Sodium bicarbonate is sometimes used to neutralize acid spills. If 2.0 L of 1.0 M H2SO4 is spilled, what mass of sodium bicarbonate (NaHCO3) is required to neutralize all the H2SO4?

Answer:

Explanation:

Volatile organic compounds are the  organic chemicals that get easily vaporized in air.

The secondary chemicals or the secondary pollutants are the photochemical substances that are formed in the presence of sunlight.

The amount of gases and small particles present in the atmosphere, responsible for influencing ecosystem and the wellness of human beings is known as the Air Quality

Air pollution refers to the high concentrations of gases or small particles that are present in the atmosphere, which can cause harm to the humans and other living organisms and structures established by humans.

Aerosols are the tiny particles present in liquid or solid state, that are suspended in air.

Answer:

M of Ba(OH)₂ = 0.818 M.

Explanation:

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O,

It is clear that every 1.0 mol of Ba(OH)₂ needs 2 mol of HNO₃ to be neutralized completely.

∴ (nMV) of Ba(OH)₂ = (nMV) for HNO₃.

where, n is the no. of producible H⁺ or OH⁻ of the acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

For Ba(OH)₂:

n = 2, M = ??? M, V = 20.0 mL.

For HNO₃:

n = 1, M = 0.85 M, V = 38.5 mL.

∴ M of Ba(OH)₂ = (nMV) of HNO₃ / (MV) for Ba(OH)₂ = (1)(0.85 M)(38.5 L)/(2)(20.0 mL) = 0.818 M.

Answer:

The final temperature of the combined metals is 49.2314 °C

Explanation:

Heat gain by gold = Heat lost by iron

Thus,  

[tex]m_{gold}\times C_{gold}\times (T_f-T_i)=-m_{iron}\times C_{iron}\times (T_f-T_i)[/tex]

Where, negative sign signifies heat loss

Or,  

[tex]m_{gold}\times C_{gold}\times (T_f-T_i)=m_{iron}\times C_{iron}\times (T_i-T_f)[/tex]

For gold:

Mass = 11.4 g

Initial temperature = 14.5 °C

Specific heat of gold = 0.129 J/g°C

For iron:

Mass = 18.4 kg

Initial temperature = 55.4 °C

Specific heat of water = 0.450 J/g°C

So,

[tex]11.4\times 0.129\times (T_f-14.5)=18.4\times 0.450\times (55.4-T_f)[/tex]

[tex]1.4706\times (T_f-14.5)=8.28\times (55.4-T_f)[/tex]

[tex]1.4706\times T_f-1.4706\times 14.5=8.28\times 55.4-8.28\times T_f[/tex]

[tex]1.4706\times T_f-21.3237=458.712-8.28\times T_f[/tex]

[tex]1.4706\times T_f+8.28\times T_f=458.712+21.3237[/tex]

[tex]T_f=49.2314[/tex]

Thus,

The final temperature of the combined metals is 49.2314 °C

The final temperature of the combined metals is approximately 49.31°C.

Given data:

Step 1: Calculate the heat gained/lost by each metal

Step 2: Set up the equation

Step 3: Solve for T_f

And there you have it! The final temperature of the combined metals is approximately 49.31°C.

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess [tex]AgNO_3[/tex] then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles[/tex]

[tex]\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles[/tex]

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = [tex]\frac{x}{58.5}moles[/tex]

Moles of chloride ions in KCl = [tex]\frac{(0.9440-x)}{74.5}moles[/tex]

The total moles of chloride ions = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

[tex]\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles[/tex]

Now we have to determine the value of 'x'.

Moles of AgCl = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

0.0133 mole = [tex]\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles[/tex]

By solving the term, we get the value of 'x'.

[tex]x=0.171g[/tex]

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

[tex]\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%[/tex]

[tex]\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%[/tex]

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Answer : The enthalpy of reaction [tex](\Delta H_{rxn})[/tex] is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of [tex]AgNO_3[/tex] and [tex]HCl[/tex].

[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole[/tex]

[tex]\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole[/tex]

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

[tex]AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)[/tex]

As, 1 mole of [tex]AgNO_3[/tex] react with 1 mole of [tex]HCl[/tex] to give 1 mole of [tex]AgCl[/tex]

So, 0.005 mole of [tex]AgNO_3[/tex] react with 0.005 mole of [tex]HCl[/tex] to give 1 mole of [tex]AgCl[/tex]

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

[tex]q=m\times C\Delta T=m\times C \times (T_2-T_1)[/tex]

where,

q = heat

C = specific heat capacity = [tex]4.18J/g^oC[/tex]

m = mass = 100 g

[tex]T_2[/tex] = final temperature = [tex]24.21^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]23.40^oC[/tex]

Now put all the given values in the above expression, we get:

[tex]q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC[/tex]

[tex]q=338.58J[/tex]

Now  we have to calculate the enthalpy of the reaction.

[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]

where,

[tex]\Delta H_{rxn}[/tex] = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

[tex]\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole[/tex]

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction [tex](\Delta H_{rxn})[/tex] is, 67.716 KJ/mole

For this reaction, the heat of reaction is -67.8 KJ/mol.

The equation of the reaction is;

AgNO3(aq) + HCl(aq) -----> AgCl(s) + HNO3(aq)

Number of moles of AgNO3 = 50/1000 L ×  0.100 M = 0.005 M

Number of moles of HCl = 50/1000 L  ×  0.100 M = 0.005 M

Temperature change =  24.21 °C - 23.40 °C = 0.81°C

Total volume of solution =  50.0 mL +  50.0 mL = 100 mL

Since the density of solution= 1.00 g/mL

Total mass of solution = 100g

Heat absorbed by solution = mcθ

m = mass of solution

c = specific heat capacity of solution

θ = temperature change

Heat absorbed by solution = 100g ×  4.18 J/g ∙ °C  × 0.81°C = 0.339 KJ

ΔHrxn = -( 0.339 KJ)/ 0.005 M

ΔHrxn = -67.8 KJ/mol

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Answer:

A) it produces a relatively small fraction of the maximum number of possible hydronium ions.

Explanation:

A weak acid is one that ionizes slightly in aqueous solutions. They set up an equilibrium in the process. Weak acids like hydrocyanic acid only gives a very small fraction of the possible hydronium ions they ought to give in solution. They are not capable of ionizing completely like the strong acids. This makes hydrocyanic acid a weak electrolyte.

"The correct answer is A) it produces a relatively small fraction of the maximum number of possible hydronium ions. The correct statement about hydrocyanic acid is that it produces a relatively small fraction of the maximum number of possible hydronium ions when dissolved in water, which aligns with option A.

Hydrocyanic acid (HCN) is indeed classified as a weak acid in water. The classification of an acid as weak or strong is based on its ability to dissociate in water. A weak acid, such as hydrocyanic acid, only partially dissociates into its ions when dissolved in water. This means that only a small fraction of the HCN molecules ionize to release hydronium ions (H3O+) and cyanide ions (CN-).

In contrast, a strong acid would dissociate completely, producing the maximum number of hydronium ions possible from the acid. For example, hydrochloric acid (HCl) is a strong acid that dissociates completely in water to form H3O+ and Cl- ions.

The degree of dissociation of a weak acid is described by its acid dissociation constant (Ka), which is a measure of the strength of the acid in solution. The Ka value for hydrocyanic acid is relatively small, indicating that the equilibrium lies more towards the reactant side (undissociated HCN) rather than the product side (H3O+ and CN-).

Therefore, the correct statement about hydrocyanic acid is that it produces a relatively small fraction of the maximum number of possible hydronium ions when dissolved in water, which aligns with option A."

Answer:

[tex]\boxed{\text{-4.25 kJ}}[/tex]

Explanation:

ΔE = q + w

By convention, anything leaving the system is negative and anything entering the system is positive.

Data:

q = -3.30 kJ

w = -950 J = -0.950 kJ

Calculation

ΔE = -3.30 - 0.950 = -4.25 kJ

[tex]\text{The change in E is }\boxed{\textbf{-4.25 kJ}}[/tex]

Final answer:

The change in internal energy (ΔE) for the decomposition of NI3 is -4250 J, which indicates the system lost energy.

Explanation:

The change in internal energy (ΔE) of the system can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system (q) minus the work done by the system (w): ΔE = q - w. In this scenario, the sample of NI3 decomposes, releasing heat (q = -3.30 kJ, since the system loses heat) and doing work (w = 950 J).

First, it is important to make sure both q and w are in the same units. We can convert kilojoules to joules by multiplying by 1000: -3.30 kJ = -3300 J.

Therefore, the change in internal energy ΔE is:

ΔE = q - w
ΔE = (-3300 J) - (950 J)
ΔE = -4250 J

The negative sign indicates that the total energy of the system decreased by 4250 joules.

Answer: The temperature of the unknown sample is 64.65 °C

Explanation:

To calculate the temperature of the unknown sample, we use the following equation:

[tex]T=\frac{(V-V_1)}{(V_2-V_1)}\times (T_2-T_1)+T_1[/tex]

where,

T = temperature of the unknown sample = ?

V = thermistor reading of the unknown sample = 116 mV

[tex]T_1[/tex] = melting point of ice = 0°C

[tex]V_1[/tex] = thermistor reading of ice = -45 mV

[tex]T_1[/tex] = boiling point of water = 100°C

[tex]V_1[/tex] = thermistor reading of water = 204 mV

Putting values in above equation, we get:

[tex]T=\frac{(116-(-45))}{(204-(-45))}\times (100-0)+0\\\\T=64.65^oC[/tex]

Hence, the temperature of the unknown sample is 64.65 °C

The temperature of the unknown sample in C° is approximately 45.00°C.

To find the temperature of the unknown sample, we first need to establish a relationship between the voltage read by the thermistor and the temperature in degrees Celsius. We have two calibration points: ice water and boiling water, which correspond to 0°C and 100°C, respectively.

Let's denote the voltage read by the thermistor as \( V \) and the temperature as \( T \). We can then set up two equations based on the calibration points:

1. For ice water (0°C), the thermistor reads -45 mV:

[tex]\[ V_{ice} = -45 \text{ mV} \] \[ T_{ice} = 0°C \][/tex]

2. For boiling water (100°C), the thermistor reads 204 mV:

[tex]\[ V_{boiling} = 204 \text{ mV} \] \[ T_{boiling} = 100°C \][/tex]

We can assume a linear relationship between voltage and temperature for the thermistor, which allows us to use a simple linear interpolation formula:

[tex]\[ T = T_{ice} + \frac{(V - V_{ice}) \times (T_{boiling} - T_{ice})}{V_{boiling} - V_{ice}} \][/tex]

Now, we need to find the temperature \( T \) when the thermistor reads 116 mV for the unknown sample:

[tex]\[ T = 0°C + \frac{(116 \text{ mV} - (-45 \text{ mV})) \times (100°C - 0°C)}{204 \text{ mV} - (-45 \text{ mV})} \] \[ T = 0°C + \frac{(116 \text{ mV} + 45 \text{ mV}) \times 100°C}{204 \text{ mV} + 45 \text{ mV}} \] \[ T = 0°C + \frac{161 \text{ mV} \times 100°C}{249 \text{ mV}} \] \[ T = 0°C + \frac{161}{2.49} \times 100°C \] \[ T = 0°C + 64.6586345 \times 100°C \] \[ T = 64.6586345°C \][/tex]

Rounding to two decimal places, the temperature of the unknown sample is approximately 45.00°C.

Answer:

addition of a neutron

Explanation:

Nuclear fission is a radioactive decay process in which a heavy nucleus spontaneously disintegrates into lighter ones accompanied by the release of energy.

An atom whose neutron/proton ratio is the same as that of the stability ratio of that atom is said to be stable. When such an atom is bombarded with a neutron particle, the stability ratio is offset. What results is a radioactive decay of such a nucleus. This form of decay is a nuclear fission.

A series of chain reaction is produced though this until the stability ratio is reached.

Answer:

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.

Explanation:

For first solution of sulfuric acid :

C₁ = 40% , V₁ = ?

For second solution of sulfuric acid :

C₂ = 10% , V₂ = ?

For the resultant solution of sulfuric acid:

C₃ = 28% , V₃ = 20L

Also,

V₁ + V₂ = V₃ = 20L ......................................(1)

Using

C₁V₁ + C₂V₂ = C₃V₃

40×V₁ + 10×V₂ = 28×20

So,

40V₁ + 10V₂ = 560........................................(2)

Solving 1 and 2 as:

V₂ = 20 - V₁

Applying in 2

40V₁ + 10(20 - V₁)  = 560

40V₁ + 200 - 10V₁ = 560

30V₁ = 360

V₁ = 12 L

So,

V₂ = 20 - V₁ = 8L

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.

Answer:

The concentration of the potassium ions in the original potassium chromate solution is 4.2927 mol/L.

The concentration of the chromate ions in the final solution is 1.0731 mol/L.

Explanation:

[tex]K_2CrO_4+2AgNO_3\rightarrow Ag_2CrO_4+2KNO_3[/tex]

Volume of solution A i.e. solution of silver nitrate = 466.0 mL = 0.466 L

Volume of solution B i.e. solution of potassium chromate = 466.0 mL = 0.466 L

Moles of silver chromate =[tex]\frac{331.8}{331.73 g/mol}=1.0002 mol[/tex]

According to reaction , 1 mol of silver chromate is produce from 2 moles of silver nitrate.

Then, 1.0002 moles of silver chromate will be formed from:

[tex]\frac{1}{2}\times 1.0002 mol=0.5001 mol[/tex] of silver nitrate.

According to reaction , 1 mol of silver chromate is produce from 1 mole of potassium chromate.

Then, 1.0002 moles of silver chromate will be formed from:

[tex]\frac{1}{1}\times 1.0002 mol=1.0002 mol[/tex] of potassium chromate

[tex]Concentration =\frac{Moles}{Volume (L)}[/tex]

a) The concentration of the potassium ions in the original potassium chromate solution.

Volume of the original solution = 0.466 L

1 mol of potassium chromate dissociates into 2mol of potassium ions and 1 mol of chromate ions:

Moles of potassium ions = 2 × 1.0002 mol = 2.0004 mol

[tex][K^+]=\frac{2.0004 mol}{0.466 L}=4.2927 mol/L[/tex]

b) The concentration of the chromate ions in the final solution

Volume of the final solution = 0.466 L + 0.466 L

Moles of chromate ions = 1 × 1.0002 mol = 1.0002 mol

[tex][CrO_4^{2+}]=\frac{1.0002 mol}{0.466 L+0.466L}=1.0731 mol/L[/tex]

Answer:

With thermodynamics, one cannot determine the speed of a reaction.

Explanation:

Chemical thermodynamics, a branch of chemistry that deals with study of  interrelation of the heat and the work with the chemical reactions or with the physical changes of the state within confines of laws of thermodynamics.

Chemical thermodynamics' structure is derived from first two laws of chemical thermodynamics. From fundamental equations of Gibbs, a multitude of some equations which relates thermodynamic properties of thermodynamic system can be derived and can be used to calculate the reaction spontaneity, equilibrium constant, etc.

Thermodynamics predicts about the direction, feasibility and the extent of a chemical process, it does not tell anything about the rate at which a chemical process may proceed.

Answer:

B

Explanation:

It has 2 NH4 molecules and 1 CO3 molecule.

NH4 has a molar mass of 18g/mol. Since there are two NH4s, it makes up 36g of the (NH4)2CO3.

CO3 has a molar mass of 60g/mol. Since there is one CO3, it makes up 60g of the (NH4)2CO3.

If you add up 36 and 60, you will get 96g.

Answer:

Explanation:

Formula

density = mass / volume

Givens

mass = 26.94

volume = 2.568

Solution

density = 26.94 / 2.568

density = 10.49 grams / cc^3

Answer

This is very likely silver.

The givens in the  table except for the first 2 are wrong.

silver is as above

copper is a bit over 8.

Answer: The given material is silver.

Explanation:

To calculate density of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

We are given:

Mass of object = 26.94 g

Volume of object = [tex]2.568cm^3[/tex]

Putting values in above equation, we get:

[tex]\text{Density of object}=\frac{26.94g}{2.568cm^3}=10.49g/cm^3[/tex]

The calculated value of density corresponds to the density of silver.

Hence, the given material is silver.

Answer:

527.68 mL

Explanation:

We will assume that nitrogen is behaving as ideal gas here.

For ideal gas the gas law is:

[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]

Where

P1= initial pressure = 740 torr

V1= initial volume = 500mL

T1= initial temperature = 25⁰C = 298 K

P2= final pressure = 760 torr

V2= final volume = ?

T2= final temperature = 50⁰C = 323 K

Putting values in the gas law

Final volume = [tex]\frac{740X500X323}{298X760}= 527.68 mL[/tex]

Answer:

C. Sr (NO3)2 •2 H2O

Explanation:

A 30.7 g sample of Strontium nitrate, Sr(NO3)2⋅nH2O, is heated to a contstent mass of 22.9.

Therefore, the calculated hydration number is Sr (NO3)2 •2 H2O.

Answer:

C

Explanation:

Answer:

According to Le Chatelier’s Principle, a stress placed on a system at equilibrium will cause the equilibrium to shift to counteract the stress.  For example, a temperature increase in the above reaction will favor the reverse reaction to use the excess heat and form brown NO2 gas.  A temperature decrease in the above reaction favors the forward reaction to produce heat and form colorless N2O4 gas.

Answer:

The equilibrium shifts to the left and more of NO₂ is formed.

Explanation:

The Le Chatelier's principle states that when a form of stress is applied to a system in equilibrium, the system shifts so as to relieve that stress.

This principle explains that addition of one of the reactants to a system in equilibrium leads to the equilibrium shifting in such a way that the reaction occurring reduces the concentration of the added reactant.

In this case the addition of dinitrogen tetraoxide increases its concentration and therefore equilibrium shifts to reduce this concentration.

Answer: The percent yield of the given reaction is 82 %.

Explanation:

The chemical equation for the reaction of zinc and hydrochloric acid follows:

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

To calculate the percent yield of the reaction, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of hydrogen gas = 164 g

Theoretical yield of hydrogen gas = 200 g

Putting values in above equation, we get:

[tex]\%\text{ yield of hydrogen gas}=\frac{164g}{200g}\times 100\\\\\% \text{yield of hydrogen gas}=82\%[/tex]

Hence, the percent yield of the reaction is 82 %.

An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a producer.

An organism that can make its own nourishment by the use of light, water, carbon dioxide, or other substances is an autotroph. Autotrophs are also referred to as producers because they make their own nourishment.

Producers are living things with the ability to grow their own nourishment. They frequently contain green vegetation. They use the process of photosynthesis to capture solar energy and use it to produce food. Since they are unable to create food on their own, all other creatures rely on producers for nourishment.

Green plants, phytoplankton, and cyanobacteria all have cells that can synthesize oxygen. Photosynthetic cells are highly diverse. Cells create sugar molecules and oxygen during the process of photosynthesis by using carbon dioxide and energy from the sun.

Thus, An organism gets carbon by using carbon dioxide in the atmosphere to make sugar molecules. This organism is a producer.

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Using Faraday's laws of electrolysis, it takes approximately 344.91 minutes to electroplate 29.6 grams of nickel using a current of 4.7 A.

The student is asking how long it would take to electroplate 29.6 grams of nickel using a current of 4.7 amps. To calculate the time required for electroplating, we'll use Faraday's laws of electrolysis, which relate the chemical amount of substance produced at an electrode to the amount of electricity used.

The half-reaction for nickel plating is [tex]Ni^{+2}[/tex] + 2e-
ightarrow Ni(s) and the molar mass of nickel is approximately 58.69 g/mol. Nickel has a valency of 2, which means it requires 2 moles of electrons (2 Faradays) to deposit one mole of nickel. Based on Faraday's law, the charge (Q) in coulombs required to deposit a substance is Q = n x F, where n is the number of moles and F is Faraday's constant (96485 C/mol).

First, we need to convert the mass of nickel to moles using the molar mass: 29.6 g / 58.69 g/mol
gives approximately 0.504 moles of Ni. Thus, the total charge needed will be 0.504 moles x 2 x 96485 C/mol, which equals 97262.2 C.

Next, we find how long it takes for this charge to pass through the solution using the current (I) given: Time (t) = Q / I, where
t = time in seconds. So, t = 97262.2 C / 4.7 A
gives approximately 20694.51 seconds. Converting that to minutes, we divide by 60, resulting in about 344.91 minutes.

Answer:

99.24%.

Explanation:

NaCl + AgNO₃ → AgCl(↓) + NaNO₃,

1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.

no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.

using cross multiplication:

1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.

∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.

mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.

∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.

An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.

Let's consider the reaction between NaCl and AgNO₃ to produce AgCl and NaNO₃.

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

We can calculate the mass of NaCl that produced 2.044 g of AgCl using the following relations.

[tex]2.044 g AgCl \times \frac{1molAgCl}{143.32 g AgCl} \times \frac{1molNaCl}{1molAgCl} \times \frac{58.44gNaCl}{1molNaCl} = 0.8335gNaCl[/tex]

An impure sample of mass 0.8421 g contains 0.8335 g of NaCl. The percentage of NaCl in the impure sample is:

[tex]\% NaCl = \frac{0.8335g}{0.8421g} \times 100\% = 98.98\%[/tex]

An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.

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Answer:

C₄H₈O₂.

Explanation:

n = mass/atomic mass,

mol C = mass/(atomic mass) = (54.5 g)/(12.0 g/mol) = 4.54 mol.

mol H = mass/(atomic mass) = (9.3 g)/(1.0 g/mol) = 9.3 mol.

mol O = mass/(atomic mass) = (36.2 g)/(16.0 g/mol) = 2.26 mol.

∴ C: H: O = (4.54 mol/2.26 mol) : (9.3 mol/2.26 mol) : (2.26 mol/2.26 mol) = 2: 4: 1.

∴ Empirical formula mass of (C₂H₄O) = 2(atomic mass of C) + 4(atomic mass of H) + 1(atomic mass of O) = 2(12.0 g/mol) + 4(1.0 g/mol) + (16.0 g/mol) = 44.0 g/mol.

∴ Number of times empirical mass goes into molecular mass = (88.0 g/mol)/(44.0 g/mol) = 2.0 times.

∴ The molecular formula is, 2(C₂H₄O), that is; (C₄H₈O₂)

Answer: 0.14 Liters

Explanation:

[tex]Q=I\times t[/tex]

where Q= quantity of electricity in coloumbs

I = current in amperes = 1.45 A

t= time in seconds = 13 min=[tex]13\times 60 =780s[/tex]

[tex]Q=1.45A\times 780s=1131C[/tex]

[tex]HNO_3\rightarrow H^++NO_3^-[/tex]

[tex]2H^++2e^-\rightarrow H_2[/tex]

[tex]96500\times 2=193000Coloumb[/tex] of electricity deposits  1 mole of [tex]H_2[/tex]

1131 C of electricity deposits =[tex]\frac{1}{193000}\times 1131=5.86\times 10^{-3}moles[/tex] of [tex]H_2[/tex]

According to the ideal gas equation:'

[tex]PV=nRT[/tex]

P = Pressure of the gas = 1.03 atm

V= Volume of the gas = ?

T= Temperature of the gas = 25°C = 298 K       (0°C = 273 K)

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= [tex]5.86\times 10^{-3}moles[/tex]

[tex]V=\frac{nRT}{P}=\frac{5.86\times 10^{-3}\times 0.0821\times 298}{1.03}=0.14L[/tex]

Thus the volume of hydrogen gas at [tex]25^0C[/tex] and 1.03 atm will be 0.14 Liters.

To calculate the volume of hydrogen gas produced at the cathode, Faraday's laws of electrolysis and the Ideal Gas Law are applied. The charge passed through the circuit is determined by the product of current and time, and the volume of H2 is calculated using the number of moles of hydrogen, pressure, and temperature.

Explanation:

Calculation of Hydrogen Gas Volume

To calculate the volume of hydrogen gas (H2) produced at the cathode during electrolysis, we use Faraday's laws of electrolysis which state that the amount of substance altered at an electrode during electrolysis is proportional to the amount of electricity that passes through the circuit. Here, we need to know the charge passed through the circuit, which can be calculated by multiplying the current (I) by the time (t), where I = 1.45 A and t = 13.00 min (converted to seconds).

The reaction at the cathode for the electrolysis of aqueous HNO3 will produce hydrogen gas according to the reaction:

2H2O(l) + 2e- → H2(g) + 2OH-

This means that 2 moles of electrons are required to produce 1 mole of H2. By using the Faraday constant (96,485 C/mol e-), we can calculate the moles of hydrogen produced:

Moles of electrons (n) = It/F, where F is the Faraday constant.

Finally, we can find the volume of H2 gas using the Ideal Gas Law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of hydrogen, R is the Ideal Gas Constant (0.0821 L·atm/mol·K) and T is the temperature in Kelvin.

With the given temperature (25.0°C) and pressure (1.03 atm), we convert the temperature to Kelvin, then rearrange the Ideal Gas Law to solve for V. Remember to use the number of moles of H2, not the number of moles of electrons.

Answer :

(a) The concentration of hydrogen ion and hydroxide ion are, [tex]1.708\times 10^{-7}M[/tex].

(b) The pH of pure water is, 6.78

(c) The pH of solution is, 13

Explanation :

(a) First we have to calculate the concentration of hydrogen ion and hydroxide ion.

As we know that,

[tex]K_w=[H^+][OH^-][/tex]

In pure water, the concentration of hydrogen ion and hydroxide ion are equal. So, let the concentration of hydroxide ion and hydrogen ion be, 'x'.

[tex]2.92\times 10^{-14}=(x)\times (x)[/tex]

[tex]2.92\times 10^{-14}=(x)^2[/tex]

[tex]x=1.708\times 10^{-7}M[/tex]

The concentration of hydrogen ion and hydroxide ion are, [tex]1.708\times 10^{-7}M[/tex].

(b) Now we have to calculate the pH of pure water.

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

[tex]pH=-\log (1.708\times 10^{-7})[/tex]

[tex]pH=6.78[/tex]

The pH of pure water is, 6.78

(c) In this, first we have to calculate the pOH when the concentration of hydroxide ion is, 0.10 M.

[tex]pOH=-\log [OH^-][/tex]

[tex]pOH=-\log (0.10)[/tex]

[tex]pOH=1[/tex]

Now we have to calculate the pH.

[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1=13[/tex]

The pH of solution is, 13

At 40.8°C, the [H+] and [OH-] concentrations in pure water are both 1.71 × 10^−7 M. The pH of pure water at 40.8°C is 6.77. If the [OH-] concentration in a solution is 0.10 M, the pH at 40.8°C is 13.00.

At 40.8°C, the value of Kw is given as 2.92 × 10^−14.

a. To calculate the [H+] and [OH-] concentrations in pure water at 40.8°C, we can use the fact that [H+] × [OH-] = Kw. Since water is neutral, the concentrations of [H+] and [OH-] are equal.

Using the given value of Kw, we have:

[H+] × [H+] = 2.92 × 10^−14

Solving for [H+], we find that [H+] = [OH-] = 1.71 × 10^−7 M.

b. The pH of pure water at 40.8°C can be calculated using the formula pH = -log[H+]. Substituting the value of [H+], we find that the pH is equal to 6.77.

c. If the hydroxide ion concentration in a solution is 0.10 M, we can calculate the pH using the formula pH = 14 - pOH. Since the [OH-] concentration is given as 0.10 M, the pOH can be calculated as -log(0.10) = 1.00. Substituting this value into the pH formula, we find that the pH is equal to 13.00.

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Answer: Option (b) is the correct answer.

Explanation:

In liquid state, particles do have kinetic energy that helps in partially overcoming the intermolecular forces between the molecules. But still the particles are close together and they are able to slide past each other.

So, when we apply pressure on a liquid then its molecules partially gets compressed.

On the other hand, molecules of a solid are held together by strong intermolecular forces of attraction. Hence, they have definite shape and volume. As a result, solids do not get compressed.

In gases and plasma state of matter, molecules are gar away from each other. So, they are able to get completely compressed when a pressure is applied.

Thus, we can conclude that liquid is the state of matter which consists of particles that can be partially compressed.

Answer:

liquid

Explanation:

The equilibrium concentration of N2O4 in the given reaction can be calculated by setting up an expression for the equilibrium constant in terms of the change in concentration and then solving for the unknown.

This question involves examining a chemical equilibrium problem for the reversible reaction: N2O4(g) ⇄ 2 NO2(g), with the given equilibrium constant, Kc = 5.8 × 10-3. The initial concentration of N2O4 is given as 0.040 M, while the initial concentration of NO2 is given as 0 M. To solve this, we will let 'x' be the amount of N2O4 that decomposes into NO2 at equilibrium. Therefore, the equilibrium concentration of N2O4 would be given as N2O4 = 0.040 - x, and for NO2 it would be 2x (because the stoichiometry of the reaction shows that for each mole of N2O4 decomposed, 2 moles of NO2 are produced).

Now we can set up an expression for the equilibrium constant such that: Kc = [NO2]^2 / [N2O4] = (2x)^2 / (0.040 - x). By substituting 5.8 x 10^-3 for the equilibrium constant and solving for x, we will be able to find the equilibrium concentration of N2O4, which will be 0.040 - x.

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Answer:

  c.  0 °C

Explanation:

Standard temperature is 273.15 K, or 0 °C, or 32 °F, as defined by the International Union of Pure and Applied Chemistry (IUPAC) in 1982.

Answer:

Standard temperature is exactly (C) 0ºC

Explanation:

STP means standard temperature pressure at which

T = 0℃ or 273K and Pressure = 1 atm

NTP means Normal temperature pressure which means room temperature and  pressure at which

T = 20℃ or 293 K and Pressure = 1 atm

SATP means standard Ambient temperature pressure at which

T = 25℃ or 298K and P = 1atm

In tropical countries, the climate will be warm and hence room temperature usually  considered as

T = 25℃ or 298K and P = 1 atm  

Lab temperature is same as that of room temperature.

Answer:

The equation that represents how the energy of an electromagnetic wave is related to the frequency of the wave and to the wavelength of the wave is known as the Planck - Einstein equation.

Such equation states that the energy of a photon of light (electromagnetic radiation) is proportional to its frequency:

Where:

Since the frequency and the wavelength are inversely related as per the equation λ = c / f, where λ is the wavelength and c is the speed of light, you can derive the equation that relates the energy of an electromagnetic wave with the wavelength:

Where  λ is the wavelength and c is the speed of light.

The value of c in vacuum is a constant and is equal to 299,792,458 m/s m/s, which is usually approximated to 3.0 × 10⁸ m/s

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Answer:

The order of reaction is 2.

Rate constant is 0.0328 (M s)⁻¹

Explanation:

The rate of a reaction is inversely proportional to the time taken for the reaction.

As we are decreasing the concentration of the reactant the half life is increasing.

a) For zero order reaction: the half life is directly proportional to initial concentration of reactant

b) for first order reaction: the half life is independent of the initial concentration.

c) higher order reaction: The relation between half life and rate of reaction is:

Rate = [tex]\frac{1}{k[A_{0}]^{(n-1)}}[/tex]

Half life =[tex]K\frac{1}{[A_{0}]^{(n-1)} }[/tex]

[tex]\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }[/tex]

where n = order of reaction

Putting values

[tex]\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}[/tex]

[tex]0.472=(0.472)^{(n-1)}[/tex]

Hence n = 2

[tex]halflife=\frac{1}{k[A_{0}]}[/tex]

Putting values

[tex]231=\frac{1}{K(0.132)}[/tex]

K = 0.0328

The reaction order is 2 (option B), which is determined by observing the change in half-life with concentration. The rate constant for this second order reaction, calculated using the provided formula, is approximately 0.032 M^-1s^-1.

The reaction order refers to how the rate of a chemical reaction is affected by the concentration of its reactants. When we see that the half-life of the reaction changes as the concentration of the reactant changes, this signifies that the order of the reaction is not zero. In zero order reactions, the half-life is dependent on the initial concentration of reactants. In this case, we see that the half-life of the reaction increases as the concentration decreases, which is characteristic of a second order reaction. Therefore, the answer to the first part is (B) 2.

For a second order reaction, the rate constant can be calculated using the formula t1/2 = 1 / [A]0k rearranging gives k = 1 / [A]0t1/2. Substituting for t1/2 = 109 s, and [A]0 = 0.280M, we get k = 1 / (0.280M * 109 s) which is approximately 0.032 M-1s-1. The units for the rate constant k for a second order reaction are M-1s-1.

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