Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Answer : The final concentration of
is, 2.9 M
Explanation :
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = 
t = time passed by the sample = 3.5 min
a = initial concentration of the reactant = 3.0 M
a - x = concentration left after decay process = ?
Now put all the given values in above equation, we get


Thus, the final concentration of
is, 2.9 M
The molar mass for PCL3 is 137.33 g/mol
4 moles of water are produced
Explanation:
- 4 moles of water are produced when 5 moles of hydrogen is reacted with 2 moles of oxygen gas
- The balanced equation given is when 2 moles of hydrogen reacts with 1 mole of oxygen and it forms 2 moles of water.
- The equation we have to solve is the 5 moles of hydrogen is reacting with 2 moles of oxygen gas, we can write the equation as
- This is the balanced equation when 5 moles of hydrogen reacts with 2 moles of oxygen. The balanced equation means the number of hydrogen atoms and oxygen atoms on both sides would be equal in number.
Answer:
47.2 g
Explanation:
Let's consider the following double displacement reaction.
3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl
The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:
44.3 g × (1 mol / 357.48 g) = 0.124 mol
The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:
3 × 0.124 mol = 0.372 mol
The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:
0.372 mol × (126.75 g/mol) = 47.2 g