Answer:
1
[tex]k = 32.5 \% [/tex]
2
[tex]z =63.56 \%[/tex]
3
[tex]m_t = 3.40 \ g[/tex]
4
[tex]j = 96.05 \%[/tex]
5
[tex]e = 3.954 \%[/tex]
Explanation:
From the question we are told that
The mass of the sample is [tex]m = 3.54 \ g[/tex]
The mass of the AgNO3 is [tex]m_s = 1.15 g[/tex]
The mass of Mg(OH)2 is [tex] m_n= 2.25 g[/tex]
Generally the percentage of AgNO3 in the mixture is
[tex]k = \frac{1.15}{3.54} * 100[/tex]
=> [tex]k = 32.5 \% [/tex]
Generally the percentage of Mg(OH)2 in the mixture is
[tex]z = \frac{m_n}{m} * 100[/tex]
=> [tex]z = \frac{2.25}{3.54} * 100[/tex]
=> [tex]z =63.56 \%[/tex]
Generally the total mass of the AgNO3 and Mg(OH)2 recovered is
[tex][tex]m_t = m_s +m_n[/tex]
=> [tex]m_t = 1.15 +2.25[/tex]
=> [tex]m_t = 3.40 \ g[/tex]
Generally the percentage recovery of the components is mathematically represented as
[tex]j = \frac{m_t}{m} * 100[/tex]
=> [tex]j = \frac{3.40}{3.54} * 100[/tex]
=> [tex]j = 96.05 \%[/tex]
Generally the percentage error for the separation of the components of the mixture is mathematically represented as
[tex]e = \frac{m- m_t}{m} * 100[/tex]
=> [tex]e = \frac{3.54 - 3.40}{3.54} * 100[/tex]
=> [tex]e = 3.954 \%[/tex]
