Given:
7.20 g sample of Al2(SO4)3
Required:
Mass of oxygen
Solution:
Since you are not given a
chemical reaction, just base your solution to the chemical formula given.
Molar mass of Al2(SO4)3 = 342.15 g/mol
7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol
O2)(32g O2/1mol O2) = 4.04 g O2
<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
18g is the most reasonable mass after the reaction
Answer:
The absorbance of the solution is 0.21168.
Explanation:
Given that,
Wavelength = 500 nm
Molar absorptivity = 252 M⁻¹ cm⁻¹
Number of moles = 0.00140
Volume of solution = 500.0 mL
Length = 3.00 mm
We need to calculate the molar concentration
Using formula of the molar concentration

Where, N = number of moles
V = volume
Put the value into the formula


We need to calculate the absorbance of the solution
Using formula of absorbance

Put the value into the formula


Hence, The absorbance of the solution is 0.21168.
The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
= 0.2916 g
To get the empirical formula we first get the number of moles of each element;\
Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon : Hydrogen : Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996 : 3.995 : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O