Answer:
The mass percentage of CaCO3(s) in the sample is 83.4%
Explanation:
CaCO₃ + 2HCl → CaCl₂ + H2O + CO₂
Let's find out the moles of each reactant:
[HCl] = 0.150 mol/L
Molarity . volume = moles
0.150 m/L . 0.050L = 7.5x10⁻³ moles HCl
Molar mass of CaCO₃ =100.08 g
Moles of CaCO₃ = Mass / Molar mass
0.450 g/100.08 g = 4.49x10⁻³ moles of salt
Ratio is 1:2
If 2 moles of HCl react with 1 mol of salt
7.5x10⁻³ moles of HCl react with 7.5x10⁻³ / 2 = 3.75x10⁻³ moles
We must find out the mass
Moles of CaCO₃ . Molar mass CO₃ = 0.3753 g
So now, the mass percentage of salt will be
0.450g ___ 100%
0.3753 g ____ (0.3753 .100 )/0.450 =83.4 %
Answer:
4121 years
Explanation:
From;
0.693/t1/2 = 2.303/t log No/N
t1/2= half life of the carbon-14
No= count rate of the living tissue
N= count rate of the sample
t = age of the sample
0.693/5730 =2.303/t log (13.5/8.2)
1.21 * 10^-4 = 2.303/t * 0.2165
1.21 * 10^-4 = 0.4986/t
t = 0.4986/1.21 * 10^-4
t = 4121 years
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
The width of the cuvette is important in the spectrophotometric analysis since wider cuvette<span> translates to more absorbing species present in the path where light passes, hence more absorbance is read in the analysis. other factors include the concentration of the sample and the species itself present.</span>
Answer:
Explanation:
False
Because " like dissolve like " polar solute dissolve in polar solvent and vice versa
