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Doss [256]
3 years ago
12

Find the pH of a solution whose H+ concentration is 1.8 x 10-12​

Chemistry
1 answer:
frutty [35]3 years ago
8 0

Answer:

11.74

Explanation:

The formula to find the pH of a solution is..

pH = -log [H⁺]

pH = -log [1.8×10¹²]

pH = 11.74

The answer would be to 2 decimal places because the concentration is two significant digits.

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How is crystallization used in the pharmaceutical industry (simple answer please)
Olegator [25]

Answer:

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6 0
3 years ago
Read 2 more answers
How many moles of NaCl must be dissolved in 0.5 L of water to make a 4 mole/L
kvv77 [185]
C = 4 mol/l
v = 0.5 l

n(NaCl)=cv

n(NaCl) = 4 mol/l · 0.5 l = 2 mol

2 moles of NaCl must be dissolved

5 0
3 years ago
Determine the mass of 9.2 x 10^18 molecules of dinitrogen tetroxide.
steposvetlana [31]

Answer: 1.4x10-3 g N2O4

Explanation: First convert molecules of N2O4 to moles using Avogadro's Number. Then convert moles to mass using the molar mass of N2O4.

9.2x10^18 molecules N2O4 x 1 mole N2O4 / 6.022x10²³ molecules N2O4

= 1.53x10-5 moles N2O4

1.53x10-5 moles N2O4 x 92 g N2O4/ 1 mole N2O4

= 1.4x10-3 g N2O4

4 0
3 years ago
An unknown metal displaces cadmium (Cd) from solution but does not displace chromium (Cr). Using the activity series, determine
adell [148]
I found this is the order of activity in Internet

Mn>Zn>Cr>Fe>Cd>Co>Ni>Pb

You can see that Fe is between Cr and Cd

Cr is more active than Fe and Fe is more active than Cd.

The Fe will displace Cd but not Cr.

Answer:Fe
3 0
3 years ago
Read 2 more answers
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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