Which types of electron orbitals will have higher energy than a 4d orbital?
A. 4p
B. 3s
C. 4f
D. 5s​

Answer:

Water has a low specific heat capacity and so large bodies of water moderate temperatures on Earth.

Explanation:

Water has a very high specific heat capacity, meaning that it has to absorb a lot of energy to raise the temperature by one degree. Because water has a high specific heat capacity, large bodies of water can moderate the temperature of nearby land.

Hope this helps.

Answer:

H20 is an acid and OH is its conjugate base.​

Explanation:

Chemical reactions involving acids and bases occur. An acid is a substance that dissociates in water i.e. lose an hydrogen ion/proton. According to the Bronsted-Lowry acid-base theory, when an acid dissociates in water and loses its hydrogen ion, the resulting substance that forms is the CONJUGATE BASE. A conjugate base is the compound formed as a result of the removal of an H+ ion from an acid.

Based on the chemical reaction in the question, NaHCO3 + H20 = H2CO3 + OH- + Na+

The H20 loses its hydrogen ion (H+) to form an anion OH-. This anion formed is the conjugate base while H20 is its acid.

Answer:

Hexane

Explanation:

You have a carbon structure with only single bonds.  This means that the name will end in -ane.

There are 6 carbon atoms.  This means that the name will begin with hex-.

The structure is hexane.

Answer:

[tex]\large \boxed{29.7 \,^{\circ}\text{C}}[/tex]

Explanation:

There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the Al be Component 1 and the H₂O be Component 2.

Data:  

For the Al:

[tex]m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

For the water:

[tex]m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

Calculations

(a) The relative temperature changes

[tex]\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}[/tex]

(b) Final temperature

[tex]\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}[/tex]

[tex]\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}}$}[/tex]

Check:

[tex]\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}[/tex]

The second term has only two significant figures because ΔT₂ has only two.

It agrees to two significant figures

Answer:

1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis

2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis

3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis

4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis

5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis

Explanation:

Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.

Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.

Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.

The following shows the various regulatory methods and their effects on both processes:

1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.

Increased levels of fructose-2,6-bisphosphatase  activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.

2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.

3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.

4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.

5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the  activation of gluconeogenesis and inhibition of glycolysis.

Answer:

Cu₂S

Explanation:

From the question,

                     Cu                                 S

Mass:            40.80 g                      51.09-40.80 = 10.29 g

Mole ratio:     40.80/63.5                 10.29/32.1

                         0.64          :                0.32

Divide by the smallest,

                         0.64/0.32   :                0.32/0.32

                           2                :                  1

   Therefore,

Empirical formula = Cu₂S.

Answer:

BaCl2

Explanation:

Barium = Ba

Chloride => Cl-

Chemical Equation:

Ba + Cl => BaCl2

Note:

The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).

Answer:

100.2g of CuI₂ you must add

Explanation:

Molality, m, is defined as the ratio between moles of solute and kg of solvent.

In the problem, you have a 0.694m of copper (II) iodide -CuI₂, molar mass: ‎317.35 g/mol-. That means there are 0.694 moles of CuI₂ per kg of water.

As you have 455g = 0.455kg of water -solvent-, moles of CuI₂ are:

0.455kg ₓ (0.694 moles CuI₂ / kg) = 0.316 moles of CuI₂

Using molar mass, grams of CuI₂ in the solution are:

0.316moles CuI₂ ₓ (317.35g / mol) =

Answer:

The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.

Explanation:

1s^2

2s^2

2p^6

3s^2

3p^6

4s^2

3d^10

4p^5

Answer:

The new volume is 808L.

Explanation:

First, you need to set up your proportion, using Charles's law.

V1         V2

       =        

T1         T2

Then, substitute for the ones you already know.

625 L            V2

            =              

273 K          353 K

Then, you need to cross multiply.

625 x 353 = 273 x V2

Then, solve for V2.

220625 = 273 x V2

220625 ÷ 273 = V2

808 = V2

The final volume is 808L.

Answer:

[tex]\boxed{Butyne}[/tex]

Explanation:

Triple Bonds => So it is an alkyne

The suffix used will be "-yne"

4 Carbons => The prefix used will be "But-"

Combining the prefix and suffix, we get:

=> Butyne

Answer:

[tex]\boxed{\mathrm{Butyne}}[/tex]

Explanation:

Alkynes have triple bonds ≡. The molecule has one triple bond.

Suffix ⇒ yne

The molecule has 4 carbon atoms and 6 hydrogen atoms.

Prefix ⇒ But (4 carbons)

The molecule is Butyne.

[tex]\mathrm{C_4H_6}[/tex]

Answer:

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :

[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]

[tex]K_p = \dfrac{1}{ (22.20)}[/tex]

[tex]K_p[/tex] = 0.045

[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]

where;

R = gas constant = 8.314 × 10⁻³ kJ

[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]

[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]

[tex]\Delta G =207.6+ (-7.687048037)[/tex]

[tex]\Delta G =[/tex] 199.912952  kJ

ΔG ≅ 199.91 kJ

Answer:

Explanation:

Atmospheric pressure = partial pressure of nitrogen + partial pressure of oxygen + partial pressure of argon + partial pressure of trace element

putting the given values

Atmospheric pressure = 592 + 160 + 7 + 1

= 760 mm of Hg .

Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

Where 1 mole of the acid reacts per mole of the base.

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = 1.40x10⁻³ moles of H₃C-CH(OH)-COOH

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

Answer:

Atomic mass of Ag-107 = 106.94 amu

Explanation:

Let the mass of Ag-107 be y

Since the mass ratio of Ag-109/Ag-107 is 1.0187, the mass of Ag-109  is 1.0187 times heavier than the mass of Ag-107

Mass of Ag-109  = 1.0187y

Relative atomic mass of Ag = sum of (mass of each isotope * abundance)

Relative atomic mass of Ag = 107.87

Abundance of Ag-107 = 51.839% = 0.51839

Abundance of Ag-109 = 41.161% = 0.48161

107.87 = (y * 0.51839) + (1.018y * 0.48161)

107.87 = 0.51838y + 0.49027898y

107.87 = 1.00865898y

y = 107.87/1.00865898

y = 106.94 amu

Therefore, atomic mass of Ag-107 = 106.94 amu

Answer: B. The period number tells which is the highest energy level occupied by the electrons

Explanation:

Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.

hence, the correct option is B. The period number tells which is the highest energy level occupied  by the electrons.

The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.

The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.

When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.

Therefore, option B , second option is correct.

Learn more about   energy levels at;

https://brainly.com/question/20561440

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Answer:

There is insufficient information to know direction of these systems

Explanation:

Delta H of a reaction is defined as the amount of energy involved when it occurs. The ΔH < 0 represents the reaction will release energy and ΔH > 0 the reaction will absorb energy.

As you can see, ΔH doesn't give information about the direction of a reaction (Spontaneity). In fact, to know spontaneity of a reaction you must know ΔG involved in this reaction.

As the reactions have ΔH but not ΔG,

Answer:

The correct answer is - alternating and direct, in order.

Explanation:

Alternating current is is type of electric current that is characterized by the direction of the flow of electrons in continuously switches its directs in opposite manner at regular cycles. While direct current or DC is flow of the electrons that move from starting to end in one direction.

Spinning turbines always leads to the alternating electric current while only solar energy produces the direct current with the help of the solar panels.

Thus, the correct answer is -  alternating and direct, in order.

Answer:

1. alternating

2. direct

3. The sun heats up the atmosphere as Earth spins, creating areas of high and low temperature. This temperature difference causes wind to start moving through convection, which can then drive a wind turbine to produce electricity.

Explanation:

From Penn

Answer:

Produces Wind Currents

Explanation:

Answer:

produces wind currents

Explanation:

i just took the test and got it right :}

Answer:

0.641 g of Nitrogen are present in the mixture.

Explanation:

We use the Ideal Gases Law, to solve this question.

For the mixture:

P mixture . V mixture = mol mixture . R . T

We convert the T° to K →  23°C + 273 = 296 K

R = Ideal gases constant → 0.082 L.atm/mol.K

1 atm . 2L = mol mixture . 0.082 L.atm/mol.K  . 296K

2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles

We know that sum of partial pressure = 1

Partial pressure N₂ + Partial pressure O₂ = 1

1 - 0.722 atm = Partial pressure N₂ → 0.278 atm

We apply the mole fraction concept:

Partial pressure N₂ / Total pressure = Moles N₂ / Total moles

Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles

Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles

We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g

641 mg

Answer:

Kc = [H₂CO₃] / [CO₂]

Explanation:

Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.

Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.

Also, each specie must be powered to its reactant coefficient.

For example, for the reaction:

aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)

The equilibrium constant, kc is:

Kc = [D]ⁿ / [B]ᵇ[E]ˣ

You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants

Thus, for the reaction:

CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)

The Kc is:

Answer:

we wear cotton clothes because it helps to cool us down and remove the excess heat that causes us to feel hot.

Answer:

[tex]\boxed{\mathrm{view \: explanation}}[/tex]

Explanation:

We wear cotton clothes in the summer beacuse cotton absorbs and removes body moisture caused by the sweat and allows better air circulation than fabric clothes.

Answer:

⇒ Percent yield = 80 %

Explanation:

Given:

Actual yield = 36 g

Theoretical yield = 45 g

Find:

Percent yield

Computation:

⇒ Percent yield = [Actual yield / Theoretical yield] 100%

⇒ Percent yield = [36 / 45] 100%

⇒ Percent yield =[0.8] 100%

⇒ Percent yield = 80 %

Answer:

2 electrons

Explanation:

Answer:

Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable

Explanation:

Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.

The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).

So, decreasing order of stability of alkenes from most stable to least stable:

2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane

Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.

Please find the attached file.

Learn more about the 3-Bromo-2,3-dimethylpentane:

brainly.com/question/7139334

Answer:

See explanation

Explanation:

We have to start, remembering the mechanism behind the McLafferty rearrangement. The hydrogen in the gamma carbon (in this case, carbon 5) would be removed by a heterolytic rupture due to the cation-radical placed in the oxygen of the carbonyl group. Then we will have several heterolytic ruptures. Between carbons alpha and beta (in this case, 4 and 3) and a rupture in the carbonyl group. Due to these ruptures, two double bonds would be formed. One double bond in the alcohol cation-radical and the other one in the alkene.

See figure 1

I hope it helps!

Answer:

[tex]4.36~g~XY[/tex]

Explanation:

In this case, we can start with the reaction:

[tex]2X + Y_2~->~2XY[/tex]

If we check the reaction, we will have 2 X and Y atoms on both sides. So, the reaction is balanced. Now, the problem give to us two amounts of reagents. Therefore, we have to find the limiting reagent. The first step then is to find the moles of each compound using the molar mass:

[tex]3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X[/tex]

[tex]4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2[/tex]

Now, we can divide by the coefficient of each compound (given by the balanced reaction):

[tex]\frac{0.04~mol~X}{1}=~0.04[/tex]

[tex]\frac{0.0875~mol~Y_2}{2}=0.04375[/tex]

The smallest value is for "X", therefore this is our limiting reagent. Now, if we use the molar ratio between "X" and "XY" we can calculate the moles of XY, so:

[tex]0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY[/tex]

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol [tex]Y_2[/tex] = 48 g [tex]Y_2[/tex] (therefore 1 mol Y = 24 g Y). With this in mind the molar mass of XY would be 85+24 = 109 g/mol. With this in mind:

[tex]0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY[/tex]

I hope it helps!

Answer:

Heat = 7375J

Final temperature of the mixture = 27.35°C

Explanation:

In the reaction:

2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l) ΔH = –118 kJ

When 2 moles of HCl reacts with excess of Ba(OH)₂ there are released 118kJ.

In the reaction, moles of HCl and Ba(OH)₂ that reacts are:

Moles HCl = 0.250L ₓ (0.500 moles / L) = 0.125 moles HCl

Moles Ba(OH)₂ = 0.500L ₓ (0.500 moles / L) = 0.250 moles Ba(OH)₂

For a complete reaction of 0.125 moles of HCl you need:

0.125 mol HCl ₓ (1 mole Ba(OH)₂ / 2 moles HCl) = 0.0625 moles Ba(OH)₂

As you have 0.250 moles of Ba(OH)₂, this reactant is in excess

2 moles of HCl that react release 118kJ, 0.125 moles of HCl release:

0.125 moles HCl ₓ (118kJ / 2 moles) = 7.375kJ =

The heat released can be obtained with the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution, m its mass and ΔT change in temperature.

Replacing:

Q = C×m×ΔT

7375J = 4.18J/g°C×750.0g×ΔT

2.35°C = ΔT

As ΔT = Final T - Initial T:

2.35°C = Final T - 25.0°C

Answer:

0.167M

Explanation:

Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (NaOH in this case) and volume of the solution in liters.

To find molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution we need to convert mass of NaOH to moles (Using its molar mass: 40g/mol) and the mililiters of solution to liters (1L = 1000mL), thus:

Moles NaOH = 5.00g × (1mol/ 40g) = 0.125 moles NaOH = Moles solute

Liters solution = 750.0mL × (1L / 1000mL) = 0.7500L solution

And molariy is:

0.125 moles NaOH  / 0.7500L solution =

0.167M

Answer:

The correct answer is 51.2 degree C.

Explanation:

The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.  

Now the dHreaction = dHf (product) - dHf (reactant)  

= -909.3 - (-814)

dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.  

The heat change in calorimetry can be determined by using the formula,  

q = mass * specific heat capacity * change in temperature -----------(i)

Based on the given information, the density of H₂SO₄ is 1.060 g/ml

The volume of H₂SO₄ is 1 Liter

Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams

The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.  

ΔT = T₂ -T₁ = T₂ - 298.2 K

Now putting the values in equation (i) we get,  

95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)

(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K

T₂ = 298.2 K + 26 K

T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.