of zinc chloride (ZnCl2) is,
= mass of zinc + 2 mass of chlorine
= 65.38 + 2 (35.45)
=65.38 + 70.90
= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)
Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine
Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8
= 3.537 g of chlorine
70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537
= 0.300 * 10^23 ions of chlorine.
Answer:
A.0.20M
Explanation:
c 1 V 1 = c 2 V 2
Initial Volume, V1 = 200 mL
Final Volume, V2 = 200 + 400 = 600 mL
Initial Concentration, c1 = 0.60 M
Final Concentration, c2= ?
Solving for c2;
c2 = c1v1 / v2
c2 = 0.60 * 200 / 600
c2 = 0.20M
The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below
write the equation for the dissociation of Ba(OH)2
Ba(OH)2 = Ba^2+ +2OH^-
calculate the OH- concentration
by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M
by use of the formula ( H3O+)(OH-) = 1 x10 ^-14
by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-
substitute for OH-
H3O+ = (1 x10^-14 )/0.1
= 1 x10^-3 M