The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed
Answer:
$1246.90
Explanation:
Since the bike lost % of it's value and it now currently at $, we have to do 20% * $1039 to find the amount of money lost. 20%*1039=207.8. We have to add it up to find the original value so 1039+207.8=$1246.8
As it dissolves a large number of solid, liquid and gaseous substances it's called a universal solvent. It's important for living thing because all of them depend on water to survive. For example, many organisms live in the water and they find some of the nutrients that are necessary for live dissolved in water - gases included.
One molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.
<h3>What is an equation?</h3>
The term chemical equation has to do with the presentation of a chemical reaction on paper in a way that it can be easily understood. It is easy to write an equation to show what is going on in a reaction system.
Now we have the reactions as shown in the question. In this reaction which is the synthesis of ammonia and occurs industrially in the Haber process. The statement that is not true is that; one molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.
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There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87