Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.
<h3>What is mass percentage</h3>
The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.
In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
<h3>Mass percentage of CaCO₃</h3>
In this case, you know:
- mass of CaCO₃: 2.62 grams
- mass of limestone: 3.82 grams
Replacing in the definition of mass percentage:
<u><em>mass percentage= 68.59 %</em></u>
Finally, the mass percentage of CaCO₃ is 68.59%.
Learn more about percentage by mass:
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Because the cold could freeze them, and they could get hypothermia.
Answer:
36.693
Explanation:
From the question given, the following were obtained:
Let 35Cl be isotope A and 37Cl be isotope
For isotope A (35Cl),
Mass number = 34.9700
Abundance = 13.85 %
For isotope B (37Cl),
Mass number = 36.9700
Abundance = 86.15 %
Atomic mass =
[( Mass of AxAbundance A)/100] + [(Mass of BxAbundance B)/100]
[(34.97x13.85)/100] + [(36.97x86.15)/100]
= 36.693
Answer:
Explanation:
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl = 195 g
Volume of solution = 4.05 L
Molarity of solution is
The molarity of the solution is
Moles of NaCl = 4.63 mol
Volume is given by
The volume of the solution that contains the required amount NaCl is
Volume of solution = 
Moles of solution is given by
The number of moles of NaCl is .
Answer:
pH = 7.233
Explanation:
Initially, the buffer contains 0.208 moles of NaHSO₃ and 0.134 moles of Na₂SO₃.
NaHSO₃ reacts with NaOH thus:
NaHSO₃ + NaOH → Na₂SO₃ + H₂O
50.0 mL of 1.00 M NaOH are:
0.0500L × (1mol / 1L) = 0.0500moles of NaOH added. That means after the addition are produced 0.0500moles of Na₂SO₃ and consumed 0.0500moles of NaHSO₃. That means final moles of the buffer are:
NaHSO₃: 0.208 mol - 0.050 mol = <em>0.158 mol</em>
Na₂SO₃: 0.134 mol + 0.050 mol = <em>0.184 mol</em>
<em> </em>
As pKa of this buffer is 7.167, it is possible to use H-H equation to find pH, thus:
pH = pKa + log₁₀ [Na₂SO₃] / [NaHSO₃]
pH = 7.167 + log₁₀ [0.184] / [0.158]
<em>pH = 7.233</em>
