The element silver has an atomic weight of 108 and consists of two stable isotopes silver-107 and silver-109. The isotope silver-107 has a mass of 107 amu and a percent natural abundance of 51.8 %. The isotope silver-109 has a percent natural abundance of 48.2 %. What is the mass of silver-109

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibbs free energy  

[tex]\Delta H[/tex] = enthalpy change

[tex]\Delta S[/tex] = entropy change  

T = temperature in Kelvin

When [tex]\Delta G[/tex] = +ve, reaction is non spontaneous

[tex]\Delta G[/tex]= -ve, reaction is spontaneous

[tex]\Delta G[/tex]= 0, reaction is in equilibrium

For the given reaction 1: [tex]glucose+Pi\rightarrow glucose-6-phosphate+H_2O[/tex] [tex]\Delta G=+13.8kJ/mol[/tex]

As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

Answer: HZ > HX > HY in order of decreasing strengths.

Explanation: Generally, the rule is that the stronger the acid, the weaker its conjugate base and vice versa; same rule applies for bases and their conjugate acids.

So the weakest base Z- would have the strongest conjugate acid. Consequently, the strongest base Y- would have the weakest conjugate acid.

I hope this was MORE helpful as this is the correct answer.

The ranking of the conjugate acids in order of decreasing strength (i.e from strongest to weakest) is; HZ < HX < HY

First we must know that the stronger a base is, the weaker is it's conjugate acid and the weaker a base is, the stronger is it's conjugate acid.

Therefore, the order of decreasing strength of the conjugate acid is; HZ < HX < HY

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Answer:

5) 1.147 moles Cl2

6) 12.57 grams ClF3

7)  339.10 grams ClF3

8) 5.63 grams F2

Explanation:

Step 1: Data given

Number of moles F2 = 3.44 moles

Molar mass F2 = 38.00 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2

Step 1: Data given

Number of moles F2 = 0.204 moles

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles ClF3

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles

Step 4: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 0.136 moles * 92.448 g/mol

Mass ClF3 = 12.57 grams ClF3

Step 1: Data given

Mass of Cl2 = 130.0 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 130.0 grams / 70.9 g/mol

Moles Cl2 = 1.834 moles

Step 4: Calculate moles

For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3

For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3

Step 5: Calculate mass ClF3

Mass ClF3 = Moles ClF3 * molar mass ClF3

Mass ClF3 = 3.668 moles * 92.448 g/mol

Mass ClF3 = 339.10 grams ClF3

Step 1: Data given

Mass of Cl2 = 3.50 grams

Molar mass F2 = 38.00 g/mol

Molar mass ClF3 = 92.448 g/mol

Step 2: The balanced equation

Cl2 + 3F2 → 2ClF3

Step 3: Calculate moles Cl2

Moles Cl2 = Mass Cl2 / molar mass Cl2

Moles Cl2 = 3.50 grams / 70.9 g/mol

Moles Cl2 = 0.0494  moles

Step 4: Calculate moles F2

For 1 mol Cl2 we need 3 moles F2

For 0.0494 moles we need 3*0.0494 = 0.1482 moles

Step 5: Calculate mass F2

Mass F2 = moles F2 * molar mass F2

Mass F2 = 0.1482 moles * 38.00 g/mol

Mass F2 = 5.63 grams F2

Answer:

certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 T 1 P 2 T 2 ... Ideal Gas Equation 5.4 Charles' law: V T (at constant n and P ) ... Consider a case in which two gases, A and B , are in a container of volume V.

Explanation:

Answer and Explanation:

The computation of the energy or wavelength gamma rays observed is shown below:

Since the energy of gamma rays is higher than 0.10 MeV.

Now We have to calculate transitions in between the given levels of energy that correspond to this energy.

As per the given question, we have the following information

Ground state = E where E < 0.075 MeV

For  Level 1 = 0.075 MeV

For Level 2 = 0.320 MeV

For Level 3 = 0.475 MeV

Now we have to take the below transitions:

1. [tex]3 \rightarrow 2[/tex]

Difference of energy is

= 0.475 - 0.320

= 0.155 MeV

This represents  a gamma radiation

2. [tex]3 \rightarrow 1[/tex]

Difference of energy is

= 0.475 - 0.075

= 0.4 MeV

This represents  a gamma radiation

3. [tex]3 \rightarrow ground[/tex]

Difference of energy is

= 0.475 - E > 0.155 MeV

This represents  a gamma radiation

4. [tex]2 \rightarrow 1[/tex]

Difference of energy is

= 0.320 - 0.075

= 0.245 MeV

This represents  a gamma radiation

5. [tex]2 \rightarrow ground[/tex]

Difference of energy is

= 0.320 - E > 0.245 MeV

This represents  a gamma radiation

6. [tex]1 \rightarrow Ground[/tex]

Difference of energy is

= 0.075 - E < 0.10 MeV

This represents not a gamma radiation

We can see that there are 5 transitions that contain gamma rays

Answer:

Consider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.

Answer:

0°C.

Explanation:

Hello,

In this case, given the heating curve of water on the attached document, we can notice that at 0 °C the solid starts melting, which means that the melting point is reached. Melting point is known as a physical change whereby a solid changes to liquid by the addition of heat as it allows the molecules to separate to each other.

Best regards.

Answer:

0 degrees celcius

Explanation:

I took the test

Answer:

Zero

Explanation:

Hello,

The question require us to calculate the mass of nitrogen present in aluminium carbonate.

This can easily be calculated using Avogadro's number as a constant with some minor calculations but however in this case, we can't because there's no single atom of nitrogen present in aluminium carbonate hence we can't calculate the mass of nitrogen present in it.

Chemical formula of aluminium carbonate = Al₂(CO₃)₃.

From the above chemical formula, we can see that there's no single atom of nitrogen present in the formula hence the mass of nitrogen present in aluminium carbonate that contains 1.23×10²³ carbon atoms is zero.

Answer:

20mL of the stock solution and dilute to 100mL

Explanation:

20mM = 0.020M is the concentration of the diluted solution. As you want to prepare this solution from a 0.1M solution, dilution factor is:

0.1M / 0.020M = 5

That means you need to dilute the stock solution 5 times to obtain the 20mM sucrose solution.

As you want to prepare 100mL, you need to add:

100mL / 5 =

Answer:

[tex]Kb=6.16x10^{-7}[/tex]

Explanation:

Hello,

In this case, given the pH of the base, we can compute the pOH as shown below:

[tex]pOH=14-pH=14-10.781=3.219[/tex]

Next, we compute the concentration of the hydroxyl ions when the triethanolamine is dissociated:

[tex]pOH=-log([OH^-])[/tex]

[tex][OH^-]=10^{-pOH}=10^{-3.219}=6.04x10^{-4}M[/tex]

Then, by writing the equilibrium expression for the dissociation of triethanolamine we have:

[tex]Kb=\frac{[OH^-][C6H14O2N^+]}{[C6H15O3N ]}[/tex]

That is suitable for the direct computation of Kb, knowing that based on the ICE procedure, [tex]x[/tex] equals the concentration of hydroxyl ions that was previously, computed, therefore, we have:

[tex]Kb=\frac{6.04x10^{-4}M*6.04x10^{-4}M}{0.592M-6.04x10^{-4}M}\\ \\Kb=6.16x10^{-7}[/tex]

Regards.

Answer:

Explanation:

Stereoisomers are two or more atoms that  have the same bonding order of atoms but there is a difference  spatial arrangement of  the  atoms in space.

A plane of symmetry divides a molecule into two equal halves.

A chiral stereoisomer are not superimposed on a mirror image , Hence they do not posses a plane of symmetry.

As a result to that. these non-superimposable mirror images are said to be Enantiomers.

However, a Fischer Projection emanates from a two - dimensional figure which is used for presenting a three - dimensional organic molecules.

From the given question;

Fischer projection for an enantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right.

we can sketch the way the enantiomer of 2-bromo-2,3-dihydroxypropanal can  be seen like the one shown below:

              CH₂OH

                   |

                   |

                   |

Br -------------|----------------OH

                   |

                   |

                   |

                 CHO

The objective of this question is to drawn the perspective formula of the molecule.

So , from the attached file below; we can see the perspective formula of the molecule in a well structured 3-D format.

Answer:

Explanation:

Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6,

0.5 atm for O2, and

2.0 atm for CO2, and

25 oC:

C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced)

ΔGfo C2H6 (g) = - 32.89 kJ/mol;

ΔGfo CO2 (g) = - 394.4 kJ/mol;

ΔGfo H2O (l) = - 237.13 kJ/mol

The balance equation of this reaction is

[tex]2C_2H_6 (g) + 7O_2 (g) ---> 4CO_2 (g) + 6H_2O (l)[/tex]

[tex]\Delta G_{rxn}=\sum G^o_f(product)-\sum G^o_f(reactant)[/tex]

[tex]=4G^o_f(CO_2)+6G^o_f(H_2O)-7G^o_f(O_2)-2G^o_f(C_2H_6)\\\\[/tex]

[tex][4(-394.4)+6(-237.13)-7(0)-2(-32.89)]kJ/mol\\\\=-1577.6-1422.78+65.78\\\\=-3000.38+65.78\\\\=-2934.6kJ/mol[/tex]

Answer:

a) Fe(s) + Ni^2+(aq) ----> Fe^2+(aq) + Ni(s)

b) no reaction

c) no reaction

d) 2Mg(s) + 2H2O(l)-----> 2Mg^2+(aq) + O2(g) +4H^+(aq)

e) no reaction

Explanation:

It is important to say here that the ability of a particular chemical specie to displace another chemical specie is dependent on the relative standard reduction potentials of the species involved.

All the reactions stated above are redox reactions. Let us take reaction E as an example. Mg^2+ has a reduction potential of -2.37 V while Cr^3+ has a reduction potential of -0.74V. Since the reduction potential of magnesium is more negative than that of chromium, there is no reaction when a piece of chromium metal is dipped into a solution of Mg^2+.

Similarly, though metals displace hydrogen gas from dilute acids, metals that are less than hydrogen in the reactivity series cannot do that. This explains why there is no reaction when copper and silver are dipped into dilute acid solutions.

Reaction occurs when iron is dipped into a nickel solution because the reduction potential of Fe^2+ is far more negative than that of Ni^2+.

Answer:

n=2 to n=4 < n=6 to n=8 < n=10 to n=12 < n=14 to n=16

Explanation:

According to Neils Bohr, electrons in an atom are found in specified energy levels. Transitions are possible from one energy level to another when the electron receives sufficient energy usually in the form of a photon of electromagnetic radiation of appropriate frequency and wavelength. The energy of this photon corresponds to the energy difference between the two energy levels. Thus the higher the energy difference between energy levels, the greater the energy of the photon required to cause the transition and the shorter the wavelength of the photon.

High energy photons have a very short wavelength. It should be noted that as n increases, the energy of successive energy levels decreases and transitions between them now occurs at longer wavelengths. Hence, the highest energy and shortest wavelength of photons are required for transition involving lower values of n because such electrons are closer to the nucleus and are more tightly bound to it than electrons found at a greater distance from the nucleus.

Hence transition involving electrons at higher energy levels occur at a longer wavelength compared to transition involving electrons closer to the nucleus. This is the basis for the arrangement of wavelengths required to effect the various electronic transitions shown in the answer.

Answer:

i hope it will help you

Explanation:

electronic configuration 1s²,2s,²2p^6,3s²3p^6,4s^1

as it has one electron in its valence shell so it is the member of group 1A(ALKALI METALS)  and the number of shells is 4 so it is in period 4

Answer:

D. rate = k [(CH3)3CCl]

Explanation:

(CH3)3CCl + OH- → (CH3)3COH + Cl-

The mechanisms are;

Step 1)

(CH3)3CCl → (CH3)3C+ + Cl- (slow)

Step 2)

(CH3)3C+ + OH- → (CH3)3COH (fast)

In kinetics, the slowest step is the ratee determining step.

For a given reaction;

A → B + C, the rate law expression is given as;

rate = k [A]

In this problem, from step 1. The rate expression is;

rate = k [(CH3)3CCl]

Answer:

The lock-and-key model:

c. Enzyme active site has a rigid structure complementary

The induced-fit model:

a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.

Common to both The lock-and-key model and The induced-fit model:

b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.

d. Substrate binds to the enzyme through non-covalent interactions

Explanation:

Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.

The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.

The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.

Answer:

See detailed answer with explanation below.

Explanation:

Valence electrons are electrons found on the outermost shell of an atom. They are the electrons in an atom that participate in chemical combination. Recall that the outermost shell of an atom is also referred to as its valence shell. Let us consider an example; if we look at the atom, sodium-11, its electronic configuration is 2,8,1. The last one electron is the valence electron of sodium which is found in its outermost or valence shell.

Positive ions are formed when electrons are lost from the valence shell of an atom. For instance, if the outermost electron in sodium is lost, we now form the sodium ion Na^+ which is a positive ion. Positive ions possess less number of electrons compared to their corresponding atoms.

Negative ions are formed when one or more electrons is added to the valence shell of an atom. A negative ion possesses more electrons than its corresponding atom. For example, chlorine(Cl) contains 17 electrons but the chloride ion (Cl^-) contains 18 electrons.

In molecular compounds, a bond is formed when two electrons are shared between the bonding atoms. Each bonding atom may contribute one of the shared electrons (ordinary covalent bond) or one of the bonding atoms may provide the both shared electrons (coordinate covalent bond). The shared pair may be located at an equidistant position to the nucleus of both atoms. Similarly, the electron may be drawn closer to the nucleus of one atom than the other (polar covalent bond) depending on the electro negativity of the two bonding atoms.

The electrons are shared in order to complete the octet of each atom by so doing, the both bonding atoms now obey the octet rule. For example, two chlorine atoms may come together to form a covalent bond in which each chlorine atom has an octet of electrons on its outermost shell.

Answer:

- Both energy and matter cannot be neither created nor destroyed.

- An equilibrium temperature will be reached.

Explanation:

Hello,

In this case, the law of conservation is applied to both matter and energy, and it states that both energy and matter cannot be neither created nor destroyed. Specifically, in chemical reactions, it states that in closed systems, the mass of the reactants equals the mass of the products even when the number of moles change. Moreover, for energy, if two substances at different temperatures come into contact, the hot one will cool down and the cold one will heat up until an equilibrium temperature so the energy lost by the hot one is gained by the cold one, which accounts for the transformation of energy.

Best regards.

Answer:

8

Explanation:

When you balance the entire equation, you should get:

C5H12 + 8O2 ---> 5CO2 + 6H2O

Answer:

It was because the vaccine generated actively acquired immunity, that is, inoculation of a portion of the measles virus so that the body forms the antibodies for a second contact and thus can destroy it without triggering the pathology.

Explanation:

Vaccines are methods of active acquired immunity since the antibody is not passively inoculated, it is manufactured by the body with a physiological process once part of the virus is inoculated.

The measles virus most of all affected the lives of infants or newborn children with severe rashes and high fevers that led to death.

Answer:

[tex]6.68~g~NO_2[/tex]

Explanation:

We have to start with the combustion reaction:

[tex]NO~+~O_2~->~NO_2[/tex]

Then we can balance the reaction:

[tex]2NO~+~O_2~->~2NO_2[/tex]

If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:

[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]

I hope it helps!

Answer:

128 gram of CaC2O4 contains 40 gram of Calcium

40.3 gram of CaC2O4 cotnains = 40*40.3/128 = 12.59 gram of Calcium

out of 15 gram 12.59 gram is Calcaim that means around 50% of orginal sample has Calcium

Answer:

D

Explanation:

[tex]F=ma \\\\F=2000\cdot 5=10,000N[/tex]

Hope this helps!

Answer:

[Ne]3s²

Explanation:

Mg

1s2 2s2 2p6 3s2   or [Ne]3s²

Abbreviated electronic configuration of magnesium is [Ne]3 s² and in complete form it is 1 s² 2 s² 2 p⁶ 3 s².

Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.

Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.

Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.

Learn more about electronic configuration,here:

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Explanation:

Electron affinity is defined as the energy released by the addition of an electron to any gaseous atom. Electron affinity of an atom depends on the electronic configuration.

a).The carbon has vacant p-orbital and nitrogen has half-filled configuration which is more stable. Therefore, one electron can be easily added to carbon whereas nitrogen having more stable configuration releases more amount of energy on adding one electron. Therefore, nitrogen has more electron affinity than carbon.

The bromine has vacant p-orbital whereas argon has filled orbital which is most stable. Therefore, one electron can be easily added to bromine whereas argon having  more stable configuration releases more amount of energy on adding one electron.Therefore, argon has more electron affinity than bromine.

Answer:

1. a. C; b. N; c. C; 2. a. Br; b. Ar; c. Br

Explanation:

Use your Periodic Table and follow the trends in atomic properties (Fig. 1).

Electron affinity increases from left to right and from bottom to top.

The elements with the most exothermic EA are at the upper right corner

Exceptions are the noble gases (group 18) and the pnictogens (group 18).

The elements of Group 18 have a complete octet and have no tendency to accept electrons.  

The elements of Group 15 have half-filled p subshells. They are more stable than the elements immediately preceding them, so they are less exothermic when adding an electron.  

Ionization energy increases from left to right and from bottom to top.  

The atoms with the highest IE are at the upper right corner.

Atomic size increases from right to left and from bottom to top.  

The largest atoms are on the lower-left corner.

1. C vs N

(a)   EA: C. N is a Group 15 element

(b)    IE: N. N is further to the right.

(c) Size: C. C is further to the left.

2. Ar vs Br

(a)   EA: Br. Ar is a noble gas.

(b)    IE: Ar. Ar is further to the right.

(c) Size: Br. Br is nearer to the bottom.

Answer:

C ) The rate will triple

Answer:

Question 1.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

[H₃O⁺] = 3.70 * 10⁻¹¹ M

Explanation:

The ion product constant of water  Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.

A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]

[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M

[H₃O⁺] = 3.70 * 10⁻¹¹ M

Answer:

H = 4,034,250 J

Explanation:

Mass, m = 165kg = 165,000g (Converting to grams)

Initial temperature = 15.7°C

Final temperature = 95.7°C

Temperature change, ΔT = 95.7 - 15.7 = 50°C

Specific heat capacity, c = 0.489 J/g·°C

Heat = ?

All the parameters are related with the equation below;

H = m * c * ΔT

H = 165000 * 0.489  * 50

H = 4,034,250 J