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3 years ago

What is the pOH of a base with a OH- concentration of 0.0500 x 10-11 M? PLEASEEEEEE ASAP

Chemistry

1 answer:

Ksivusya [100]3 years ago

4 0

Answer:

poH= -log(oh^-)

= - log (0.500* 10^-11)

= =- log 0.5 + log (10^-11)

= -( -0.30 +(-11))

=+ 11.3

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2 years ago

Read 2 more answers

Neon lamps, is composed of 90.483% of 20ne, 0.271% of 21ne, and 9.253% of 22ne. the 20ne atoms have a mass of 19.992 u, 21ne ato

WITCHER [35]

20.181 u

The average atomic mass of Ne is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).

Thus,

Avg. at. mass

= (0.904 83× 19.992 u) + (0.002 71 × 20.994) + (0.092 53× 21.991 u)

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3 years ago

The solubility of a compound with a molecular weight of 233.2 is determined to be 27.9 g in 125.0 g of water at 398 K. Express t

sp2606 [1]

Answer:

0.960 m

Explanation:

Given data

Mass of the solute: 27.9 g

Molar mass of the solute: 233.2 g/mol

Mass of the solvent: 125.0 g = 0.1250 kg

First, we will calculate the moles of solute.

27.9 g × (1 mol/233.2 g) = 0.120 mol

The molality of the compound is:

m = moles of solute / kilograms of solvent

m = 0.120 mol / 0.1250 kg

m = 0.960 m

5 0

3 years ago

Find the pH. What are the pH values for the following solutions? (a) 0.1 M HCl (b) 0.1 M NaOH (c) 0.05 M HCl (d) 0.05 M NaOH

slega [8]

Answer:

(a) $pH=1$

(b) $pH=1.3$

(d) $pH=12.7$

Explanation:

Hello,

In this case, we define the pH in terms of the concentration of hydronium ions as:

$pH=-log([H^+])$

Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:

(a)

$[H^+]=[HCl]=0.1M$

$pH=-log(0.1)=1$

(b)

$[H^+]=[HCl]=0.05M$

$pH=-log(0.05)=1.3$

Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

$[OH^-]=[NaOH]$