Calculate the molality of a solution prepared by dissolved 19.9 g of kcl in 750ml of water

Answer:

(d) 2

Explanation:

Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer

Answer:

A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Explanation:

Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.

Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C

However, when ethylene glycol is mixed with water in a ratio of  50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.

This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl

Answer:

5.81 g/L

Explanation:

Let's apply the Ideal Gases Law to determine this:

P . V = n . R . T

Pressure = 1.6 atm

Volume = ?

Mol = 1 mol

Temperature = 96 K

In order to find the density, we should know the volume of the atmosphere which is a mixture of gases so, we consider all the atmosphere as a unique ideal gas.

1. 6 atm . V = 1 mol . 0.082 L.atm/mol.K . 96K

V = (1 mol . 0.082 L.atm/mol.K . 96K) / 1.6 atm

V = 4.92 L → As this is the volume for the whole atmosphere and the mass of 1 mol is 28.6 g, density should be:

28.6 g / 4.92L = 5.81 g/L

Density → mass / volume

Answer:

Mass = 26.53 g

Explanation:

Heat of combustion = −1380.0 kJ/mol

This means 1 mol of the compound releases 1380 kJ

Molar mass = 44.53 g/mol

This means 1 mol of the compound has a mass of 44.53 g

How many grams would release 822kJ..?

First, we have to obtain the number of moles

1 mol = 1380

x = 822

x = 0.5957 moles

Moles = Mass / Molar mass

Mass = Molar mass * moles

Mass = 44.53 * 0.5957

Mass = 26.53 g

Answer:

I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.

Explanation:

Answer:

1120.62 kJ

Explanation:

In order to find how much heat is released for 7.70 mol, we have to compare it with the heat released from one mole.

So from the question, we have;

196.6 kJ = 1 mol

x = 5.70

x = 5.70 * 196.6 / 1

x = 1120.62 kJ

Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

Answer:

Option B. 30 KJ.

Explanation:

The following data were obtained from the question:

Temperature (T) = 5000 K

Enthalpy change (ΔH) = – 220 kJ/mol

Change in entropy (ΔS) = – 0.05 KJ/mol•K

Gibbs free energy (ΔG) =...?

The Gibbs free energy, ΔG can be obtained by using the following equation as illustrated below:

ΔG = ΔH – TΔS

ΔG = – 220 – (5000 x – 0.05)

ΔG = – 220 – (– 250)

ΔG = – 220 + 250

ΔG = 30 KJ

Therefore, the Gibbs free energy, ΔG is 30 KJ.

The name of the molecule which is given below is 2-pentene.

Alkenes are the organic compounds which are composed of carbon and hydrogen atoms, in which double bond is present.

In the given diagram:

So the compound is 2 pentene.

Hence, 2 pentene is the name of the compound.

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Answer:

pH = 5.22

Explanation:

As you can see, your initial concentration of sodium acetate (NaCH₃COO) is 0.02M (0.02mol /L). When you add HCl, the reaction is:

NaCH₃COO + HCl → CH₃COOH + NaCl.

If you add HCl, and final concentration of NaCH₃COO is 0.015M, the concentration of CH₃COOH is 0.005M.

You can know the pH of this solution using H-H equation:

pH = pKa + log {NaCH₃COO} / {CH₃COOH}

pH = 4.74 +  log {0.015M} / {0.005M}

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.

Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.

CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+

Answer:

(B)

Explanation:

edg 2020

The four seasons in earth is originating from the earth's revolution around the sun. Therefore, the most suitable model for Amanda is  Four earth models, each with a different tilt to represent the earth’s position relative to the sun, lined up in order of season.

Seasons in earth is originating from the difference in the distance from the sun over each time period. Hence, revolution of earth around sun make these seasons.

The time period at which earth comes closer to sun more hot will be earth's surface and we experience summer season. When we far from sun winter season occurs.

Therefore, different season are coming based on the distance of earth from sun at each revolution point. This is also affected by the tilt of earth's in its own axis.

Hence, different earth models with different distance from sun is most suitable model here for Amanda. Thus option B is correct.

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Answer:

A.∆s>0contribute to spontaneity.

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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Answer:

91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Explanation:

The formula for sodium phosphate is Na₃PO₄

Molar mass of sodium phosphate = 164 g/mol

The dissociation of one mole of sodium phosphate produces 3 moles of sodium ions;

Na₃PO₄ ------> 3Na⁺ + PO₄³

Number of moles of Na₃PO₄ that will produce 1.67 moles of Na⁺ = 1/3 * 1.67 = 0.556 moles  of Na₃PO₄

Mass of 0.556 moles of Na₃PO₄ = 0.556 moles * 164 g/mol = 91.29 g

Therefore, 91.29 g of Na₃PO₄ are needed to produce 1.67 moles of sodium ions.

Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.  

Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

Note the following:

Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.

Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

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Answer:

0.312g

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules

1mole of sulphur = 32g

If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules

Then, Xg of sulphur will contain 5.87x10^21 molecules i.e

Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g

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Answer:

244.7 g of acetylene

Explanation:

The balanced reaction equation is shown below;

2H20 (l) + CaC2 (s) → Ca(OH)2 (s) + C2H2 (g)

Number of moles of were reacted = reacting mass/molar mass = 358g/18gmol-1 = 19.89 moles of water

From the balanced reaction equation;

2 moles water yields 1 mole of acetylene

19.89 moles of water will yield 19.89 × 1/2 = 9.945 moles of acetylene

Theoretical yield of acetylene = 9.945 moles of acetylene × molar mass of acetylene

Molar mass of acetylene = 26.04 g/mol

Theoretical yield of acetylene = 9.945 moles of acetylene × 26.04 g/mol

Theoretical yield of acetylene = 258.9678 g of acetylene

% yield = actual yield/ theoretical yield × 100

94.5 = actual yield/258.9678 g × 100

Actual yield= 94.5 × 258.9678 g/100

Actual yield = 244.7 g of acetylene

Answer:

1. 6.92 g of H2O

2i. - 46 KJ of energy.

ii. Option A. Evolved.

Explanation:

1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.

This can be obtained as follow:

The equation for the reaction is given below

2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ

Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.

Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.

This is illustrated below:

From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.

Therefore, Xg of H2O will react to 110 kJ of energy i.e

Xg of H2O = (36 x 110)/572

Xg of H2O = 6.92 g

Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.

2i. Determination of the energy.

The balanced equation for the reaction is given below:

CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ

Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.

This is illustrated below:

Molar mass of CO = 12 + 16 = 28 g/mol

Mass of CO from the balanced equation = 1 x 28 = 28 g

From the balanced equation above,

28 g of CO reacted to produce -206 kJ of energy.

Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:

From the balanced equation above,

28 g of CO reacted to produce -206 kJ of energy.

Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.

Therefore, - 46 KJ of energy were produced from the reaction.

2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.

Therefore, the heat is evolved.

Answer:

In the other of increasing wavelength we arrange as

X-rays

ultraviolet

green light

red light

infrared

radio waves

Explanation:

In the electromagnetic spectrum, wavelength decreases with increase in the energy of the electromagnetic wave. Since the e-m wave spectrum is arranged in the order of increasing energy (decreasing wavelength) as: Radio wave; infrared; visible light; ultraviolet; x-rays; gamma rays. Within the visible light, the green light has more energy than the red light. Therefore, the arrangement should be in the reverse direction of their increasing energy.

Answer:

b. Al3+ + 3Br– → Al + (3/2)Br2

Explanation:

If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.

However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.

b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]

Electrolytic  Cell v/s Electrochemical Cell:

Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]  is carried out in an electrolytic cell rather than in an electrochemical cell.

As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells

In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.

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Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

Answer:

58.45g is the answer

Explanation:

took the test

The mass of chromium that can be deposited is equal to 58.45 g. Therefore, option B is correct.

For a steady flow of charge through a conductor, the current can be determined with the following equation:

[tex]{\displaystyle I={Q \over t}[/tex]

where Q is the electric charge while time t. If Q and t are measured in coulombs (C) and seconds then I will be in amperes.

Electric charge flows by electrons, from lower potential to higher electrical potential. Any stream of charged objects can constitute an electric current.

Given, the amount of electric current flowing through the solution:

I = 45.2 A

The time for which the current flows, t = 2 hrs = 2 × 60 ×60 = 7200 sec

The charge flowing through the solution, Q = I × t

Q = 45.2 × 7200

Q = 325440 C

The number of moles of electrons in 325440 C charge = 325440/96500 = 3.37 mol

We know Cr³⁺ + 3e⁻ →  Cr (s)

3 moles of electrons deposit of chromium  = 1 mol

3.37 mol of electrons deposit of chromium  = 3.37/3 = 1.12 mol

The mass of chromium in 1.12 mol = 1.12 × 52 = 58.45 g

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Answer:

The reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C

Explanation:

Hello,

In this case, for the given chemical reaction, we can write the law of mass action (equilibrium expression) as shown below:

[tex]Kc=\frac{[CH_3C_5H_9]}{[C_6H_{12} ]}[/tex]

Thus, since Kc < 1, we can conclude there are more moles of cyclohexane at equilibrium (denominator is greater than numerator), therefore, the reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C.

Best regards.

Explanation:

The mass to charge ratio =136

Answer:

Explanation:

The given formula is empirical formula

Let the molecular formula of first be

[tex]( CCl )_n[/tex]

molecular weight = n x ( 12 + 35.5 )

= 47.5 n

Given molecular weight = 189.83 so

47.5 n = 189.83

n = 3.99 or 4 approx

Molecular formula =

[tex]( CCl )_4[/tex]

= C₄ Cl₄

Let the molecular formula of second compound  be

[tex]( C_3H_2N)_n[/tex]

molecular weight = n x ( 3 x 12 +2+14 )

= 52 n

Given molecular weight = 156.23  so

52 n = 156.23

n = 3.0044 or 3 approx

Molecular formula =

[tex]( C_3H_2N )_3[/tex]

= C₉H₆ N₃

Answer:

Poop Buttt.

Explanation:

Answer:

Approximately [tex]10.88[/tex].

Explanation:

[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:

[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].

(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)

However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.

At equilibrium:

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].

As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:

[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].

In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].

Again, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:

[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].

Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:

[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].

Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At equilibrium:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].

Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].

That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:

[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 3.35\times 10^{-4}[/tex].

In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:

[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].

(Rounded to two decimal places.)