Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given solution is as follows.
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
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Answer:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
Explanation:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
The element Sodium (Na) has 11 protons and 1 valence electron.
4p, 3d, 3p, 3s, 2p, 2s and 1s orbitals may be occupied during de-excitation.<span />