Answer:
The value is [tex]L = 0.985 \pm 0.00801 \ g[/tex]
Explanation:
From the question we are told that
The molar mass of [tex]CaCO_3[/tex] is [tex]MW = 100.085 \ g/mol[/tex]
The total mass is [tex]m_g = 12.3 \ g[/tex]
The uncertainty of the total mass is [tex]\Delta g = 0.1[/tex]
Generally the molar weight of calcium is [tex]M_c = 40 g/mol[/tex]
The percentage of calcium in calcite is mathematically represented as
[tex]C = \frac{40.07}{100.085} * 100[/tex]
[tex]C = 40.03 \% [/tex]
Generally the mass of each sample is mathematically represented as
[tex]m= \frac{m_g}{5}[/tex]
[tex]m= \frac{12.3}{5}[/tex]
[tex]m= 2.46 \ g [/tex]
Generally mass of calcium present in a single sample is mathematically represented as
[tex]m_c = 2.46 * \frac{40.04}{100}[/tex]
[tex]m_c = 0.985 \ g [/tex]
The uncertainty of mass of a single sample is mathematically represented as
[tex]k = \frac{\Delta g }{5}[/tex]
[tex]k = \frac{0.1 }{5}[/tex]
[tex]k = 0.02\ g [/tex]
The uncertainty of mass of calcium in a single sample is mathematically represent
[tex]G = \frac{0.02 * 40.04}{ 100}[/tex]
[tex]G = 0.00801 \ g [/tex]
Generally the average mass of calcium in each sample is
[tex]L = 0.985 \pm 0.00801[/tex]
