Lose electrons - electrons want to fill their outer valence shell, so sometimes instead of gaining it is easier to lose some and have a filled outer shell
Answer:
The answer to your question is 177.8 g of NaOH
Explanation:
Balanced chemical reaction
2 Na + 2 H₂O ⇒ 2 NaOH + H₂
Data
Grams of H₂ = ?
Mass of Na = 120 g
Mass of H₂O = 80
Process
1.- Calculate the limiting reactant
Molar mass of Na = 2 x 23 = 46 g
Molar mass of H₂O = 2 x 18 = 36 g
Theoretical proportion Na/H₂O = 46/36 = 1.28
Experimental proportion Na/H₂O = 120/80 = 1.5
As the experimental proportion increases, we conclude that the limiting reactant is the water.
2.- Calculate the mass of hydrogen
Molar mass of NaOH = 2[23 + 16 + 1] = 80 g
36 g of H₂O ------------------ 80 g of NaOH
80 g of H₂O ----------------- x
x = (80 x 80) / 36
x = 6400 / 36
x = 177.8 g of NaOH
P1/T1 = P2/T2
125⁰C = 398.15 k
182⁰C = 455.15 k
1.22/398.15 = p2/455.15
p2= 1.39atm
the pressure of the gas be after the temperature change is 1.39 atm
Answer:
3.41 g
Explanation:
<em>A chemist adds 260.0 mL of a 0.0832 M potassium permanganate solution to a reaction flask. Calculate the mass in grams of potassium permanganate the chemist has added to the flask.</em>
Step 1: Given data
- Volume of the solution (V): 260.0 mL (0.2600 L)
- Molar concentration of the solution (C): 0.0832 M (0.0832 mol/L)
Step 2: Calculate the moles (n) of potassium permanganate added
We will use the following expression.
n = C × V
n = 0.0832 mol/L × 0.2600 L = 0.0216 mol
Step 3: Calculate the mass corresponding to 0.0216 moles of potassium permanganate
The molar mass of potassium permanganate is 158.03 g/mol.
0.0216 mol × 158.03 g/mol = 3.41 g