Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr ( s ) ( K sp = 6.3 × 10 − 9 ) in 0.76 M NH 3 ( aq ) .

Answers

Answer 1

Answer:

53.18 gL⁻¹

Explanation:

Given that:

[tex]Cu^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] [tex]------>[/tex]  [tex][Cu(NH_3)_2]^+_{(aq)}[/tex]      ------equation (1)

where;

Formation Constant  [tex](k_f) =[/tex] [tex]6.3*10^{10}[/tex]

However, the Dissociation of [tex]CuBr_{(s)[/tex] yields:

[tex]CuBr_{(s)}[/tex]      ⇄    [tex]Cu^{+}_{(aq)}[/tex]  [tex]+[/tex] [tex]Br^-_{(aq)}[/tex]      -------------- equation (2)

where;

the Solubility Constant [tex](k_{sp})[/tex]  [tex]= 6.3 *10^{-9[/tex]

From equation (1);

[tex](k_f) =[/tex] [tex]\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}[/tex]            ---------  equation (3)

From equation (2)

[tex](k_{sp})[/tex]  [tex]= [Cu^+][Br^-][/tex]           ---------  equation (4)

In [tex]NH_3[/tex], the net reaction for [tex]CuBr_{(s)[/tex] can be illustrated as:

[tex]CuBr_{(s)[/tex]   [tex]+[/tex] [tex]2NH_{3(aq)}[/tex]  ⇄  [tex][Cu(NH_3)_2]^+_{(aq)}[/tex]  [tex]+ Br^-_{(aq)}[/tex]

The equilibrium constant (K) can be written as :

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]

If we multiply both the numerator and the denominator with  [tex][Cu^+][/tex] ; we have:

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}[/tex]

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}[/tex]

[tex]K = k_f *k_{sp}[/tex]

[tex]K= (6.3*10^{10})*(6.3*10^{-9})[/tex]

[tex]K= 3.97*10^2[/tex]

[tex]K[/tex] ≅ [tex]4.0*10^2[/tex]

Now; we can re-write our equilibrium constant again as:

[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]

[tex]4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}[/tex]

[tex]4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}[/tex]

[tex]4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2[/tex]

By finding the square of both sides, we have

[tex]\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2[/tex]

[tex]2.0*10 = \frac{x}{(0.76-2x)}[/tex]

[tex]20(0.76-2x) =x[/tex]

[tex]15.2 -40x=x[/tex]

[tex]15.2 = 40x +x[/tex]

[tex]15.2 = 41x[/tex]

[tex]x = \frac{15.2}{41}[/tex]

[tex]x = 0.3707 M[/tex]

In gL⁻¹; the solubility of [tex]CuBr_{(s)[/tex] in 0.76 M [tex]NH_3[/tex] solution will be:

[tex]= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}[/tex]

=  53.18 gL⁻¹

Answer 2
Final answer:

The problem involves finding the solubility of copper(I) bromide in a solution of ammonia. The solution entails expressing the formation of complex ions and dissolution of CuBr in terms of equilibrium expressions. Solving the system of expressions for solubility 's' can eventually give the solubility in g·L-1.

Explanation:

In this problem, we are looking for the solubility of copper(I) bromide (CuBr) in a solution with 0.76 M NH3. Since CuBr does not have high solubility in water, we are using ammonia which forms complex ions with copper, enhancing its solubility. We will utilize the given equilibrium constant (Kf) which helps us to understand how far the forward reaction (complex ion formation) proceeds.

To start, we'll write the reaction of Cu+ ion with NH3 and the corresponding Kf expression:

Cu+ (aq) + 2NH3 (aq) <=> Cu(NH3)2+ (aq); Kf = [Cu(NH3)2+] / [Cu+] [NH3]^2

Next, we need to find an expression for [Cu+] in terms of solubility and solve for solubility. Let's denote the solubility of CuBr as 's'. The dissolution of CuBr can be represented as:

CuBr (s) <=> Cu+ (aq) + Br- (aq)

We can see that for every mole of CuBr that dissolves, one mole of Cu+ ion is produced. Thus, [Cu+] = s.

Now using the Kf expression, we can solve the system of equations to find s, which effectively gives us the solubility of CuBr in 0.76 M NH3. After that, by taking the molecular weight of CuBr into account, we can finally express the solubility in g·L−1.

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Related Questions

A 100.0g sample of tin is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 25.0 oC. After the metal cools, the final temperature of the metal and the water is 27.4 oC. Calculate the specific heat capacity of tin from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. Specific heat of water

Answers

Explanation:

It is known that specific heat of water is 4.184 [tex]J/g^{o}C[/tex] and atomic mass of tin is 118.7 g/mol. For the given situation,

                 [tex]Q_{lost} = Q_{gained}[/tex]

Let us assume that,

               [tex]m_{1}[/tex] = mass of Sn

               [tex]m_{2}[/tex] = mass of [tex]H_{2}O[/tex]  

Therefore, heat energy expression for heat lost and gained is as follows.

           [tex]Q_{lost} = Q_{gained}[/tex]

      [tex]m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})[/tex]

   [tex]100 g \times C_{1} \times (100^{o}C - 27.4^{o}C) = 150 g \times 4.184 /g^{o}C \times (27.4^{o}C - 25^{o}C)[/tex]

           [tex]7260C_{1} = 150 \times 4.184 \times 2.4[/tex]

                 [tex]C_{1} = \frac{1506.24}{7260}[/tex]

                              = 0.207 [tex]J/g^{o}C[/tex]

For, 118.7 g the specific heat of tin will be calculated as follows.

               [tex]C_{1} = 0.207 J/g^{o}C \times 118.7 g[/tex]

                          = 24.5 [tex]J/mol^{o}C[/tex]

Thus, we can conclude that specific heat of tin is 24.5 [tex]J/mol^{o}C[/tex].

Final answer:

To calculate the specific heat capacity of the metal, we can use the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. By using the known temperatures and masses of the metal and water, we can solve for the specific heat capacity of the metal.

Explanation:

To calculate the specific heat capacity of the metal, we can use the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we know the initial and final temperatures of the metal and water, as well as the masses of the metal and water. We can rearrange the equation to solve for the specific heat capacity of the metal, c.

We can start by calculating the heat transferred from the metal to the water. The heat transferred to the water can be calculated using the equation q_water = m_water * c_water * ΔT_water, where m_water is the mass of the water, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.

Next, we can calculate the heat transferred from the metal to the water using the equation q_metal = m_metal * c_metal * ΔT_metal, where m_metal is the mass of the metal, c_metal is the specific heat capacity of the metal, and ΔT_metal is the change in temperature of the metal.

Since the metal and water reach the same final temperature, we can set q_water equal to q_metal and solve for the specific heat capacity of the metal, c_metal.

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How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.

Answers

Here is the full question

How many moles of sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00? Ignore the volume change due to the addition of sodium acetate.  Ka of acetic acid is 1.7×10-5.

Answer:

0.3396 moles

Explanation:

pH = pKA + log[tex]\frac{[acetate]}{[aceticacid]}[/tex]

[acetic acid] = 0.1 M

pH = 5.0

[tex]K_a[/tex] = [tex]1.7*10^{-5[/tex]

5.0 = -log ([tex]1.7*10^{-5[/tex]) + log [tex]\frac{[acetate]}{0.1}[/tex]

5.0 = 4.77 + log [acetate] + 1

5.0 = 5.77 + log [acetate]

- log [acetate] = 5.77 - 5.0

- log [acetate]  = 0.77

log [acetate]  = -0.77

[acetate]  = log⁻¹ (-0.77)

[acetate]  =  0.1698

∴ for 2L = 2 × 0.1698

= 0.3396 moles

The sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles

Given:

The concentration of acetic acid = 0.10 M

pH = 5.0  

[tex]K_a[/tex] = [tex]1.7\times10^{-5[/tex]

Solution:

According to Hasselbalch equation:

[tex]pH=pKa + log\frac{[A^-]}{[HA]}\\\\or\\\\pH = pKa + log\frac{(acetate)}{(acetic acid)}[/tex]

5.0 = -log [tex](1.7\times10^{-5})[/tex] + log [tex]\frac{(acetate)}{(0.1)}[/tex]

5.0 = 4.77 + log [acetate] + 1

5.0 = 5.77 + log [acetate]

- log [acetate] = 5.77 - 5.0

- log [acetate]  = 0.77

log [acetate]  = -0.77

[acetate]  = log⁻¹ (-0.77)

[acetate]  =  0.1698

∴ for 2L = 2 × 0.1698

= 0.3396 moles

Thus, the sodium acetate must be added to 2.0 L of 0.10 M acetic acid to give a solution that has a pH equal to 5.00 - 0.3396 moles

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Argon, which comprises almost 1 percent of the atmosphere, is approximately 27 times more abundant than CO2, but does not contribute to global warming. Which explanation accounts for this fact

Answers

Transparent to the wavelengths of light

Explanation:

Argon, an element comprises of almost 1 percent in the Earth’s atmosphere, is approximately 27 times more abundant than CO2, is not a greenhouse gas. The reason that it is not a greenhouse gas or not contribute in global warming is that it is largely transparent to the wavelengths of light that is responsible for trapping heat that is similar to that of oxygen, nitrogen and other gasesArgon is not able to form large and complex molecules that are enough to block infrared light, which is generally done by greenhouse gases such as carbon dioxide and methane also.

Consider the reaction of hydrogen peroxide with iodine: H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) The reaction is first order in I2 and second order overall. What is the rate law?

Answers

Answer:

Consider the reaction of hydrogen peroxide with iodine: H2O2(aq)+I2(aq)⇌OH−(aq)+HIO(aq) The reaction is first order in I2 and second order overall.

What is the rate law?

If the concentration of H2O2 is increased by half and the concentration of I2 is quadrupled, by what factor does the reaction rate increase?

Rate law = k[H₂O₂] [I₂]

The new rate is six times that of the old rate

Explanation:

Rate law = k[H₂O₂] [I₂]

lets input

[H₂O₂] = x

[I₂] = y

Rate = kxy

[H₂O₂] = x + x/2

= 3x/2

[I₂] = 4y

New rate = [tex]k * \frac{3x}{2} *4y[/tex]

New rate = 6

The new rate is six times that of the old rate

At what temperature in K will 4.00 moles of gas occupy a volume of 12.0 L at a pressure of 5.60 atm?

Answers

Answer:

The answer to your question is  T = 204.9°K

Explanation:

Data

Temperature = T = ?

number of moles = n = 4

Volume = V = 12 L

Pressure = P = 5.60 atm

constant of gases = 0.082 atm L /mol °K

Process

To solve this problem, use the Ideal Gas Law and solve it for T

Formula

            PV = nRT

            T = PV / nR

-Substitution

             T = (5.60)(12)/(4)((0.082)

-Simplification

              T = 67.2 / 0.328

-Result

               T = 204.9°K

Considering the ideal gas law, at 204.878 K will 4.00 moles of gas occupy a volume of 12.0 L at a pressure of 5.60 atm.

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances. The numerical value of R will depend on the units in which the other properties are worked.

In this case, you know:

P= 5.60 atmV= 12 Ln= 4 molesR= 0.082 [tex]\frac{atmL}{molK}[/tex]T=?

Replacing in the ideal gas law:

5.60 atm×12  L = 4 moles×0.082[tex]\frac{atmL}{molK}[/tex]×T

Solving:

[tex]T=\frac{5.60 atmx12 L}{4 moles x 0.082 \frac{atmL}{molK}}[/tex]

T=204.878 K

Finally, at 204.878 K will 4.00 moles of gas occupy a volume of 12.0 L at a pressure of 5.60 atm.

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The normal boiling point of a certain liquid is , but when of urea are dissolved in of , it is found that the solution boils at instead. Use this information to calculate the molal boiling point elevation constant of .

Answers

Answer:

The molal boiling point elevation constant (Kb) is calculated by dividing the change in boiling point (∆Tb) by the molality of the solution (m)

Explanation:

Kb = ∆Tb/m

Kb is the molal boiling point elevation constant of the liquid

∆Tb is the change in boiling point. It is calculated by subtracting the initial boiling point of the liquid from the final boiling point of the solution.

m is the molality of the solution. It is calculated by dividing the number of moles of urea (number of moles of urea is calculated by dividing the mass in grams of urea by its molecular weight) by the mass of the liquid in kilograms.

If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

Answers

Here is the full question.

Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L

If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

Answer:

7.90%

Explanation:

The equation for the reaction can be written as:

[tex]Na_2CO_3+Ca^{2+} ---------> CaCO_3(s) +2Na^+[/tex]

[tex]Na_2CO_3+Mg^{2+} ---------> MgCO_3(s) +2Na^+[/tex]

Molar mass of [tex]Na_2CO_3[/tex] = 106 g/mol

1.00 gallon of water = 3.785 L

∴ 24.5 gallons of water = 24.5 × 3.785

= 92.7325 L

mass precipitate = [tex][(3.8*10^{-3})+(1.1*10^{-3})]\frac{mol}{L} *92.7325*\frac{106g}{mol}[/tex]

mass precipitate = 48.162605

mass precipitate = 48.2 g

mass % = [tex]\frac{mass ofprecipitate}{mass of detergent} *100%[/tex]%

mass % = [tex]\frac{48.2g}{0.61kg}*\frac{1kg}{1000g}*100[/tex]%

mass % = 7.901639344

mass % = 7.90 %

The percentage (by mass) of the detergent should be 7.90%.

Calculation of the percentage:

Since 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L

The molar mass of Na_2CO_3 is 106g/mol

Since 1.00 gallon of water = 3.785 L

so, for 24.5 gallons of water, it is

= 24.5 × 3.785

= 92.7325 L

Now mass precipitate is

= ((3.8*10^-3) * (1.1*10^-3) * 92.7325 * 106g/mol

= 48.162605

= 48.2g

Now the mass percentage is

= mass of precipitate / mass of detergent

= 48.3g / 0.61 * 1kg / 1000g

= 7.90%

This is an incomplete question. Please find the full question below.

Sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.8×10−3M in Ca2+ and 1.1×10−3M in Mg2+ and that the average capacity of a washing machine is 24.5 gallons of water. 1gallon=3.785L

If the detergent requires using 0.61 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

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As their name suggests, calcium ion channel blockers block calcium ion channels in the sarcolemma of pacemaker and contractile cardiac muscle cells and slow calcium ion entry into the cell during an action potential. What effects would these drugs have on the myocardial cells? View Available Hint(s)

Answers

Answer:

Effects of Calcium channel blockers on the myocardial cells include lower excitability of the cells due lower influx of calcium ions during an action potential. This effect helps to regulate abnormal rapid heart rhythms (arrhythmias).

Calcium channel blockers also reduce blood pressure by dilating the arteries thereby reducing pressure in the arteries. This in turn also reduces the workload on the heart and therefore allows for adequate oxygen supply to heart in the treatment of Angina.

Examples of calcium ion channel blockers are amlodipine, nimodipine, verapamil, diltiazem etc.

The most abundant elements in the universe are hydrogen and helium, but there are also small but significant amounts of heavier elements in stars and planets and in our own bodies. Where did these heavy elements originate?

Answers

Answer:

in nuclear fusion deep in the interiors of stars

Explanation:

Nuclear fusion -

It is the type of reaction , where two or more atomic nuclei of the atom merges together to release two or more different nuclei along with some subatomic particles , is referred to as a nuclear fusion reaction .

The reaction can very well be done on stars , because of very high energy .

Hence , a nuclear fusion occurs deep inside the stars .

During digestion, hydrochloric acid (HCl) lowers the stomach's pH and this causes the enzyme pepsinogen to be converted to its active form, called pepsin. Pepsin is a proteolytic enzyme, which means it cleaves (breaks) peptide bonds assocciated with food particles. The enzyme pepsin can also cleave pepsinogen into pepsin, and this causes the rate at which pepsinogen is converted to pepsin to increase. This is an example of:

Answers

Answer:

Catalysis

Explanation:

Pepsin is able to break peptide bonds, turning large protein molecules into small peptide chains.

When pepsin acts to break down pepsinogen (inactive form of pepsin), it is accelerating pepsinogen → pepsin reactions, acting as a catalyst, reducing activation energy and favoring proteolytic reactions at a higher rate.

This process of accelerating reactions is characteristic of enzymes and is known as catalysis.

Determine the volume of SO2 (at STP) formed from the reaction of 96.7 g of FeS2 and 55.0 L of O2 (at 398 K and 1.20 atm). The molar mass of FeS2 is 119.99 g/mol. 4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)

Answers

Answer:  32.9 Liters

Explanation:

To calculate the moles, we use the equation:

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

1. moles of [tex]FeS_2=\frac{96.7g}{119.99g/mol}=0.806mol[/tex]

2. moles of [tex]O_2[/tex]

[tex]PV=nRT[/tex]

P = pressure of the gas = 1.20  atm

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]398K[/tex]

[tex]1.20\times 55.0=n\times 0.0821\times 398[/tex]

[tex]n=2.02[/tex]

[tex]4FeS_2(s)+11O_2(g)\rightarrow 2Fe_2O_3(s)+8SO_2(g)[/tex]

According to stoichiometry:

11 moles of oxygen reacts with 4 moles of [tex]FeS_2[/tex]

Thus 2.02 moles of oxygen reacts with =[tex]\frac{4}{11}\times 2.02=0.73[/tex] moles of [tex]FeS_2[/tex]

Thus oxygen acts as limiting reagent and [tex]FeS_2[/tex] is excess reagent.

As 11 moles of oxygen gives = 8 moles of [tex]SO_2[/tex]

2.02 moles of oxygen gives =[tex]\frac{8}{11}\times 2.02=1.47[/tex] moles of [tex]SO_2[/tex]

[tex]PV=nRT[/tex]

P = pressure of the gas = 1  atm (at STP)

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]273K[/tex]   (at STP)

[tex]1\times V=1.47\times 0.0821\times 273[/tex]

[tex]V=32.9L[/tex]

Thus volume of [tex]SO_2[/tex] (at STP) formed from the reaction is 32.9 L

The volume of SO₂ formed from the reaction at STP is 32.92 L

From the question,

We are to determine the volume of SO₂ formed from the reaction

The given balanced chemical equation for the reaction is

4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)

This means,

4 moles of FeS₂ reacts with 11 moles of oxygen to produce 2 moles of Fe₂O₃ and 8 moles of SO₂

First, we will determine the number of moles of each reactant present

For FeS₂

Mass = 96.7 g

From the formula

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

Molar mass of FeS₂ = 119.99 g/mol

∴ Number of moles of FeS₂ present =[tex]\frac{96.7}{119.99}[/tex]

Number of moles of FeS₂ present = 0.8059 mole

For O₂

Using the formula

PV = nRT

[tex]n =\frac{PV}{RT}[/tex]

Putting the given parameters into the formula, we get

[tex]n = \frac{1.2 \times 55.0}{0.08206 \times 398}[/tex]

[tex]n = \frac{66}{32.65988}[/tex]

n = 2.0208 moles

Since,

4 moles of FeS₂ reacts with 11 moles of oxygen to produce 8 moles of SO₂

Then,

[tex]\frac{2.0208 \times 4}{11}[/tex] moles of FeS₂ will react with 2.0208 moles of oxygen to produce [tex]\frac{2.0208 \times 4}{11} \times 2[/tex]  moles of SO₂

That is,

0.7348 moles of FeS₂ will react with 2.0208 moles of oxygen to produce 1.4697 moles of SO₂

∴ Number of moles of SO₂ formed is 1.4697 moles

Now for the volume of SO₂ formed at STP

Since

1 mole of a gas has a volume of 22.4 L at STP

Then

1.4697 moles of the SO₂ will have a volume of 1.4697 × 22.4 L

1.4697 × 22.4 L = 32.92L

Hence, the volume of SO₂ formed from the reaction at STP is 32.92 L

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Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide:
4HCL(aq) + MnO2(s) ----> MnCl2(aq) + 2H2O(l) + Cl2(g)
You add 42.5 g of MnO2 to a solution containing 47.7 g of HCl.
(a) What is the limiting reactant? MnO2 or HCL?
(b) What is the theortical yield of CO2?
(c) If the yield of the reaction is 79.9%, what is the actual yield of chlorine?

Answers

Answer:

a) HCl

b) 22.9g

c) 18.11g

Explanation:

MMn = 54.94g/mol

MO2 = 2(16) = 32g/mol

MH = 1g/mol

MCl = 35.45g/mol

Molar Mass of MnO2:

54.94 + 2(16)= 86.94

Molar Mass of HCl:

1+35.45=36.45

Mols of MnO2:

42.7 /86.94 = 0.49

Mols of HCl:

47.1 /36.45 = 1.29

Molar Mass of Cl2:

35.45 ×2 = 70.9

Mols of Cl2

1.29/4=0.323

Mass of Cl2 (Theoretical yield)

0.323 ×70.9= 22.9

To calculate the actual yield, we multiply the theoretical yield by the final percentage:

22.9 ×0.791=18.11

Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) BrN3 amu (b) C2H6 amu (c) NF2 amu (d) Al2S3 amu (e) Fe(NO3)3 amu (f) Mg3N2 amu (g) (NH4)2CO3 amu

Answers

Answer:

Explanation:

The atomic unit of mass (a.m.u) is the unit used to compare the relative masses of atoms. An atomic mass unit is a twelfth of the mass of a carbon-12 atom.

The calculation of the molecular mass of a compound is done by adding the relative atomic masses of the atoms of the formula of said substance, taking into account its abundance in the compound. The atomic mass of the atoms that make up the molecule are obtained in the Periodic Table.

So,  in these cases you have:

BrN₃:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

Br: 79.9 amu

N: 14 amu

I take into account the abundance of each element in the compound you get:

BrN₃=79.9 amu + 3* 14 amu= 121.9 amu

C₂H₆:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

C: 12 amu

H: 1 amu

I take into account the abundance of each element in the compound you get:

C₂H₆= 2*12 amu + 6*1 amu= 30 amu

NF₂:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

N: 14 amu

F: 19 amu

I take into account the abundance of each element in the compound you get:

NF₂= 14 amu + 2*19 amu= 52 amu

Al₂S₃:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

Al: 27 amu

S: 32 amu

I take into account the abundance of each element in the compound you get:

Al₂S₃= 2*27 amu + 3*32 amu= 150 amu

Fe(NO₃)₃:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

Fe: 55.85 amu

N: 14 amu

O: 16 amu

I take into account the abundance of each element in the compound you get:

Fe(NO₃)₃= 55.85 amu + 3*(14 amu + 3*16 amu)= 241.85 amu

Mg₃N₂:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

N: 14 amu

Mg: 24 amu

I take into account the abundance of each element in the compound you get:

Mg₃N₂= 3*24 amu + 2*14 amu= 100 amu

(NH₄)₂CO₃:

You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:

N: 14 amu

H: 1 amu

C: 12 amu

O: 16 amu

I take into account the abundance of each element in the compound you get:

(NH₄)₂CO₃: 2*(14 amu + 4*1 amu) + 12 amu+3*16 amu= 96 amu

Final answer:

The molecular mass or formula mass of the listed substances have been calculated using atomic masses from the periodic table.

Explanation:

The molecular mass or formula mass of a substance can be calculated by summing up the atomic masses of all atoms in the substance. Using the atomic masses from the periodic table, we have:

BrN3 = 1(Br) + 3(N) = 79.9 + 3(14.0) = 122.9 amu C2H6 = 2(C) + 6(H) = 2(12.0) + 6(1.0) = 30.0 amu NF2 = 1(N) + 2(F) = 14.0 + 2(19.0) = 52.0 amu Al2S3 = 2(Al) + 3(S) = 2(26.98) + 3(32.06) = 150.16 amu Fe(NO3)3 = 1(Fe) + 3(1(N) + 3(O)) = 55.85 + 3(1(14.0) + 3(16.0)) = 241.85 amu Mg3N2 = 3(Mg) + 2(N) = 3(24.3) + 2(14.0) = 100.9 amu (NH4)2CO3 = 2(1(N) + 4(H)) + 1(C) + 3(O) = 2(1(14.0) + 4(1.0)) + 12.0 + 3(16.0) = 96.0 amu

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Write a statement that reads a floating point (real) value from standard input into temperature. Assume that temperature. has already been declared as an double variable.

Answers

Answer:

temperature = stdin.nextDouble();

Explanation: since the temperature has already been declared as a double variable, the above statement will read so.

As temperature equal standard input dot next double variable.

Final answer:

To read a floating point value from standard input into a double variable, use the scanf function with %lf format specifier.

Explanation:

To read a floating point (real) value from standard input into a double variable named temperature, you can use the scanf function in the C programming language. The scanf function allows you to read formatted input, and the %lf format specifier is used to read a double value. Here's an example:

scanf("%lf", &temperature);

This line of code will read a floating point value from the standard input and store it in the temperature variable declared as a double.

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In order to get high-quality energy, you must use high-quality energy. This rule is the result of which of the following? a.the law of gravityb.the universal law of energyc.the second law of thermodynamicsd.the law of diminishing returnse.the first law of thermodynamics

Answers

Answer: C. The Second Law of Thermodynamics.

Explanation:

The law of gravity is related to the force interaction due to gravitation between two bodies (i.e. a person and planet Earth, a star and a planet).

The law of diminishing return is not used in Physics, but in Economics and describes the diminishing of marginal returns in time.

The first law of Thermodynamics analyses the energy interactions of a system in a quantitative manner and it is a generalized form of the Principle of Energy Conservation. It is oriented to the analyses of energy inflows and outflows.

The second law of Thermodynamics analyses the energy interactions of a system in a qualitative manner and it is based on thermodynamic property named Entropy, which may helpful to measure irreversibilities associated with system and irreversibility generation as well, provided that a comparison with another equivalent system exists.

Irreversibilities depends on the characteristics of system and nature of energy sources. Empirically, it is known that heat offers a lower quality than electricity in order to get a available work. That is to say, there is a lower of obtaining available work from heat than from electricity.

Hence, the statement is a consequence of using the second law of Thermodynamics and, therefore, the correct answer is C.

A compound is 92.2% Carbon and 7.76% Hydrogen. The formula mass of the compound is 78.1 g. Calculate the empirical formula and molecular formula of the compound.

Answers

Answer:

The answer to your question is empirical formula = CH

                                                    molecular formula = C₆H₆

Explanation:

Data

Carbon  92.2%

Hydrogen  7.76%

Formula mass = 78.1 g

Process

1.- Express the percents as grams

Carbon            92.2 g

Hydrogen          7.76 g

2.- Convert the grams to moles

Carbon                 12 g ---------------- 1 mol

                            92.2 g -------------  x

                            x = (92.2 x 1)/12

                            x = 7.68 moles

Hydrogen             1 g ------------------ 1 mol

                            7.76 g -------------  x

                            x = 7.76 moles

3.- Divide by the lowest number of moles

Carbon         7.68 / 7.68 = 1

Hydrogen     7.76 / 7.68 = 1.01

4.- Write the empirical formula

                                      CH

5.- Calculate the molecular weight of the empirical formula

CH = 12 + 1 = 13

6.- Divide the molecular weight by the molecular weight of the empirical formula

                78.1 / 13 = 6

7.- Write the molecular formula

               6(CH) = C₆H₆        

The empirical formula of the compound is CH. The molecular formula of the compound is [tex]\rm \bold C_6H_6[/tex]

A molecular formula has been the one that represents the exact number of atoms in a compound. The empirical formula has been able to represent the whole number ratio of atoms present in a compound.

The given compound has 92.2% Carbon and 7.76% Hydrogen. The weight will be:

Carbon = 92.2 g

Hydrogen = 7.76 g

From the weight, the moles of elements in the compound can be calculated.

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of carbon = [tex]\rm \dfrac{92.2}{12}[/tex]

Moles of carbon = 7.76 moles

Moles of hydrogen = weight of hydrogen

Moles of hydrogen = 7.76 moles

For the empirical formula, divide the moles of each element to the nearest mole number;

Carbon = [tex]\rm \dfrac{7.76}{7.76}[/tex]

Carbon = 1

Hydrogen = [tex]\rm \dfrac{7.76}{7.76}[/tex]

Hydrogen = 1

Thus, the empirical formula of the compound will be CH.

For finding the molecular formula, the molecular weight from the empirical formula has been divided with the molecular weight of the compound to find the number of atoms.

Molecular weight of empirical formula = weight of carbon + weight of hydrogen

Molecular weight of empirical formula = 12 + 1

Molecular weight of empirical formula = 13 grams

Molecular mass of compound = 78.1 grams.

Number of atoms of each element = [tex]\rm \dfrac{78.1}{13}[/tex]

Number of atoms of each element = 6

Thus, the molecular formula of the compound is [tex]\rm C_6H_6[/tex].

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"In carrying out a titration of a hydrochloric acid solution with a standard sodium hydroxide solution, a student went beyond the end point before reading the volume on the burette. That is, the volume used was larger than the volume required to reach the end point. How will this error affect the calculated concentration of the hydrochloric acid?"

Answers

Answer:

The calculated concentration of HCl will be less than actual.

Explanation:

Suppose during titration, the HCl was taken in burette and the NaOH in the volumetric flask.

Now we will use equivalence formula for the calculation of concentration of HCl.

             [tex]N_{1} V_{1} = N_{2} V_{2}[/tex]

Where L.H.S is for hydrochloric acid and R.H.S is for sodium hydroxide. The terms N and V represent normality and volume respectively.

If we calculate for

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

We see that if the volume of the HCl is greater then the concentration of the HCl will be reduced.

The error will cause the concentration of the hydrochloric acid to be underestimated.

The concentration of a solution is calculated from the ratio of the number of moles of the solutes and that of the volume of the solution.

Mathematically; concentration = mole/volume

Thus, with the number of moles being constant, the higher the volume of the solution, the lower the concentration that would be derived and vice versa.

This means that any volume that exceeds that of the accurate endpoint will cause the concentration to be underestimated and below the endpoint, the concentration would be overestimated.

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N2O5 decomposes to form NO2 and O2 with first-order kinetics. How long does it take for the N2O5 concentration to decrease from its initial value of 2.75 M to its final value of 1.85 M, if the rate constant, k, equals 5.89 × 10−3?

Answers

Answer:

67.3 s

Explanation:

The equation for the reaction can be represented as:

N₂O₅               ⇄                     NO₂        +          O₂

Rate (k) = [tex]5.89 * 10^{-3[/tex]

Rate law for first order is expressed as:

In [A] = -kt + In [A]₀

Given that:

[A] = Final Concentration = 1.85 M

[A]₀ = Initial Concentration = 2.75 M

time-taken  = ???

substituting our given data; we have:

In[1.85] = -[5.89 × 10⁻³](t) + In [2.75]

t = [tex]\frac{In(2.75)-In(1.85)}{(5.89*10^{-3}}[/tex]

t = [tex]\frac{0.3964}{5.89*10^{-3}}[/tex]

t = 67.3 s

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

Answers

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

[tex]R_{in}[/tex] [tex]= \frac{0.04}{100}*2000[/tex]

[tex]R_{in}[/tex] = 0.8

The rate-out

[tex]R_{out}[/tex] = [tex]\frac{A}{6000}*2000[/tex]

[tex]R_{out}[/tex] = [tex]\frac{A}{3}[/tex]

We can say that:

[tex]\frac{dA}{dt}=[/tex] [tex]0.8-\frac{A}{3}[/tex]

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

[tex]\frac{dA}{dt} +\frac{A}{3} =0.8[/tex]

Integration of the above linear equation =

[tex]e^{\int\limits \frac {1}{3}dt } =[/tex] [tex]e^{\frac{1}{3}t[/tex]

so we have:

[tex]e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A[/tex] [tex]= 0.8e^{\frac{1}{3}t[/tex]

[tex]\frac{d}{dt}[e^{\frac{1}{3}t}A][/tex] [tex]= 0.8e^{\frac{1}{3}t[/tex]

[tex]Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C[/tex]

∴ [tex]A(t) = 2.4 +Ce^{-\frac{1}{3}t[/tex]

Since A(0) = 12

Then;

[tex]12 =2.4 + Ce^{-\frac{1}{3}}(0)[/tex]

[tex]C= 12-2.4[/tex]

[tex]C =9.6[/tex]

Hence;

[tex]A(t) = 2.4 +9.6e^{-\frac{t}{3}}[/tex]

[tex]A(0) = 2.4 +9.6e^{-\frac{10}{3}}[/tex]

[tex]A(t) = 2.74[/tex]

∴ the concentration at 10 minutes is ;

=  [tex]\frac{2.74}{6000}*100[/tex]%

= 0.0456667 %

= 0.046% to three decimal places

What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by mass? Insert subscripts as needed.

Answers

Answer : The empirical of the compound is, [tex]C_1H_2O_3[/tex]

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 19.36 g

Mass of H = 3.25 g

Mass of O = 77.39 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{19.36g}{12g/mole}=1.613moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{3.25g}{1g/mole}=3.25moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{77.39g}{16g/mole}=4.837moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{1.613}{1.613}=1[/tex]

For H = [tex]\frac{3.25}{1.613}=2.01\approx 2[/tex]

For o = [tex]\frac{4.837}{1.613}=2.99\approx 3[/tex]

The ratio of C : H : O = 1 : 2 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_1H_2O_3[/tex]

Therefore, the empirical of the compound is, [tex]C_1H_2O_3[/tex]

Final answer:

The empirical formula of a compound with 3.25% H, 19.36% C, and 77.39% O by mass is CH2O3, found by converting the mass of each element to moles and then determining the simplest whole number ratio.

Explanation:

To determine the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass, we first assume a 100 g sample of the compound. This means we would have 3.25 g of H, 19.36 g of C, and 77.39 g of O.

Next, we convert the mass of each element to moles by dividing by their respective molar masses:

Carbon: 19.36 g / 12.01 g/mol = 1.612 molesHydrogen: 3.25 g / 1.008 g/mol = 3.225 molesOxygen: 77.39 g / 16.00 g/mol = 4.837 moles

To find the simplest whole number ratio, we divide each mole value by the smallest number of moles calculated:

Carbon: 1.612 / 1.612 = 1Hydrogen: 3.225 / 1.612 = 2Oxygen: 4.837 / 1.612 = 3

The ratios indicate the empirical formula is CH2O3.

The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, whereas nitrogen gas required 48 s. The molar mass of the gas is ________

Answers

Final answer:

The molar mass of the unknown gas can be determined by comparing its effusion rate to the effusion rate of a known gas, such as nitrogen gas. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Let's set up a proportion to find the molar mass of the unknown gas.

Explanation:

The molar mass of the unknown gas can be determined by comparing its effusion rate to the effusion rate of a known gas, such as nitrogen gas. In this case, it took 60 seconds for the unknown gas to effuse, whereas nitrogen gas required 48 seconds. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. Using this information, we can set up a proportion to find the molar mass of the unknown gas.

Let's assume the molar mass of the unknown gas is represented by 'X'. The proportion can be set up as follows:

(rate of unknown gas)/(rate of nitrogen gas) = sqrt(molar mass of nitrogen gas)/sqrt(molar mass of unknown gas)

Substituting the known values, we have:

(60 s)/(48 s) = sqrt(28.01 g/mol)/sqrt(X)

Solving for X, the molar mass of the unknown gas is approximately 37.07 g/mol.

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The initial concentration of C2H6N2 in the first-order reaction C2H6N2→N2+C2H6 is 0.427 M. After 30. seconds, the concentration of C2H6N2 is 0.395 M. What is the rate constant k for the reaction?

Answers

Answer:

2.6×10^-3 mols-1

Explanation:

From

k= 2.303/t × log[A]o/[A]t

Time taken t= 30 seconds

Initial concentration [A]o= 0.427M

Concentration at time (t) [A]t= 0.395M

k= 2.303/30 ×log[0.427]/[0.395]=2.6×10^-3 mols-1


On the periodic table, calcium and bromine are located in period 4. What is one difference between the two elements?

A. Calcium atoms are more reactive than bromine atoms.

B. Bromine and calcium atoms will each combine with the same elements to form compounds.

C. Calcium atoms combine more easily with other atoms in Group 2.

D. Bromine atoms have more electrons in their outer shell than calcium atoms.

Answers

Explanation:

it's is d cause bromine can even react with water so

Bromine atoms have more electrons in their outer shell than calcium atoms. So the correct option is D.

What is valency?

In chemistry, valence, usually spelled valency, is the characteristic of an element that establishes the maximum number of other atoms that one atom of the element may combine with. The phrase, which was first used in 1868, is used to indicate both the broad power of a combination of elements as well as its numerical value.

For 19th-century chemists, the explanation and systematization of valence posed significant difficulties. The majority of the work was focused on developing empirical guidelines for finding the valences of the elements because there was no good hypothesis for its origin.

The amount of hydrogen atoms that an element's atom may mix with or replace in a compound is used to quantify the characteristic valences of the various elements. But it soon became clear that various compounds had varied valences for numerous elements.

The identification of the chemical bond of organic compounds with a pair of electrons held jointly by two atoms and acting to hold them together in 1916 by the American chemist G.N. Lewis represented the first significant step in the development of a satisfactory explanation of valence and chemical combination. The German scientist W. Kossel studied the nature of the chemical connection between electrically charged atoms (ions) in the same year.

The theory of valence was reconstructed in terms of electronic structures and interatomic forces following the advent of the precise electronic theory of the periodic system of the elements. In response to the many ways that atoms interact, new notions such as ionic valence, covalence, oxidation number, coordination number, and metallic valence were developed.

Therefore the correct option is D.

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The ionization energy of atoms ________. Group of answer choices does not change going down within a group increases going down within a group decreases going down within a group none of the above decreases going across a period

Answers

Answer:

decreases going down within a group

Explanation:

Ionization energy of an atom is defined as the energy required to remove electron from the gaseous form the atom. The energy required to remove the highest placed electron in the gaseous form of an atom is referred to as the first ionization energy.

In the periodic table, the first ionization energy decreases down the group because as the principal quantum number increases, the size of the orbital increases and the electron is easier to remove.

In addition, the first ionization energy increases across the period because electrons in the same principal quantum shell do not completely shield the increasing nuclear charge of the protons.

Final answer:

The ionization energy of atoms decreases going down a group on the periodic table and increases going across a period. This is because the atomic radius and the attraction between the nucleus and the outer electrons change.

Explanation:

The ionization energy of atoms is the energy required to remove an electron from an atom or ion. As you move down a group on the periodic table, the ionization energy generally decreases. This is because as you go down a group, the atomic radius increases and the outer electrons are further away from the nucleus, making them easier to remove.

On the other hand, as you go across a period from left to right, the ionization energy generally increases. The electrons are closer to the nucleus and thus more strongly attracted to the center, making them more difficult to remove. So, the correct choices would be 'decreases going down within a group' and 'increases going across a period'.

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When a nonvolatile solute is added to a solvent, it tends to ___________________ the freezing point and ___________________ the boiling point, compared to the pure solvent.

Answers

Answer:

decreases and increases

Explanation:

the addition of solutes can alter the freezing point and the boiling point .the changes in solute content where there is an addition of solutes results in the dropping of freezing point and increase in boiling point.this is known as freezing point depression and boiling point depression.

An aqueous solution of calcium hydroxide is standardized by titration with a 0.164 M solution of hydroiodic acid. If 17.2 mL of base are required to neutralize 14.7 mL of the acid, what is the molarity of the calcium hydroxide solution?

Answers

Answer:

0.07 M

Explanation:

We have to start with the reaction between the Calcium hydroxide and the Hydroiodic acid:

[tex]Ca(OH)_2~+~HF~->~CaF_2~+~H_2O[/tex]

Then we have to balance the reaction:

[tex]Ca(OH)_2~+~2HF~->~CaF_2~+~2H_2O[/tex]

We have to keep in mind that we have to do use the volume in "L" units (14.7 mL= 0.0147L). When we plug the values into the equation we will get:

[tex]M=\frac{#mol}{L}[/tex]

[tex]0.164~M=\frac{#mol}{0.0147}[/tex]

[tex]#mol=~0.0024[/tex]

Now, we have to use the information from the balanced reaction to finding the moles of Calcium hydroxide. When we check the reaction we found a molar ratio of 1:2 ( 1 mol of [tex]Ca(OH)_2[/tex]: 2 mol of  [tex]HF[/tex]). With this molar ratio, we can find the moles of [tex]HF[/tex].

[tex]0.0024~mol~HF\frac{1~mol~Ca(OH)_2}{2~mol~HF}[/tex]

[tex]0.0012~mol~Ca(OH)_2[/tex]

Finally, to find the molarity we have to divide the moles of [tex]Ca(OH)_2[/tex] and the volume of [tex]Ca(OH)_2[/tex] in liters (17.2 mL=0.0172L) so:

[tex]M=\frac{0.012}{0.0172}=0.07[/tex]

The molarity of [tex]Ca(OH)_2[/tex] is 0.07M.

Please Answer! Help! Will Give Brainliest. Find Keq for a 3.0 L container with 1.2 moles of both NO2 and N2O4 initially and 0.38 M NO2 at equilibrium. 2NO2(g) ⇌ N2O4(g)

Answers

Answer:

Kc → 41.9

Explanation:

This is the equilibrium:

2NO₂(g) ⇌ N₂O₄(g)

So the expression for Kc will be:

Kc = [N₂O₄] / [NO₂]²

We prospose the situations:

Initially we have 1.2 moles of NO₂ and N₂O₄

X amount has reacted. As stoichiometry is 2:1, we have produced x/2 of the product during the reaction

Finally In equilibrium we have, 0.38 NO₂

                2NO₂(g)     ⇌       N₂O₄(g)

Initially        1.2                        1.2

React            x                         x/2

Eq         (1.2 - x) = 0.38          1.2 + x/2

As we have [NO₂] in the equilibrium, we can determine x (the amount that has reacted) to solve and determine, the [N₂O₄] in the equilibrium

1.2-0.38 = x → 0.82

1.2 + 0.82/2 = 2.02 → [N₂O₄]

For Kc, we need Molar concentration, so we have to divide [N₂O₄] and [NO₂] by the volume

[N₂O₄] → 2.02 mol/3L = 0.673 M

[NO₂] → 0.38 mol/3L = 0.127 M

Now we can replace the Kc expression:

Kc →  [N₂O₄] / [NO₂]² → 0.673 / 0.127² = 41.9

Remember that Kc has no UNITS

A calorimeter contains 32.0 mLmL of water at 15.0 ∘C∘C . When 1.20 gg of XX (a substance with a molar mass of 54.0 g/molg/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq)X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 29.0 ∘C∘C . Calculate the enthalpy change, ΔHΔHDelta H, for this reaction per mole of XX. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)J/(g⋅∘C)], that density of water is 1.00 g/mLg/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Answers

The enthalpy change [tex](\( \Delta H \))[/tex] for this reaction per mole of [tex]\( X \)[/tex] is approximately 88.27 kj/mol.

To calculate the enthalpy change[tex](\( \Delta H \))[/tex] for the reaction per mole of  X , we can use the formula:

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

where:

- [tex]\( q \)[/tex] is the heat absorbed or released by the reaction,

- [tex]\( n \)[/tex]  is the moles of [tex]\( X \).[/tex]

First, let's find the moles of \( X \):

[tex]\[ \text{moles of } X = \frac{\text{mass}}{\text{molar mass}} \][/tex]

[tex]\[ \text{moles of } X = \frac{1.20 \, \text{g}}{54.0 \, \text{g/mol}} \][/tex]

[tex]\[ \text{moles of } X \approx 0.0222 \, \text{mol} \][/tex]

Now, we need to find the heat [tex](\( q \))[/tex] absorbed or released by the reaction. The heat can be calculated using the formula:

[tex]\[ q = mc\Delta T \][/tex]

where:

- m  is the mass of the solution [tex](water + \( X \)),[/tex]

- c  is the specific heat of the solution,

- [tex]\( \Delta T \)[/tex]is the change in temperature.

First, calculate the mass of the solution:

[tex]\[ \text{mass of solution} = \text{mass of water} + \text{mass of } X \][/tex]

[tex]\[ \text{mass of solution} = 32.0 \, \text{g} + 1.20 \, \text{g} \][/tex]

[tex]\[ \text{mass of solution} = 33.20 \, \text{g} \][/tex]

Now, calculate [tex]\( \Delta T \)[/tex]:

[tex]\[ \Delta T = T_f - T_i \][/tex]

[tex]\[ \Delta T = 29.0^\circ \text{C} - 15.0^\circ \text{C} \][/tex]

[tex]\[ \Delta T = 14.0^\circ \text{C} \][/tex]

Now, plug in the values into the heat formula:

[tex]\[ q = (33.20 \, \text{g}) \times (4.18 \, \text{J/(g} \cdot ^\circ \text{C})) \times (14.0 \, ^\circ \text{C}) \][/tex]

[tex]\[ q \approx 1961.12 \, \text{J} \][/tex]

Now, use the first formula to find [tex]\( \Delta H \):[/tex]

[tex]\[ \Delta H = \frac{1961.12 \, \text{J}}{0.0222 \, \text{mol}} \][/tex]

[tex]\[ \Delta H \approx 88272 \, \text{J/mol} \][/tex]

Therefore, the enthalpy change[tex](\( \Delta H \))[/tex] for this reaction per mole of X is approximately[tex]\( 88.27 \, \text{kJ/mol} \).[/tex]

Calculate the heat absorbed using q = mcΔT and convert to kJ. Determine the moles of substance X, then compute the enthalpy change per mole. The enthalpy change (ΔH) for the reaction per mole of X is 84.34 kJ/mol.

To calculate the enthalpy change (ΔH) for the reaction per mole of substance X, follow these steps:

Determine the heat absorbed by the solution using the formula q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.The mass of the solution is the volume of water (32.0 mL) converted to grams (32.0 g) and the specific heat capacity is 4.18 J/g°C. The temperature change (ΔT) is 29.0°C - 15.0°C = 14.0°C.Calculate q: q = 32.0 g * 4.18 J/g°C * 14.0°C = 1874.56 J.Convert q to kJ: 1874.56 J * (1 kJ / 1000 J) = 1.87456 kJ.Calculate the moles of substance X: Moles = 1.20 g / 54.0 g/mol = 0.02222 mol.Finally, calculate the ΔH per mole of X: ΔH = 1.87456 kJ / 0.02222 mol = 84.34 kJ/mol.

A total of 2.00 molmol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 137 gg of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

Answers

Answer : The enthalpy of this reaction is, 1.06 kJ/mol

Explanation :

First we have to calculate the heat produced.

[tex]q=m\times c\times (T_2-T_1)[/tex]

where,

q = heat produced = ?

m = mass of solution = 137 g

c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]21.00^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]24.70^oC[/tex]

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

[tex]q=137g\times 4.18J/g^oC\times (24.70-21.00)^oC[/tex]

[tex]q=2118.842J=2.12kJ[/tex]

Now we have to calculate the enthalpy of this reaction.

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

where,

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

q = heat released = 2.12 kJ

n = moles of compound = 2.00 mol

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

[tex]\Delta H=\frac{2.12kJ}{2.00mole}[/tex]

[tex]\Delta H=1.06kJ/mol[/tex]

Thus, the enthalpy of this reaction is, 1.06 kJ/mol

A sample of ethane (C2H6)was combusted completely and the water that formed has a mass of 1.61 grams.How much ethane, in grams, was in the sample?Put your answer in thespace provided belo socratic.org

Answers

Answer:

0.89 g of ethane

Explanation:

The balanced reaction equation is

C2H6(g) + 7/2 O2(g) -----------> 2CO2(g) + 3H2O(g)

From this balanced reaction

30g of ethane yields 54g of water

Therefore mass of ethane necessary to obtain 1.61g of water we have:

30 × 1.61/54 = 0.89 g of ethane

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