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antiseptic1488 [7]
3 years ago
8

Data was collected by students in an acid base titration lab. They used 1.63 M Ca(OH)2(AQ)

Chemistry
1 answer:
jenyasd209 [6]3 years ago
6 0

Answer:

3.8 M

Explanation:

Volume of acid used VA= 57.0 - 37.5 = 19.5 ml

Volume of base used VB= 67.8 - 45.0 = 22.8 ml

Equation of the reaction

2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)

Number of moles of acid NA= 2

Number of moles of base NB= 1

Concentration of acid CA= ???

Concentration of base CB= 1.63 M

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA= CBVBNA/VANB

CA= 1.63 × 22.8 × 2/ 19.5 × 1

CA= 3.8 M

HENCE THE MOLARITY OF THE ACID IS 3.8 M.

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4 years ago
What mass of NaCl is dissolved in 150 g of water in a .050<br> msolution?
mart [117]

Answer:

0.4383 g

Explanation:

Molality is defined as the moles of the solute present in 1 kg of the solvent.

It is represented by 'm'.

Thus,  

Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}

Given that:

Mass of solvent, water = 150 g = 0.15 kg ( 1 g = 0.001 g )

Molality = 0.050 m

So,

0.050=\frac {Moles\ of\ the\ solute}{0.15}

Moles = 0.050\times 0.15\ mol= 0.0075\ mol

Molar mass of NaCl = 58.44 g/mol

Mass = Moles*Molar mass = 0.0075\times 58.44\ g = 0.4383 g

6 0
4 years ago
PLEASE HELP ME FAST Name the parts of the distillation apparatus set-up. 2. 3. 5. 70 6. 10 7 8. 15 14 13. 10 09 13​
guapka [62]

Answer:

2. Distillation Flask

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8 0
3 years ago
How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh? h3po4 + koh yields h2o + k3po4 be sur
schepotkina [342]

the balanced chemical equation for the reaction is as follows

H₃PO₄ + 3KOH ---> K₃PO₄ + 3H₂O

stoichiometry of H₃PO₄ to KOH is 1:3

first we have to find which the limiiting reactant is

as the amount of product formed depends on the amount of limiting reactant present

number of H₃PO₄ moles reacted - 4.4 mol

if H₃PO₄ is the limiting reactant

1 mol of H₃PO₄ reacts with 3 mol of KOH

then 4.4 mol of H₃PO₄ reacts with - 3 x 4.4 mol = 13.2 mol of KOH

but only 3.8 mol of KOH is present

therefore KOH is the limiting reactant


stoichiometry of KOH to K₃PO₄ is 3:1

number of KOH moles reacted - 3.8 mol

therefore number of K₃PO₄ formed = number of KOH moles reacted / 3

= 3.8 mol / 3 = 1.3 mol


answer is 1.3 mol of K₃PO₄


3 0
4 years ago
A scientist has 8.5 g of silver nitrate and needs to prepare 1.5 L of a 0.020 M solution. Will there be enough silver nitrate? I
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