If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?

Answer:

There are fifteen molecular orbitals in benzene filled with electrons.

Explanation:

Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;

There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.

There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons

The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.

All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

Answer:

The pH of the solution is 9.06.

Explanation:

The reaction of the dissociation of NH₃ in water is:

NH₃(aq) + H₂O(l)  ⇄  NH₄⁺(aq) + OH⁻(aq)     (1)

[NH₃] - x                     [NH₄⁺] + x     x  

The concentration of NH₃ and NH₄⁺ is:

[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]

[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]

From equation (1) we have:

[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]

[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]

[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]

By solving the above equation for x we have:

x =  1.15x10⁻⁵ = [OH⁻]

The pH of the solution is:

[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]

[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]

Therefore, the pH of the solution is 9.06.

I hope it helps you!

Answer:

[tex]\Delta G=-97.14kJ[/tex]

Explanation:

Hello,

In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

[tex]\Delta G=-RTln(K)[/tex]

Hence, we compute it as required:

[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]

And for 2.37 moles of hydrogen bromide, we obtain:

[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]

Best regards.

Answer:

Here are four possible voltaic cells.  

Explanation:

1. Standard reduction potentials

                                         E°/V

I₂(s) + 2e⁻ ⟶ 2I⁻(aq);        0.54

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);   0.34

Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);   -0.41

Zn²⁺(aq) + 2e⁻ ⟶ Zn(s);   -0.76

2. Possible Voltaic cells

(a) Zn/I₂

                                                                       E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                 0.76

Cathode:  I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 1.30

Zn(s)|Zn²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Zn is the anode; graphite is the cathode.

(b) Zn/Cu²⁺

                                                                          E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                    0.76

Cathode:  Cu²⁺(aq) + 2e⁻ ⟶ Cu(s);                  0.34

Cell:          Zn(s) +  Cu²⁺(s) ⟶  Zn²⁺(aq) + Cu(s); 1.10

Zn(s)|Zn²⁺(aq)∥Cu²⁺(aq)|Cu(s)

Zn is the anode; Cu is the cathode.

(c) Zn/Fe²⁺

                                                                            E°/V

Anode:     Zn(s) ⟶ Zn²⁺(aq) + 2e⁻;                     0.76

Cathode:  Fe²⁺(aq) + 2e⁻ ⟶ Fe(s);                    -0.41

Cell:          Zn(s) +  Fe²⁺(s) ⟶  Zn²⁺(aq) + Fe(s);  0.35

Zn(s)|Zn²⁺(aq)∥Fe²⁺(aq)|Fe(s)

Zn is the anode; Fe is the cathode.

(d) Fe/I₂

                                                                         E°/V

Anode:     Fe(s) ⟶ Fe²⁺(aq) + 2e⁻;                   0.41

Cathode: I₂(s) + 2e⁻ ⟶ 2I⁻(aq);                     0.54

Cell:         Zn(s) +  I₂(s) ⟶  Zn²⁺(aq) + 2I⁻(aq); 0.95

Fe(s)|Fe²⁺(aq)∥I⁻(aq)|I₂(s)|C(s, graphite)

Fe is the anode; graphite is the cathode.

Answer:

Ka = 6.87x10⁻⁵

Explanation:

The equilibrium of benzoic acid in water is:

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The equilibrium constant, Ka, is:

Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]

The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.

X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.

Concentrations in equilibrium are:

pH is defined as -log [H₃O⁺]. As pH = 2.270

2.270 = -log [H₃O⁺]

10^-2.270 = [H₃O⁺]

5.37x10⁻³M = [H₃O⁺] = X.

Replacing, concentrations in equilibrium are:

[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M

[C₆H₅COO⁻] = 5.37x10⁻³M

[H₃O⁺] = 5.37x10⁻³M

Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]

Answer: [tex]\Delta H = -272.25kJ[/tex] for 1 mole of NO.

Explanation: Hess' Law of Constant Summation or Hess' Law states that the total enthalpy change of a reaction with multiple stages is the sum of the enthalpies of all the changes.

For this question:

1) [tex]N_{2}_{(g)} + 3H_{2}_{(g)}[/tex] => [tex]2NH_{3}_{(g)}[/tex]       [tex]\Delta H=-92kJ[/tex]

2) [tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex]       [tex]\Delta H=-905kJ[/tex]

Amonia ([tex]NH_{3}_{(g)}[/tex]) appeares as product in the first equation and as reagent in the 2 reaction, so when adding both, there is no need to inverse reactions. However, in the 2nd, there are 4 moles of that molecule, so to cancel it, you have to multiply by 2 the first chemical equation and enthalpy:

[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex]     [tex]\Delta H=-184kJ[/tex]

Now, adding them:

[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex]     [tex]\Delta H=-184kJ[/tex]  

[tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex]       [tex]\Delta H=-905kJ[/tex]

[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex]  [tex]\Delta H = -185-905[/tex]

[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex]  [tex]\Delta H = -1089kJ[/tex]

Note net enthalpy is for the formation of 4 moles of nitric oxide.

For 1 mole:

[tex]\Delta H = \frac{-1089}{4}[/tex]

[tex]\Delta H=-272.25kJ[/tex]

To form 1 mol of nitric oxide from nitrogen, oxygen and hydrogen, net change in enthalpy is [tex]\Delta H=-272.25kJ[/tex].

Answer:

848 torr  

Explanation:

The only variables are the pressure and the volume, so we can use Boyle's Law.

p₁V₁ = p₂V₂

Data:

p₁ = 565 torr; V₁ = 24.0 L

p₂ = ?;            V₂ =  16.0 L

Calculations:

[tex]\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}[/tex]

Answer:

Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.

Answer:

1.49e24

Explanation:

Standars temperature and pressure are 273.15K and 1atm, respectively.

Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:

PV = nRT

PV / RT = n

Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.

Replacing:

PV / RT = n

1atm*55.3L / 0.082atmL/molK*273.15K = n

2.47 moles = n

Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.

As we have 2.47 moles:

2.47 mol × (6.022x10²³ molecules / 1 mole) =

1.49x10²⁴ molecules that are in the jar

In scientific notation:

Answer:

They get more energy, so they vibrate!

Explanation:

Answer:

105/22*(1.251-0.620)

105/22*0.631

4.772*0.631

3.011132

Hope it helps

Answer:

3.0

Explanation:

First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).

The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

Answer:

The intermolecular forces present in CH_3NH_2 includes

Explanation:

The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)

It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.

Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.

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Answer:

SCN- is a weak field ligand while CN- is a strong field ligand

Explanation:

The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.

Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.

CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.

SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.

SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.

Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.

Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

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Answer:

d. HF(g)

Explanation:

Hello,

In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.

Best regards.

Answer:

- you have to do maths in all 3

- atoms make up everything even parts of a cel and theyre studied in chem and physics

- chemistry is used in biology by finding out what different substances are eg cytoplasm in a cell

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

Explanation:

this is the ans

hope this helps

Answer:

C

Explanation:

According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.

Answer:

I think it's C

Explanation:

Please, tell me if I'm incorrect.

Answer:

1.6x10¹¹ = Kc

Explanation:

For the reaction:

AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)

Kc is defined as:

Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²

Ksp of AgCl is:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Where Ksp is:

Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰

In the same way, Kf of Ag(CN)₂⁻ is:

Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻

Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹

The multiplication of Kf with Ksp gives:

[Ag⁺] [Cl⁻] *  [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf

[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf

Obtaining the same expression of the first reaction

That means Ksp*Kf = Kc

1.6x10⁻¹⁰*1.0x10²¹ = Kc

Answer:

0.17 lb

Explanation:

78 g * (1 lb/454 g)=0.17 lb

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Answer:

Water when it freezes into ice

Explanation:

Most liquids expand when heated and contract when cooled, water behaves in an anomalous fashion. Water rather expands when cooled and contracts when heated.

Water usually contracts on cooling from any temperature until 4°C, after 4°C, the water begins to expand rapidly. Hence water has its least volume at 4°C and increases rapidly afterwards.

Thus the largest volume change for water occurs during freezing since it expands when cooled.

Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

Answer:

1, C, Xe 2, B,He

Explanation:

1, cause as u go down a group the boiling point increases.

2, boiling point of single element is greater than a compound

According to  periodic trends in periodic table boiling point increases down the  group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.

Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.

It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.

Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.

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Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

Find more about activation energy here:

I hope it helps you!

delete please .....................................

Answer:

14.297 g

Explanation:

From the question;

1 mo of the compound requires 1320.0 kJ

From the molar mass;

1 ml of the compound weighs 30.55g

How many grams requires 617.30kJ?

1 ml = 1320

x mol = 617.30

x = 617.30 / 1320

x = 0.468 mol

But 1 mol = 30.55

0.468 mol = x

x = 14.297 g

Answer:

The correct option is C.

Note the full question and structure of the moleculesis found in the attachment below.

Explanation:

Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.

The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.

During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.

From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.

The correct option therefore, is C

Answer:

Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.

Explanation:

There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm.  The atomic radius of aluminium atom is 143 ppm.  The atomic radius of silicon atom is 111 ppm.  The atomic radius of phosphorus atom is 98 ppm.  The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.

Answer:

8.68

Explanation:

pOH = 8.68

all you need is contained in the sheet

Answer:

Approximately [tex]8.68[/tex].

Explanation:

The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:

[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].

On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:

[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].

Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:

[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].

Therefore, the [tex]\rm pOH[/tex] of this solution would be:

[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].

Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].

For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.