Sodium metal and water form aqueous sodium hydroxide and hydrogen gas:
Na(s) + 2H2O(l) → (Na+)(aq) + 2(OH-)(aq) + H2(g)
Looks like your entries didn't translate well on screen, so find the solution with the coefficients and ions which match this one.
Answer:
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Explanation:
The answer would be:
A. Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.
In this question, there are two half-reaction equations. To merge them up, you need to add the reactant with the reactant, then the product with the product. If there is a molecule on both side, you can cancel them. The full reaction would be:
C+ 1/2 O2 + CO + 1/2O2 ==>CO+ CO2 -----> remove CO from both side
C+ O2 ==>CO2
Hmmmm i understand but i wanna know what to measure or whats about?
Answer:
V = 22.34 L
Explanation:
Given data:
Volume of O₂ needed = ?
Temperature and pressure = standard
Number of molecules of water produced = 6.0× 10²³
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of water:
1 mole contain 6.022× 10²³ molecules
6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules
0.99 mole
Now we will compare the moles of oxygen and water.
H₂O : O₂
2 : 1
0.996 : 0.996
Volume of oxygen needed:
PV = nRT
V = nRT/P
V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm
V = 22.34 L
