Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
9/4 Be +2 (the 9 and 4 are stacked next to Be). Atomic #: 4
Mass #: 9, # protons: 4, # neutrons: 5, #electrons: 2.
31/15 P (31 is stacked over 15 next to the P). Atomic #: 15,
Mass #: 31, # protons: 15, # neutrons: 16, # electrons: 15.
(g solute/g solution)*100 = % mass/mass
30 g / 400 * 100
0,075 * 100
= 7,5% w/w
hope this helps!
