You have 3.50 l of solution that contains 90.0 g of sodium chloride, nacl. what is the molar mass of nacl? what is the molarity
1 answer:
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Answer:
See explanation and image attached
Explanation:
This reaction is known as mercuric ion catalyzed hydration of alkynes.
The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.
Answer:
1) Net ionic equation :
2) 0.765 M is the molarity of the carbonic acid solution.
Explanation:
1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:
..[1]
In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:
..[2]
In aqueous potassium carbonate , potassium ions and carbonate ion is present.:
..[3]
From one:[1] ,[2] and [3]:
Cancelling common ions on both sides to get net ionic equation :
2)
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
0.765 M is the molarity of the carbonic acid solution.
57.5 % is the percent yield if 10.9 g of H₃PO₄ is formed.
Explanation:
Balanced equation for the reaction:
Ca₃(PO₄)₂ (s) + 2H₂SO₄ (aq) → H₃PO₄ (aq) + 2CaSO₄ (aq)
data given:
mass of Ca₃(PO₄)₂ = 19.9 grams
mass of H₂SO₄, = 54.3 grams
mass of H₃PO₄ produced = 10.9 grams (actual yield)
percent yield=?
atomic mass of Ca₃(PO₄)₂ = 310.17 grams/mole
atomic mass of H₂SO₄ = 98.07 grams/mole
number of moles is calculated as:
number of moles =
putting the values in the above equation:
for Ca₃(PO₄)₂ =
= 0.064 moles
moles of H₂SO₄ =
= 0.55 moles
from the reaction, it can be found that the limiting reagent is Ca₃(PO₄)₂
1 mole of Ca₃(PO₄)₂ reacts to form 1 mole of phosphoric acid
0.064 moles of Ca₃(PO₄)₂ will produce x moles of phosphoric acid
0.064 moles of phosphoric acid produced
mass = number of moles x atomic mass of H3PO4
=0.064 x 97.994
= 6.27 grams (theoretical yield)
FORMULA FOR PERCENT YIELD:
percent yield = x 100
= x 100
= 57.5 %