Mass of MnO2 = 25 g
The reaction would be 3MnO2 + 4Al --> 3Mn(s) + 2Al2O3
Molar mass of Al = 26.982 g/mol
Molar mass of MnO2 = 54.938 + 2(15.999) = 86.936 g/mol
Calculating the moles = 25 / 86.936 = 0.2876 mol.
Mole ratio MnO2 and Al considering the equation = 3 mol of MnO : 4 mol of Al
Calculating the moles of Al = 0.2876 mol MnO2 x (4 mol of Al / 3 mol of MnO)
Number of moles of Al = 0.3834
Getting the mass in grams as asked = 0.3834 mol x 26.982 g/mol = 10.34 grams.
Because 1,2-diaminoethane is base, this compound gain protons from acid (in this example hydrochloric acid) and becaume ionic compound with positive charge (cation).
1,2-diaminoethane is organic compound and hydrochloric acid is inorganic compound.<span>
Balance chemical reaction:
</span>H₂NCH₂CH₂NH₂(aq) + 2HCl(aq) → ⁺H₃NCH₂CH₂NH₃⁺(aq) + 2Cl⁻(aq).
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:
T2 = 133.333°K
Explanation:
Using Combined Gas Laws:
(600 torr)(10L)/500°K = (200 torr)(8L)/x°K
Cross multiply:
x°K (600 torr)(10L) = 500°K(200 torr)(8L)
Divide:
x°K = (500°K(200 torr)(8L))/(600 torr)(10L)
x = 400/3°K or 133.333°K
