Aspirin is usually packaged with

Why do the balls of the Newton’s cradle eventually stop?

Artemon [7]

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8 0

3 years ago

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When 25.0 mL of sulfuric acid solution was completely neutralized in a titration with 0.05 M NaOH solution, it took 18.3 mL of t

ladessa [460]

Balanced equation will be as follows

2NaOH + H2SO4=Na2SO4 + 2H2O
from MaVa/MbVb=Na/Nb where Ma is molarity of acid=?
Va is volume of acid=25.0 mL
Mb is molarity of base= 0.05 M
Vb is volume of base=18.3 mL
Nb number of moles of base= 2
Na is number of moles of acid =1

Ma*25 mL / 18.3mL*0.05M = 1/2
Ma=18.3*0.05M / 50 = 0.0183M

The molarity of sulphuric acid = 0.0183M

8 0

3 years ago

a client is to receive 2.00g of a drug. The drug comes on a solution that has 500 mg per 5.0 mL. Each vial contains 10 millilite

Gekata [30.6K]

Maybe he shouldn’t be using drugs.

7 0

3 years ago

How much 5.50M NaOH must be added to 680.0 mL of a buffer that is 0.0215 M acetic acid and 0.0270 M sodium acetate to raise the

Lelu [443]

5.50M NaOH must be added to 680.0 mL of a buffer, The amount of NaOH is mathematically given as

V=2.107mL

<h3>What is the volume of 5.50M NaOH?</h3>

Generally, the equation for Chemical Reaction is mathematically given as

CH3COOH+OH^{-} ⇄ CHCOO^{-} + H2O

Therefore

pka=-log(1.8*10{-5})

pKa=5-0.25527

pKa=4.3447

Hence

$pH=pKa+log\frac{CH3COO}{CH3COOH}\\\\0.75=4.7447+log\frac{0.01564+x}{0.01428-x}$

x=0.0115898

Volume of NaOH

V=0.0021072L

V=2.107mL