The mass of calcium hydroxide that is formed when 10.0 g of CaO reacts with 10.0 g of water is 13.024 grams
calculation
from the equation
CaO + H2O → Ca(OH)2,
1 moles of CaO reacted with 1 moles of H2O to form 1 moles of Ca(OH)2
find the moles of each reactant
moles=mass/molar mass
moles of CaO= 10 g/56 g/mol=0.179 moles
moles of H2O = 10 g/18 g/mol 0.556 moles
CaO is the limiting reagent therefore by use of mole ratio of CaO:Ca(OH)2 which is 1:1 moles of Ca(OH)2 is = 0.179 moles
mass= moles x molar mass
= 0.176 moles x 74 g/mol = 13.024 grams
For the titration we use the equation,
M₁V₁ = M₂V₂
where M is molarity and V is volume. Substituting the known values,
(0.15 M)(43.2 mL) = (2)(M₂)(20.5 mL)
We multiply the right term by 2 because of the number of H+ in H2SO4. Calculating for M₂ will give us 0.158 M. Thus, the answer is approximately 0.16M.
Question: Liquids will boil at a higher temperature when they are at a lower elevation true or false ?
Answer: True
Answer:
E - Be and O
A - Mg and N
E - Li and Br
F - Ba and Cl
B - Rb and O
Explanation:
Be and O
Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).
Mg and N
Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).
Li and Br
Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).
Ba and Cl
Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).
Rb and O
Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).
