Consider the reaction. FeCl2(aq)+(NH4)2 SO4(aq)⟶FeSO4+2NH4Cl Identify the precipitate, or lack thereof, for the reaction. (A) FeSO4 (B) no precipitate (C) NH4Cl

Answer:

[tex]\boxed{\text{0.780 atm}}[/tex]

Explanation:

Hermione is pretty smart. She realizes that, according to Dalton's Law of Partial Pressures, each gas exerts its pressure independently of the others, as if the others weren't even there.

She shows Ron how to use the Ideal Gas Law to solve the problem.

pV = nRT

She collects the data:

V = 1.00 L; n = 0.0319 mol; T = 25.0 °C

She reminds him to convert the temperature to kelvins

T = (25.0 +273.15) K = 298.15 K

Then she shows him how to do the calculation.

[tex]p \times \text{1.00 L} = \text{0.0319 mol} \times \text{L}\cdot\text{atm}\cdot\text{0.082 06 K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\\\1.00p = \text{0.7805 atm}\\\\p = \textbf{0.780 atm}\\\\\text{The partial pressure of the nitrogen is } \boxed{\textbf{0.780 atm}}[/tex]

Isn't she smart?

The partial pressure of N2 is 0.78 atm.

The following are information contained in the question;

Volume (V) =  1.00 L

Total number of moles of the gases(n) = Number of moles of N2 + Number of moles of O2 + Nuber of moles of Ar = 0.0319 mol + 0.00856 mol + 0.000381 mol = 0.041 moles

Temperature(T) = 25.0◦C + 273 = 298 K

Using PV = nRT

Where;

P = pressure of the gas (the unknown)

V = volume of the sample = 1.00 L

T = absolute temperature = 298K

R = molar gas constant = 0.082 atmLK-1Mol-1

n = total number of moles present = 0.041 moles

Making P the subject of the formula and substituting values;

P = nRT/V

P =  0.041 moles × 0.082 atmLK-1Mol-1  × 298 K/1.00 L

P = 1 atm

Recall that partial pressure = mole fraction × total pressure

Mole fraction of N2 = 0.00856 mol /0.00856 mol  + 0.0319 mol + 0.000381 mol =

Partial pressure of N2 =0.0319 mol /0.041 moles ×  1 atm

Partial pressure of N2 = 0.78 atm

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Answer:

Kc = 9.52.

Explanation:

A + 2B ⇌ C,

Kc = [C]/[A][B]²,

Concentration:     [A]                [B]              [C]

At start:               0.3 M         1.05 M        0.55 M

At equilibrium:   0.3 - x        1.05 - 2x     0.55 + x

                            0.14 M        1.05 - 2x      0.71 M

∵ 0.3 M - x = 0.14 M.

∴ x = 0.3 M - 0.14 M = 0.16 M.

∴ [B] at equilibrium = 1.05 - 2x = 1.05 M -2(0.16) = 0.73 M.

∵ Kc = [C]/[A][B]²

∴ Kc = (0.71)/(0.14)(0.73)² = 9.5166 ≅ 9.52.

Answer: The correct answer is Option D.

Explanation:

To calculate the pressure of the Helium gas, we use the equation:

[tex]p_i=\chi \times P[/tex]

where,

[tex]p_i[/tex] = partial pressure of the gas

[tex]\chi[/tex] = mole fraction of the gas

P = total pressure

We are given:

Sum of all the mole fraction is always equal to 1.

[tex]\chi_{Ne}=0.47\\\chi_{Ar}=0.23\\\chi_{He}=(1-(0.47+0.23))=0.3\\P=7.85atm[/tex]

Putting values in above equation, we get:

[tex]p_i=0.3\times 7.85=2.4atm[/tex]

Hence, the correct answer is Option D.

To find the pressure of He in the mixture, calculate the mole fraction of He by subtracting the mole fractions of Ne and Ar from 1. Then, use Dalton's law of partial pressures to find the pressure of He by multiplying the total pressure of the mixture by the mole fraction of He.

To find the pressure of He in the mixture, we need to first calculate the mole fraction of He. Since the mole fractions of Ne and Ar are given, we can calculate the mole fraction of He by subtracting their mole fractions from 1.

Therefore, the mole fraction of He is

1 - 0.47 - 0.23 = 0.3.

Next, we can use Dalton's law of partial pressures to find the pressure of He. According to Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Since we know the total pressure of the mixture is 7.85 atm, we can set up the equation:

Pressure of He = Total pressure of mixture × Mole fraction of He = 7.85 atm × 0.3 = 2.355 atm.

Therefore, the pressure of He is approximately 2.4 atm (option D).

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Answer: The mass of butane reacting with oxygen gas is 9.76 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]    .....(1)

Given mass of oxygen gas = 35.1 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of oxygen gas}=\frac{35.1g}{32g/mol}=1.096mol[/tex]

For the given chemical reaction:

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

By stoichiometry of the reaction:

13 moles of oxygen gas is reacting with 2 moles of butane.

So, 1.096 moles of oxygen gas will react with = [tex]\frac{2}{13}\times 1.096=0.168moles[/tex] of butane.

Now, calculating the mass of butane from equation 1, we get:

Molar mass of butane = 58.12 g/mol

Moles of butane = 0.168 moles

Putting values in equation 1, we get:

[tex]0.168mol=\frac{\text{Mass of butane}}{58.12g/mol}\\\\\text{Mass of butane}=9.76g[/tex]

Hence, the mass of butane reacting with oxygen gas is 9.76 grams.

Answer : The equilibrium constant [tex]K_c[/tex] for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of [tex]Br_2[/tex].

[tex]\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}[/tex]

[tex]\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M[/tex]

Now we have to calculate the dissociated concentration of [tex]Br_2[/tex].

The balanced equilibrium reaction is,

                              [tex]Br_2(g)\rightleftharpoons 2Br(aq)[/tex]

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of [tex]Br_2[/tex] = [tex]\alpha[/tex] = 1.2 %

So, the dissociate concentration of [tex]Br_2[/tex] = [tex]C\alpha=1.731M\times \frac{1.2}{100}=0.2077M[/tex]

The value of x = 0.2077 M

Now we have to calculate the concentration of [tex]Br_2\text{ and }Br[/tex] at equilibrium.

Concentration of [tex]Br_2[/tex] = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of [tex]Br[/tex] = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

[tex]K_c=\frac{[Br]^2}{[Br_2]}[/tex]

Now put all the values in this expression, we get :

[tex]K_c=\frac{(0.4154)^2}{1.5233}=0.1133[/tex]

Therefore, the equilibrium constant [tex]K_c[/tex] for the reaction is, 0.1133

The equilibrium constant (Kc) for the dissociation reaction of Br₂ into 2Br at high temperature, given an initial amount of 1.35 moles in a 0.780 L flask and 3.60% dissociation, is calculated to be approximately 0.0093.

You've been tasked with calculating the equilibrium constant (Kc) for the dissociation of bromine into bromine atoms at high temperature using the given data: An initial amount of 1.35 moles of Br₂ in a 0.780 L flask with 3.60 percent dissociation.

Determine the initial concentration of Br₂ by dividing moles by volume: CBr₂(initial) = moles / volume.

Calculate the amount dissociated by multiplying the initial concentration by the percentage dissociated.

Determine concentrations at equilibrium using the stoichiometry of the reaction.

Use the formula Kc = [Br]² / [Br₂] to find the equilibrium constant.

Let's calculate it:

CBr₂(initial) = 1.35 moles / 0.780 L = 1.731 moles/L.

Amount dissociated = 1.731 moles/L x 3.60% = 0.06232 moles/L.

At equilibrium, [Br₂] = 1.731 - 0.06232 = 1.6687 moles/L, and [Br] = 2 x 0.06232 moles/L = 0.12464 moles/L.

The equilibrium constant Kc = (0.12464)^2 / 1.6687 = 0.00930628672.

The equilibrium constant Kc for the given reaction is approximately 0.0093.

Temperature is defined as  a measure of the average kinetic energy of the individual atoms or molecules composing a substance.

Answer: The [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

[tex]CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)[/tex]

We are given:

[tex]\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol[/tex]

To calculate [tex]\Delta G^o_{rxn}[/tex] for the reaction, we use the equation:

[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)][/tex]

For the given equation:

[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})][/tex]

Putting values in above equation, we get:

[tex]\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J[/tex]

Conversion factor used = 1 kJ = 1000 J

The expression of [tex]K_p[/tex] for the given reaction:

[tex]K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}[/tex]

We are given:

[tex]p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm[/tex]

Putting values in above equation, we get:

[tex]K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85[/tex]

To calculate the gibbs free energy of the reaction, we use the equation:

[tex]\Delta G=\Delta G^o+RT\ln K_p[/tex]

where,

[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?

[tex]\Delta G^o[/tex] = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = [tex]8.314J/K mol[/tex]

T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

[tex]\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol[/tex]

Hence, the [tex]\Delta G[/tex] for the reaction is 54.425 kJ/mol

The change in free energy for the reaction is 54.4  kJ/mol.

The change in free energy under nonstandard conditions can be obtained from the chage in free energy under standard conditions using the formula;

ΔG = ΔG° + RTlnK

Now ΔG° is obtained from;

ΔG°reaction = 2[−204.9kJ/mol] - [(−394.4kJ/mol) + (−62.3kJ/mol)]

ΔG°reaction =(-409.8) + 456.7

ΔG°reaction = 46.9 kJ/mol

Q =PCOCl2^2/ PCO2 * PCCl4

Q = (0.735)^2/ 0.140 * 0.185

Q = 20.8

ΔG = 46.9 * 10^3 + [8.314 * 298 * ln(20.8)]

ΔG = 54.4  kJ/mol

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Answer: The correct answer is Option B.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ....(1)

Given mass of nitrogen gas = 10 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of iron oxide}=\frac{10g}{28g/mol}=0.357mol[/tex]

The given chemical reaction follows:

[tex]N_2+O_2\rightarrow 2NO[/tex]

As, oxygen gas is present in excess. Thus, it is considered as an excess reagent and nitrogen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of nitrogen gas produces 2 moles of nitrogen oxide.

So, 0.357 moles of nitrogen gas will produce = [tex]\frac{2}{1}\times 0.357=0.714mol[/tex] of nitrogen oxide.

Now, calculating mass of nitrogen oxide by putting values in equation 1, we get:

Moles of nitrogen oxide = 0.714 mol

Molar mass of nitrogen oxide = 30 g/mol

Putting values in equation 1, we get:

[tex]0.714mol=\frac{\text{Mass of nitrogen oxide}}{30g/mol}\\\\\text{Mass of nitrogen oxide}=21.4g[/tex]

Hence, the correct answer is Option B.

Answer:

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Explanation:

Cu + 2AgNO₃  =>  Cu(NO₃)₂ + 2Ag

? moles Cu + 5.8 moles AgNO₃ =>    Cu(NO₃)₂ + 2Ag

Note coefficient of Cu° vs coefficient of AgNO₃

=> 1 mole Cu° < 2 moles AgNO₃ ...

=> Since moles of Cu° are smaller than moles of AgNO₃ in the given equation, the moles of Cu° needed to react with 5.8 moles AgNO₃ will be smaller than the 5.8 by the ratio of coefficients that will make 5.8 smaller. That is ...

moles of Cu° needed =  ½ (5.8 moles) = 2.9 moles Cu° needed.

Note: Using 2/1(5.8) will make value greater than 5.8 and incorrect.

One can also set up a ratio relationship as follows...

1 mole Cu° <=> 2 moles AgNO₃ (fm equation)

? mole Cu° <=> 5.8 moles AgNO₃ (problem)

=> (1 mole Cu°)/X=(2 mole AgNO₃)/(5.8 moles AgNO₃)

=> X = (1 mole Cu°)(5.8 mole AgNO₃)/2 mole AgNO₃

        = ½(5.8) mole Cu° = 2.9 mole Cu°

Final answer:

2.9 moles of Cu are needed to react with 5.8 moles of AgNO3 according to the stoichiometry of the balanced chemical reaction.

Explanation:

To determine how many moles of Cu (copper) are required to react with 5.8 moles of AgNO3 (silver nitrate), we use the stoichiometry of the balanced chemical reaction:

Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

This balanced equation tells us that one mole of Cu reacts with two moles of AgNO3. If we have 5.8 moles of AgNO3, then the amount of Cu required is calculated by dividing the moles of AgNO3 by 2:

(5.8 moles AgNO3) / (2 moles AgNO3/moles Cu) = 2.9 moles of Cu

Therefore, 2.9 moles of Cu are needed to react with 5.8 moles of AgNO3.

Answer: The molarity of calcium hydroxide in the solution is 0.1 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HBr[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]

We are given:

[tex]n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL[/tex]

Putting values in above equation, we get:

[tex]1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M[/tex]

Hence, the molarity of [tex]Ca(OH)_2[/tex] in the solution is 0.1 M.

Answer: The correct answer is Option D.

Explanation:

Atomic mass of an atom is defined as the sum of number of neutrons and number of protons that are present in an atom. It is represented as 'A'.

Atomic number = Number of protons + Number of neutrons

We are given:

Number of protons = 6

Number of neutrons = 6

Number of electrons = 6

Atomic mass = 6 + 6 = 12

Hence, the correct answer is Option D.

Answer: The correct answer is Option B.

Explanation:

Precipitate is defined as insoluble solid substance that emerges when two different aqueous solutions are mixed together. It usually settles down at the bottom of the solution after sometime.

For the given chemical equation:

[tex]CaCl_2(aq.)+K_2CO_3(aq.)\rightarrow CaCO_3(s)+2KCl(aq.)[/tex]

The products formed in the reaction are calcium carbonate and potassium chloride. Out of the two products, one of them is insoluble which is calcium carbonate. Thus, it is considered as a precipitate.

Hence, the correct answer is Option B.

In the double displacement reaction between CaCl₂ and K₂CO₃, CaCO₃ is the precipitate (B).

Let's consider the following double displacement reaction.

CaCl₂(aq) + K₂CO₃(aq) ⟶ CaCO₃(s) + 2 KCl(aq)

Regarding solubility rules, we know that:

With this information, we can conclude that CaCO₃ is a precipitate.

In the double displacement reaction between CaCl₂ and K₂CO₃, CaCO₃ is the precipitate (B).

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Answer:

C₆H₄O₂

Explanation:

Given parameters:

Mass of the unknown compound = 0.315g

Atoms contained in the compound C, H and O

Mass of CO₂ produced = 0.771g

Mass of H₂O = 0.105g

Molar mass of compound = 108gmol⁻¹

The complete combustion of most hydrocarbon compounds like this would produce carbon dioxide and water only.

Solution

We first find the mass of the Carbon, hydrogen and Oxygen in the compounds given.

Mass of carbon in compound = [tex]\frac{Molar mass of Carbon}{Molar mass of CO_{2} }[/tex] x mass of CO₂

Molar mass of C = 12

Molar mass of CO₂ = 12 + (16x2) = 44

Mass of CO₂ produced = 0.771g

Mass of carbon in compound =  [tex]\frac{12}{44 }[/tex] x 0.771=0.2103g

Mass of H in compound =  [tex]\frac{Molar mass of H}{Molar mass of H_{2}O }[/tex] x mass of H₂O

Molar mass of H in H₂O = 2

Molar mass of H₂O = 2 + 16 = 18

Mass of H₂O = 0.105g

Mass of H in compound =  [tex]\frac{2}{18}[/tex] x 0.105 = 0.012g

Now, to find the mass of oxygen in the compound, we sum the mass H and C and subtract from the mass of the compound given:

Mass of oxygen = 0.315 - (0.2103 + 0.012)= 0.0927g

We then continue to derive the empirical formula of the compound.

                                    C                       H                         O

mass of

atoms                      0.2103                0.012                   0.0927

Moles                  0.2103/12             0.012/1                 0.0927/16

                               0.018                   0.012                    0.006

Dividing by

the smallest       0.018/0.006       0.012/0.006           0.006/0.006

                                   3                           2                                1

The empirical formula of the compound is C₃H₂O

From the empirical formula, we can find the compound from the given molar mass:

Molecular formula = (Empirical formula)n

Where n is the number of repeating times of the empirical formula present in one mole of the molecule.

Therefore n = [tex]\frac{molar mass of the compound}{molar mass of the empirical formula of the compound}[/tex]

 Molar mass of the empirical formula C₃H₂O = (12x3) + (2x1) + (16) = 54g/mol

               n = [tex]\frac{108}{54}[/tex] = 2

The molecular formula of the compound is = 2(C₃H₂O) = C₆H₄O₂

Final answer:

To identify the unknown organic compound from the combustion analysis data, one must calculate the moles of carbon and hydrogen from the CO2 and H2O produced, determine the empirical and eventually the molecular formula, and verify which compound options match the calculated formula and given molar mass.

Explanation:

When an unknown organic compound undergoes complete combustion, the products are CO2 and H2O. The amounts of these products can be used to determine the empirical formula of the compound.

First, from the 0.771 g of CO2 produced, we can calculate the moles of carbon, since each mole of CO2 contains one mole of carbon atoms. Similarly, from the 0.105 g of H2O produced, we can determine the moles of hydrogen, as each mole of H2O contains two moles of hydrogen atoms.

To find the identity of the unknown compound, we need to calculate the moles of each element and then the empirical formula. After finding the empirical formula, we can use the given approximate molar mass to find the molecular formula, which, in turn, will help us identify the compound. The compounds given as the options for the identity of the unknown must be checked against the calculated molecular formula.

Answer:

See Explanation

Explanation:

a. pH of 1M HOAc(aq)

                HOAc      ⇄      H⁺   +      OAcˉ

C(eq)         1.0M                 x               x

Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵

=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M

=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37

b. pH of 0.10M CH₃NH₃OH(aq)

                CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴

C(eq)             0.10M                x              x

=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M

=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M

=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18

=> pH = 14 – pOH = 14 – 2.18 = 11.82

c. pH of 0.30M HOAc/0.10M OAcˉ(aq)

                HOAc      ⇄      H⁺   +      OAcˉ

C(eq)        0.30M               x           0.10M

=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]

          = 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M

=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26

To calculate pH for acetic acid, protonated methylamine, and a formic acid/formate mixture, we utilize an ICE table for the weak acids and the Henderson-Hasselbalch equation for the buffer solution, considering their respective pKa values.

To calculate the pH values of the given solutions, we apply the pH formula and utilize ICE (Initial, Change, Equilibrium) tables for weak acids and bases. Using the provided pKa values, we can determine the pH for each solution:

These calculations showcase the principles of acid-base equilibrium and buffer systems in solution chemistry.

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Try this suggested solution (the figures are not provided).

The answer is marked with red colour.

Answer: The specific heat of substance A is 1.1 J/g°C

Explanation:

When substance A is mixed with substance B, the amount of heat released by substance B (initially present at high temperature) will be equal to the amount of heat absorbed by substance A (initially present at low temperature)

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]       ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of substance A = 6.07 g

[tex]m_2[/tex] = mass of substance B = 26.1 g

tex]T_{final}[/tex] = final temperature = 47.0°C

[tex]T_1[/tex] = initial temperature of substance A = 20.7°C

[tex]T_2[/tex] = initial temperature of substance B = 52.8°C

[tex]c_1[/tex] = specific heat of substance A = ?

[tex]c_2[/tex] = specific heat of substance B = 1.17 J/g°C

Putting values in equation 1, we get:

[tex]6.07\times c_1\times (47-20.7)=-[26.1\times 1.17\times (47-52.8)][/tex]

[tex]c_1=1.1J/g^oC[/tex]

Hence, the specific heat of substance A is 1.1 J/g°C

Answer:

E = 1.602v

Explanation:

Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …

             Zn⁰(s) => Zn⁺²(aq) + 2 eˉ

2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)          

_____________________________

Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)

Given E⁰ = 1.562v

Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044

E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v

Answer:

E = 1.602 V

Explanation:

Let's consider the following galvanic cell.

Zn(s) | Zn²⁺(aq, 1.00 × 10⁻³ M) || Ag⁺(aq, 0.150 M) | Ag(s)

The corresponding half-reactions are:

Zn(s) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 e⁻

2 Ag⁺(aq, 0.150 M) + 2 e⁻ → 2 Ag(s)

The overall reaction is:

Zn(s) + 2 Ag⁺(aq, 0.150 M) → Zn²⁺(aq, 1.00 × 10⁻³ M) + 2 Ag(s)

We can find the cell potential (E) using the Nernst equation.

E = E° - (0.05916/n) . log Q

where,

E°: standard cell potential

n: moles of electrons transferred

Q: reaction quotient

E = E° - (0.05916/n) . log [Zn²⁺]/[Ag⁺]²

E = 1.562 V - (0.05916/2) . log (1.00 × 10⁻³)/(0.150)²

E = 1.602 V

Answer: The amount of heat released is 56 MJ.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

Given mass of [tex]C_2H_6[/tex] = 12 kg = 12000 g    (Conversion factor: 1 kg = 1000 g)

Molar mass of [tex]C_2H_6[/tex] = 30 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }C_2H_6=\frac{12000g}{30g/mol}=400mol[/tex]

The chemical reaction for hydrogenation of ethene follows the equation:

[tex]C_2H_4+H_2\rightarrow C_2H_6[/tex]

By Stoichiometry of the reaction:

When 1 mole of ethane releases 140 kJ of heat.

So, 400 moles of ethane will release = [tex]\frac{140}{1}\times 400=56000kJ[/tex] of heat.

Converting this into Mega joules, using the conversion factor:

1 MJ = 1000 kJ

So, [tex]\Rightarrow 56000kJ\times (\frac{1MJ}{1000kJ})=56MJ[/tex]

Hence, the amount of heat released is 56 MJ.

Answer:

The value of the dissociation constant will be:[tex]3.94\times 10^{-4}[/tex].

Explanation:

                [tex]AsH\rightleftharpoons As^++H^+[/tex]

At ,t= 0      c              0     0

At eq'm   (c-x)            x     x

Concentration of aspirin = c

[tex]c=\frac{2.00 g}{180 g/mol\times 0.600 L}=0.01851 M [/tex]

Expression for dissociation constant will be given as:

[tex]K_a=\frac{[H^+][As^+]}{[AsH]}=\frac{x\times c}{(c-x)}=\frac{x^2}{(c-x)}[/tex]..(1)

The pH of the solution  = 2.60

The pH of the solution is due to free hydrogen ions whcih come into solution after partial dissociation of aspirin.

[tex]pH=2.60=\log[H^+]=-\log[x][/tex]

[tex]x=0.002511 M[/tex]

Putting value of x in (1).

[tex]K_a=\frac{x^2}{(c-x)}=\frac{(0.002511 M)^2}{(0.01851 M-0.002511 M)}[/tex]

[tex]K_a=3.94\times 10^{-4}[/tex]

The value of the dissociation constant will be:[tex]3.94\times 10^{-4}[/tex].

Answer:

-667Kj see attached

Explanation:

Final answer:

To find the lattice energy for CsCl, we sum the given energies related to sublimation, bond dissociation, ionization, and electron affinity, then subtract the overall energy of formation. The magnitude of lattice energy for CsCl is calculated to be 624 kJ/mol.

Explanation:

To calculate the lattice energy for CsCl, we can use the provided enthalpy values in a Born-Haber cycle. We have:

Heat of sublimation for Cs: +76 kJ/mol

Bond dissociation energy for 1/2Cl₂: +121 kJ/mol

Ionization energy (Ei1) for Cs: +376 kJ/mol

Electron affinity (Eea) for Cl:
349 kJ/mol

Overall energy for the formation of CsCl:
443 kJ/mol

The lattice energy is the remaining term that balances the Born-Haber cycle equation, which summarizes the energy changes that occur when an ionic solid forms. Thus, we calculate the lattice energy (U) using the equation:

U = Sublimation energy + Bond dissociation energy  + Ionization energy + Electron affinity + Formation energy

U = 76 kJ/mol + 121 kJ/mol + 376 kJ/mol - 349 kJ/mol - 443 kJ/mol

U = 624 kJ/mol, which is the magnitude of the lattice energy for CsCl.

Answer:

A compressed spring

Explanation:

A compressed spring has potential energy only and no kinetic energy.

This is because kinetic energy is only possessed by particles in motion.

Energy in a compressed spring= -1/2kx² where x is the displacement.

In this equation there is no velocity so there is no kinetic energy.

Answer:

A compressed spring

Explanation:

Option C is correct. The potential energy is the energy stored in a compressed spring. This potential energy depends on the spring constant and the distance traveled by the spring as it is stretched. The work that is done in stretching a spring gets stored in the compressed spring as potential energy.

Option A is incorrect. As sound wave is a mechanical wave that carries both potential and kinetic energy.  

Option B is incorrect. Arrow shot from a bow has kinetic energy.  

Option D is incorrect. A vibrating atom has vibrational energy.  

Answer: The correct answer is Option B.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]        (at constant volume)

where,

[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.

[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.

We are given:

Conversion factor:  [tex]T(K)=T(^oC)+273[/tex]

[tex]P_1=1.85atm\\T_1=25^oC=(25+273)K=298K\\P_2=?atm\\T_2=7.5^oC=(7.5+273)K=280.5[/tex]

Putting values in above equation, we get:

[tex]\frac{1.85atm}{298K}=\frac{P_2}{280.5K}\\\\P_2=1.74atm[/tex]

Hence, the correct answer is Option B.

Final answer:

Using Gay-Lussac's Law, after converting the temperatures to Kelvin and applying the given initial pressure and temperatures, the new pressure of the sports ball when the temperature drops to 7.5 °C is calculated to be B) 1.74 atm.

Explanation:

The problem you've presented involves the concept of gas laws, specifically Gay-Lussac's Law, which states that the pressure of a gas varies directly with its absolute temperature, provided the volume does not change. This law can be mathematically represented as P1/T1 = P2/T2, where P1 and P2 are the initial and final pressures, and T1 and T2 are the initial and final temperatures in Kelvin.

To solve for the new pressure (P2), we first convert the temperatures from °C to Kelvin: T1 = 25 °C + 273.15 = 298.15 K, and T2 = 7.5 °C + 273.15 = 280.65 K. Then we rearrange the formula to solve for P2: P2 = P1 * (T2/T1). Substituting the given values, P2 = 1.85 atm * (280.65 K / 298.15 K) = 1.74 atm.

Answer:

[tex]\boxed{1.19 \times 10^{-4} \text{ s}^{-1}}[/tex]

Explanation:

1. Determine the order of reaction

The question gives us a hint: the units of k are s⁻¹. This suggests a first order rate law.

To confirm, we plot ln(p) vs t.  We should get a straight line with slope = -k.

Here are your data with the pressures converted to natural logarithms.

[tex]\begin{array}{rcc}\textbf{t/s} & \textbf{p/mmHg} &\textbf{ln(p)}\\0 & 400 & 5.99\\2000 & 316 & 5.76\\4000 & 248 & 5.51\\6000 & 196 & 5.28\\8000 & 155 & 5.04\\10000 & 122 & 4.80\\\end{array}[/tex]

We get the graph shown below.

2. Determine the rate constant

The points fit so well that we can just use the end points to determine the slope.

[tex]\text{slope} = \dfrac{y_{2} - y_{1}}{ x_{2} - x_{1} } = \dfrac{4.80 - 5.99}{10 000 - 0} = -1.19 \times 10^{-4} \text{ s}^{-1}}\\\\k = \boxed{1.19 \times 10^{-4} \text{ s}^{-1}}[/tex]

The value of the rate constant for the reaction based on the given pressures  =  [tex]1.19 * 10^{-4} s^{-1}[/tex]

First step : determine the order of reaction

The reaction is a first-order reaction because the S.I. unit of K = s⁻¹  in the question

next step : Plot the value of  In( p ) vs Time

Where:  P = pressure ( mmHg )

            In ( p ) =  natural Logarithm of P values

Graph is attached below

Final step : calculate the value of the slope of the graph

slope = Δ y / Δ x

         = ( 4.8 - 5.99 ) / ( 10000 - 0 )

         = -1.19 / 10000  

∴ The value of  K ( rate constant )  = 1.19  *  10^{-4} s^{-1}

Hence we can conclude that the The value of the rate constant for the reaction based on the given pressures  =  [tex]1.19 * 10^{-4} s^{-1}[/tex]

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Final answer:

The theoretical yield of MnO2, given 2 moles of Mn and 2 moles of O2, is calculated using stoichiometry and is found to be approximately 1.33 moles based on O2 being the limiting reactant.

Explanation:

To calculate the theoretical yield of MnO₂ from the reaction 2 Mn(s) + 3 O₂ (g) → MnO₂(s), with the initial quantities of 2 mol Mn and 2 mol O₂, we will use stoichiometry. First, we write the balanced chemical equation:

2 Mn(s) + 3 O₂(g) → 2 MnO₂(s)

This equation tells us that 2 moles of Mn react with 3 moles of O₂ to produce 2 moles of MnO₂. Therefore, Mn and O₂ react in a 2:3 mole ratio to produce MnO₂.

The stoichiometry shows the molar ratio of Mn to MnO₂ is 1:1. Since we have 2 moles of Mn, we can produce 2 moles of MnO₂, assuming Mn is the limiting reactant. However, we must also consider O₂. With 2 moles of O₂, the stoichiometry suggests every 3 moles of O₂ produce 2 moles of MnO₂. The limiting reactant is O₂ since it will limit the formation of MnO₂ to 4/3 moles (which is the amount of MnO₂ formed from 2 moles of O₂ based on the 3:2 ratio of O₂ to MnO₂).

Therefore, the theoretical yield of MnO₂ in moles is 4/3 moles or approximately 1.33 moles. This is based on the stoichiometric calculations from the balanced equation taking into account the initial quantities of both reactants and identifying the limiting reactant.

Answer: The correct answer is Option A.

Explanation:

We are given:

4.5 wt % of Pb means that 4.5 grams of lead is present in 100 g of alloy.

95.5 wt % of Sn means that 95.5 grams of tin is present in 100 g of alloy.

To calculate the atom percent of any compound in a mixture, we use the equation:

[tex]\text{atom }\%=\frac{\text{Moles of compound}\times N_A}{\text{Total number of moles of mixture}\times N_A}\times 100[/tex]

where,

[tex]N_A[/tex] = Avogadro's number

Moles of a compound is given by the formula:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of lead = 4.5 g

Molar mass of lead = 207.19 g/mol

[tex]\text{Atom percent of lead}=\left(\frac{\frac{4.5g}{207.17g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of lead}=2.6\%[/tex]

Given mass of tin = 95.5 g

Molar mass of tin = 118.71 g/mol

[tex]\text{Atom percent of Tin}=\left(\frac{\frac{95.5g}{118.71g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of Tin}=97.4\%[/tex]

Hence, the correct answer is Option A.

The composition, in atom percent, of an alloy consisting of 4.5 wt% Pb and 95.5 wt% Sn is 2.6 at% Pb and 97.4 at% Sn. The correct answer is (A).

The atomic weights for Pb and Sn are 207.19 g/mol and 118.71 g/mol, respectively.

To determine the atom percent, we need to follow these steps:

Moles of Pb: 4.5 g / 207.19 g/mol = 0.0217 mol

Moles of Sn: 95.5 g / 118.71 g/mol = 0.8047 mol

Total moles = 0.0217 mol (Pb) + 0.8047 mol (Sn) = 0.8264 mol

Atom percent of Pb: (0.0217 mol / 0.8264 mol) * 100 ≈ 2.6 at% Pb

Atom percent of Sn: (0.8047 mol / 0.8264 mol) * 100 ≈ 97.4 at% Sn

Thus, the correct answer is (A) 2.6 at% Pb and 97.4 at% Sn.

Answer:

The 1st functional group is secondary amine and the 2nd fun. group is tertiary amine.

Explanation:

Lidocaine does not contain amide functional group in its composition as nitrogen is associated only with carbon.

In lidocaine, Functional group 1 is a secondary amine because it has two hydrocarbon groups attached to the nitrogen atom, and Functional group 2 is an amide due to the presence of a carbonyl group bonded to the nitrogen atom.

To classify the nitrogen-containing functional groups in lidocaine, it is important to understand the structure of amines and amides. Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom, with a primary amine having one alkyl or aryl group, a secondary amine having two, and a tertiary amine having three. An amide is a functional group with a carbonyl group (C=O) bonded to a nitrogen atom. In the structure of lidocaine, Functional group 1 is a secondary amine because it has two carbon atoms directly bonded to the nitrogen atom. Functional group 2 is an amide, identifiable by its carbonyl group bonded to the nitrogen atom.

Answer : The number of moles of [tex]H_2O[/tex] produced are, 0.66 mole.

Explanation : Given,

Given moles of [tex]NH_3[/tex] = 0.44 mole

The given balanced chemical reaction is,

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

From the given balanced chemical reaction, we conclude that

As, 4 moles of [tex]NH_3[/tex] react to give 6 moles of [tex]H_2O[/tex]

So, 0.44 moles of [tex]NH_3[/tex] react to give [tex]\frac{6}{4}\times 0.44=0.66[/tex] moles of [tex]H_2O[/tex]

Therefore, the number of moles of [tex]H_2O[/tex] produced are, 0.66 mole.

To neutralize the 0.50 M H2SO4, calculate the moles of NaHCO3 required using the mole ratio from the balanced equation and then convert the moles to grams.

To neutralize the 0.50 M H2SO4, we need to understand the stoichiometry of the reaction.

From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaHCO3.

Using the molarity and volume of the H2SO4 solution, we can calculate the moles of H2SO4, and then use the mole ratio to find the moles of NaHCO3 required to neutralize it. Finally, we can convert the moles of NaHCO3 to grams by using the molar mass of NaHCO3.

The correct answer is C) 19 g.

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Answer:

Case I => %C-14 remaining = 77% => Age of artifact = 2200 yrs

Case II => %C-14 remaining = 6.2% => Age of artifact = 23,000 yrs

Explanation:

Given:

Half-Life C-14 = 5730 yrs

=> Rate Constant = k = 0.693/t(1/2) = (0.693/5730)yrs⁻¹ = 1.2 x 10⁻⁴ yrs⁻¹

NOTE => All radioactive decay is 1st order kinetics.

=> A = A₀eˉᵏᵗ  (classic 1st order decay equation)

- A = remaining activity

-A₀ = initial activity

- k = rate constant

- t = time of decay (or, age of object of interest; i.e., not everything is organic but the 1st order decay equation is good for non-organic objects (rocks) also. Analysts just use a different decay standard => K-40 → Ar-40 + β).

Solving the decay equation for time (t) ...

t = ln(A/A₀)/-k

Applying to problem cases...

Case I => %C-14 remaining = 77%

t = ln(A/A₀)/-k = ln(77/100)/-1.2x10⁻⁴ years = 2178 yrs ~ 2200 yrs

Case II => %C-14 remaining = 6.2%

t = ln(A/A₀)/-k = ln(6.2/100)/-1.2x10⁻⁴ years = 23,172 yrs ~ 23,000 yrs

Answer:

[tex]\boxed{\text{a) 2160 yr ago; b) 23 000 yr ago}}[/tex]

Explanation:

Two important equations in radioactive decay are

[tex](1) \qquad \ln \dfrac{N_{0} }{N_{t}} = kt\\\\(2) \qquad t_{\frac{1}{2}} = \dfrac{\ln2}{k }[/tex]

We use them for carbon dating.

a) The dye

Data:

[tex]t_{\frac{1}{2}} = \text{5730 yr}\\\\N_{t} = 0.77 N_{0}[/tex]

Calculations:

[tex]\text{From Equation (2)}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{\text{5730 yr}} = 1.21 \times 10^{-4} \text{ yr}^{-1}\\\\\text{From Equation (1)}\\\\\ln \dfrac{N_{0} }{0.77N_{0}} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\\ln\dfrac{1}{0.77} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\-\ln0.77 = 0.261 = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\t = \dfrac{0.261}{ 1.21 \times 10^{-4} \text{ yr}^{-1}} = \textbf{2160 yr}\\\\\text{The cloth was painted } \boxed{\textbf{2160 yr}}\text{ ago}[/tex]

b) The wood

Data:

[tex]N_{t} = 0.062 N_{0}[/tex]

Calculations:

[tex]\text{From Equation (1)}\\\\\ln \dfrac{N_{0} }{0.062N_{0}} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\\ln\dfrac{1}{0.0.062} = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\-\ln0.062 = 2.78 = 1.21 \times 10^{-4}t \text{ yr}^{-1}\\\\t = \dfrac{2.78}{ 1.21 \times 10^{-4} \text{ yr}^{-1}} = \textbf{23 000 yr}\\\\\text{The wood was cut} \boxed{\textbf{23 000 yr}}\text{ ago}[/tex]

To find the number of moles of CO needed, we use the balanced equation and convert the mass of Fe2O3 to moles. The number of moles of CO needed is 75.12 mol. To find the number of moles of each product formed, we use the same approach and find that 50.08 moles of Fe and 75.12 moles of CO2 are formed.

To find the number of moles of CO needed to react with 4.00 kg of Fe2O3, we can use the balanced equation:

Fe2O3 + 3CO --> 2Fe + 3CO2

The molar mass of Fe2O3 is 159.69 g/mol. Converting 4.00 kg to grams gives us 4000 g. Using the molar mass of Fe2O3, we can calculate the number of moles of Fe2O3 as follows:

moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3

moles of Fe2O3 = 4000 g / 159.69 g/mol = 25.04 mol

Since the ratio of Fe2O3 to CO in the balanced equation is 1:3, we can use this ratio to find the number of moles of CO:

moles of CO = moles of Fe2O3 x (3 mol CO / 1 mol Fe2O3)

moles of CO = 25.04 mol x (3 mol CO / 1 mol Fe2O3) = 75.12 mol CO

Therefore, 75.12 moles of CO are needed to react with 4.00 kg of Fe2O3.

To find the number of moles of each product formed, we can use the same approach. Since the balanced equation tells us that the ratio of Fe2O3 to Fe is 1:2, and the ratio of Fe2O3 to CO2 is 1:3, the number of moles of Fe and CO2 will be twice the number of moles of Fe2O3:

moles of Fe = 2 x 25.04 mol = 50.08 mol

moles of CO2 = 3 x 25.04 mol = 75.12 mol

Therefore, 50.08 moles of Fe and 75.12 moles of CO2 are formed.

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(a) If 4.00 kg FeO3 are available to react then 37.43 moles of CO are needed. (b) 49.90 moles of Fe and 37.43 moles of CO₂ are formed.

a) First, we need to convert the mass of Fe₂O₃ to moles. The molar mass of Fe₂O₃ is approximately 159.69 g/mol, so 4.00 kg of Fe₂O₃ is about 24.95 moles.

The balanced chemical equation for the reaction is:

2Fe₂O₃ + 3CO → 4Fe + 3CO₂

From this equation, we can see that 2 moles of Fe₂O₃ react with 3 moles of CO. Therefore, the number of moles of CO needed is:

moles of CO = moles of Fe₂O₃ × (moles of CO / moles of Fe₂O₃) = [tex]24.95 * \frac{3}{2} = 37.43 moles[/tex]

b) From the balanced chemical equation, we can see that 2 moles of Fe₂O₃ produce 4 moles of Fe and 3 moles of CO₂. Therefore, the number of moles of each product formed is:

moles of Fe = moles of Fe₂O₃ × (moles of Fe / moles of Fe₂O₃) = [tex]24.95 * \frac{4}{2} = 49.90 moles[/tex]

moles of CO₂ = moles of Fe₂O₃ × (moles of CO₂ / moles of Fe₂O₃) = [tex]24.95 * \frac{3}{2} = 37.43 moles[/tex]

Therefore, 4.00 kg of Fe₂O₃ can react with 37.43 moles of CO to produce 49.90 moles of Fe and 37.43 moles of CO₂.

Answer: The chemical reaction is given below.

Explanation:

When solid sodium metal reacts with water molecule to produce aqueous sodium hydroxide and hydrogen gas. The equation for this follows:

[tex]2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)[/tex]

By Stoichiometry of the reaction:

2 moles of solid sodium metal reacts with 2 moles of water molecule to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.

Sodium metal is present in solid state, Water molecule is present in liquid state, Sodium hydroxide is present in aqueous state and hydrogen is present in gaseous state.

A chemical reaction between sodium (Na) and water (H₂O) yields hydrogen gas (H₂) and aqueous sodium hydroxide (NaOH). The reaction is highly exothermic as sodium hydroxide reacts vigorously with water, generating a lot of heat. The balanced chemical equation is '2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)'.

To answer your question, we need to first understand the reaction occurring in this process. This is a classic example of a reaction between a metal and water, specifically it's an alkali metal (sodium) reacting with water. Upon addition of solid sodium to liquid water, the sodium displaces hydrogen from the water, leading to the production of hydrogen gas and aqueous sodium hydroxide.

The balanced chemical equation is thus written as: 2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g). In this equation, (s) represents solid, (l) represents liquid, (aq) represents aqueous or a material dissolved in water, and (g) represents gas.

Sodium hydroxide NaOH is a strong base and it reacts vigorously with water, generating a great deal of heat and forming very basic solutions. In fact, 40 grams of sodium hydroxide can dissolve in only 60 grams of water at 25 °C.

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