Answer:
corbonization for dictionnal
Explanation:
correct
Answer:
3%
Explanation:
Substract the actual error from the final and multiply by 100
Answer:
Mass of sodium chloride decomposed = 24.54 g
Explanation:
Given data:
Mass of sodium chloride decomposed = ?
Mass of chlorine gas formed = 15 g
Solution:
Chemical equation:
2NaCl → 2Na + Cl₂
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 15 g/ 71 g/mol
Number of moles = 0.21 mol
Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.
Cl₂ : NaCl
1 : 2
0.21 : 2×0.21 = 0.42 mol
Mass of Sodium chloride decompose:
Mass = number of moles × molar mass
Mass = 0.42 mol × 58.44 g/mol
Mass = 24.54 g
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
