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elena-14-01-66 [18.8K]
3 years ago
6

Titration of a 24.0 mL sample of acid rain required 1.8 mL of 0.0882 M NaOH to reach the end point. If we assume that the acidit

y of the rain is due to the presence of sulfuric acid, what was the concentration (in M) of sulfuric acid in this sample of rain
Chemistry
1 answer:
wolverine [178]3 years ago
3 0

Answer:

The concentration of sulfuric acid is 0.0033 M

Explanation:

Step 1: Data given

Volume of the sulfuric acid (H2SO4) = 24.0 mL = 0.024 L

Volume of NaOH = 1.8 mL = 0.0018 L

Molarity of NaOH = 0.0882 M

Step 2: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 3: Calculate the concentration of H2SO4

b*Ca*Va = a * Cb * Vb

⇒b = the coefficient of NaOH = 2

⇒Ca = the concentration of H2SO4 = TO BE DETERMINED

⇒Va = the volume of H2SO4 = 0.024 L

⇒a = the coefficient of H2SO4 = 1

⇒Cb = the concentration of NaOH = 0.0882 M

⇒Vb = the volume of NaOH = 0.0018 L

2* Ca * 0.024 = 1 * 0.0882 * 0.0018

0.048 * Ca =  0.00015876‬

Ca = 0.0033 M

The concentration of sulfuric acid is 0.0033 M

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Which compound are ionic and which are covalent? (N2) (CCl4) (SiO2) (AlCl3) (CaCl2) (LiBr)
muminat
Covalent compounds: N2, CCl4, SiO2 and AlCl3.

Ionic compounds: CaCl2 and LiBr.

Hope this helps!
3 0
3 years ago
A 20.00 ml sample of a solution of sr(oh)2 is titrated to the equivalence point with 40.03 ml of 0.1159 n hcl. what is the molar
goblinko [34]
The   molarity  of Sr(OH)2  solution is  =  0.1159 M

    calculation
write the equation  for reaction
that is,  Sr(OH)2 +2HCl→ SrCl2 + 2 H2O

then finds the mole  of HCl used

moles = molarity x volume 
=40.03 x0.1159 =  4.639 moles

by  use of mole ratio between Sr(OH)2 to  HCL which is 1 :2  the moles of Sr(OH)2  is therefore =  4.639  x1/2 = 2.312  moles

molarity  of  Sr(OH)2  is =  moles  / volume

=2.312 /20 =0.1159 M
3 0
3 years ago
How do I balance a chemical reaction like one of these NaBr + Ca(OH)2 ➡️ CaBr2 + NaOH
mojhsa [17]
2NaBr + Ca(OH)2 ➡️ CaBr2 + 2NaOH
3 0
3 years ago
if a television with mass 15kg is lifted to a shelf of height 1.2m, how much gravitational potential energy is added to the tele
belka [17]

P =mgh

You have mass, g =9.8 m/s2 and height calculate the potential energy P

6 0
3 years ago
Of the following solutions, which has the greatest buffering capacity?
Goshia [24]

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[Base]}{[Acid]} )

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. \frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36

b. \frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417

c. \frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868

d. \frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959

Therefore, the d. solution has the best buffering capacity.

Regards.

4 0
3 years ago
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