Ask question

Ask question

All categories

3 years ago

7

In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 11.

30 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH, note the corresponding OH− ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L)" above the graphic. Find the pOH of the solution. Express your answer numerically to two decimal places.

1 answer:

Mila [183]3 years ago

4 0

Answer:

Hydroxyl ion concentration is $5.08\times10^{-11}M$ .

Explanation:

From the given,

pH of the solution = 11.30

pH+pOH = 14

11.30 + pOH = 14

pOH = 14-11.30 = 2.7

The pOH of the solution is 2.7

<u>Concentration of hydroxyl ion:</u>

$2.7= -log[OH^{-}]$

$log[OH^{-}]= -2.7=5.08\times10^{-11}M$

Therefore, hydroxylion concentration is $5.08\times10^{-11}M$ .

You might be interested in

Oksi-84 [34.3K]

Answer:

2K + Br2 --> 2KBr is the answer

4 0

3 years ago

Complete the nuclear equation<br><br> 42 K → 0 e <br>19____-1

kenny6666 [7]

Answer:

42 19 K→42 20 Ca+e−

Explanation:

Naturally-occurring potassium atoms have a weighted average atomic mass of 39.10 (as seen on most modern versions of the periodic table.) Each potassium atom contains 19 protons p+ and thus an average potassium atom contains about 39.10−19≈20 neutrons n0.

This particular isotope of potassium-42 contains 42 nucleons (i.e., protons and neutrons, combined;) Like other isotopes of potassium 19 out of these nucleons are protons; the rest 42−19=23 are therefore neutrons.

5 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

Read 2 more answers

How many moles of NaCl are needed to make 10.0L of a 5 M solution of<br> Salt water?*

vesna_86 [32]

Answer: 50 mol

Explanation:

$Molarity=\frac{Moles -of -solute}{liters -of -solution}$

$M=\frac{mol}{L}$

$mol=M*L$

$mol=5*10\\mol=50mol$

3 0

3 years ago

1. Oxygen<br> full electron configuration

Arada [10]

Answer:

$1s^{2} 2s^{2} 2p^{4}$

Explanation:

Oxygen has eight eletrons and six valence electrons, giving it the electron configuration of $1s^{2} 2s^{2} 2p^{4}$ .

7 0

3 years ago