Answer:
0.478 M
Explanation:
Let's consider the neutralization reaction between KOH and H₂SO₄.
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O
12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:
0.0127 L × 1.50 mol/L = 0.0191 mol
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol
0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:
M = 0.0382 mol/0.0800 L = 0.478 M
Answer;
294.13 amu
Solution;
-293nv is 293.10 amu and that of 295nv is 295.45 amu
293.05+295.2=588.25
588.25/2= 294.13 amu
an amu of 294.13 using significant figures
Answer:
The coefficient in a balanced chemical equation indicates the mole ratio of both reactants and products.
Explanation:
For example lets consider the reation between Hydrogen and Oxygen to form water:
2H2 + O2 ----------------------- 2H2O
In this reaction, the coefficients of the balanced reaction can be transformed to Mole ratio according to Avogadro's Law which states that at standard temperature and pressure, equal volume of gases contain the same number of moles.
So the mole ratio for the above equation is the ratio of the coefficient:
2moles : 1 mole : 2 moles