If a research paper appeared reporting the structure of a new molecule with formula C2H8 , most chemists would be highly skeptical. Why?

Final answer:

After 10.0 g of ammonia reacts with 10.0 g of hydrogen chloride, 31.5 g of ammonium chloride is produced with leftover hydrogen chloride remaining.

Explanation:

When ammonia is mixed with hydrogen chloride (HCl), there is a chemical reaction in which ammonium chloride (NH4Cl) is formed. This reaction can be represented by the balanced chemical equation: NH3 (g) + HCl (g) → NH4Cl (s). The molar mass of ammonia (NH3) is approximately 17 g/mol and that of hydrogen chloride (HCl) is approximately 36.5 g/mol. When 10.0 g of each reactant is mixed, stoichiometry is used to determine the products and their masses after the reaction goes to completion.

Form the given masses, we can calculate the moles of each reactant: 10.0 g NH3 ÷ 17 g/mol ≈ 0.588 mol NH3 and 10.0 g HCl ÷ 36.5 g/mol ≈ 0.274 mol HCl. Since the reaction occurs in a 1:1 molar ratio, ammonia is the limiting reactant because we have less moles of it compared to moles of HCl. Therefore, all the ammonia will react with HCl to produce 0.588 mol of NH4Cl. The mass of the produced NH4Cl can be found by multiplying the moles of NH4Cl by its molar mass (approximately 53.5 g/mol), so 0.588 mol × 53.5 g/mol ≈ 31.5 g of NH4Cl. There will be no remaining ammonia. Since HCl is in excess, after the reaction the system will contain 31.5 g of NH4Cl and some unreacted HCl.

Answer: 13.47g of MgBr2

Explanation:

MM of MgBr2 = 24 + (2 x 80) = 24 + 160 = 184g/mol

Mass conc. Of MgBr2 = 0.183 x184 = 33.672g

33.672g of MgBr2 dissolves in 1000mL

Therefore Xg of MgBr2 will dissolve in 400mL

Xg of MgBr2 = ( 33.672 x 400)/1000

Xg of MgBr2 = 13.47g

Answer:

1. 24.7g of Br2

2. 19.9g of HCl

Explanation:Please see attachment for explanation

Answer: [tex]CrF_3[/tex] has greater impact on the freezing point depression of ice.

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f-T_f^0[/tex] = Depression in freezing point

i= vant hoff factor

[tex]K_f[/tex] = freezing point constant

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

a) i = 4 for [tex]CrF_3[/tex] as it dissociates to give 4 ions in water.

[tex]CrF_3\rightarrow Cr^{3+}+3F^-[/tex]

b) i = 3 for [tex]CaF_2[/tex] as it dissociates to give 3 ions in water.

[tex]CaF_2\rightarrow Ca^{2+}+2F^-[/tex]

As the vant hoff factor is higher for [tex]CrF_3[/tex] , it has greater impact on the freezing point depression of ice.

Final answer:

CrF3 would have a greater impact on the freezing point depression of ice than CaF2 because it dissociates into more particles, four compared to CaF2 which dissociates into three.

Explanation:

In determining which substance would have a greater impact on the freezing point depression of ice, we need to consider the number of particles each substance releases into solution when dissolved. CrF3 will ionize to form one Cr3+ ion and three F- ions, for a total of four particles. Conversely, CaF2 will dissociate to form one Ca2+ and two F- ions, resulting in three particles. According to colligative properties, CrF3 will therefore have a greater effect on freezing point depression than CaF2 because it produces a larger number of dissociated particles.

Answer: C2H4S5

Explanation:

Since the total mass is 5.000g

Mass of sulphur = 5.000-(0.6357+0.1070)

Mass of sulphur = 4.2573g

Using Empirical relation

C= 0.6357 H= 0.1070 S= 4.2573

Divide through by their molar mass to obtain the smallest ratio

C= 0.6357/12 H=0.1070/1 S=4.2573/32

C= 0.053 H= 0.1070 S= 0.133

Divide through by the smallest ratio (0.053)

C=0.053/0.053 H=0.1070/0.053 S=0.133/0.053

C=1 H=2 S=2.5

1:2:2.5 ,multiply through by 2 ,to obtain whole numbera

2:4:5

Therefore the empirical formula is C2H4S5. Thus only gives the ratio

Molecular formula is the chemical formula .

(Empirical formula) n = molecular formula

(C2H4S5)n = molar mass

[(12×2) + ( 1×4) +(32×5)]n = 188.4

188n=188.4

n= 1

Molecular formula = (C2H4S5)×1

Therefore the chemical formula of

lenthionine is C2H4S5

Answer:

Kc = 4.774 * 10¹³

Explanation:

the desired reaction is

2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)

Kc =[N₂]*[O₂]² /[NO₂]²

Since

1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)

Kc₁= [NO]/(√[N₂]√[O₂]) →  Kc₁²= [NO]²/([N₂][O₂])

and

2 NO₂(g) ⇋ 2 NO(g) + O₂(g)

Kc₂= [NO]²*[O₂]/[NO₂]² →  1/Kc₂= [NO₂]²/([NO]²[O₂])

then

Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂])  = [NO₂]²/([N₂]*[O₂]²) = 1/Kc

Kc₁² /Kc₂ = 1/Kc

Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³

Final answer:

To find the Kc for the reaction 2NO²(g) ⇌ N²(g) + 2O²(g), we reverse and manipulate the given equilibria, taking reciprocals and squaring as necessary, and then multiplying the constants to get the overall Kc.

Explanation:

To determine the equilibrium constant Kc for the given reaction 2NO²(g) ⇌ N²(g) + 2O²(g), we can manipulate the given equilibria (1) and (2) to yield the desired reaction.

First, we reverse the equilibrium (1), 1/2 N²(g) + 1/2 O²(g) ⇌ NO(g), whose Kc is 4.8 x 10-10. When an equation is reversed, the equilibrium constant becomes the reciprocal of the original. Thus, the new Kc for NO(g) ⇌ 1/2 N²(g) + 1/2 O²(g) is 1 / (4.8 x 10-10). To find the Kc for N²(g) + O²(g) ⇌ 2NO(g), we need to multiply the new reaction by two, which means squaring the Kc.

Second, we consider reaction (2), which is 2NO²(g) ⇌ 2NO(g) + O²(g), with a Kc of 1.1 x 10-5.

By adding the modified equilibrium (1) with the reaction (2), we get the desired equilibrium: 2NO(g) + O2(g) ⇌ 2NO2(g). We then multiply the Kc values of these individual steps to obtain the overall Kc for the desired reaction. This is because the equilibrium constant of the overall reaction is the product of the equilibrium constants of the individual steps.

In summary:

The dissolution of ammonium nitrate is an endothermic process. In terms of calculating the final temperature of an activated cold pack containing water and ammonium nitrate, you would use a specific formula with the given data. The details of this calculation were outlined in the detailed answer.

a. The dissolution of ammonium nitrate is an endothermic process. This is because it absorbs heat from its surroundings in the process of dissolving, leading to a drop in temperature. Hence, endothermic reactions cool their surroundings. The given heat of solution for ammonium nitrate is positive (+25.4 kJ/mol), which is characteristic of endothermic reactions.

b. To calculate the final temperature of the activated cold pack, you would need to use the formula q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature (final-initial). Given data can be used to calculate the heat absorbed by the reaction, then using that value along with the given initial temperature and specific heat, we can find the final temperature. Please note that the calculations involved may require quite a few steps and it may be best to consult your chemistry teacher or textbook for the specific mathematical operation.

Learn more about Endothermic Dissolution here:

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Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

∆H is the change in the sum of the internal energy of a system.

We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

Δ H^∘ (rxn)= - 894.6 kJ/mol.

if there is no carbon dioxide your test tube will be blue

if there is a medium amount of carbon dioxide your test tube is green

if there are high amounts of CO2 it will be

yellow

In the context of bromothymol blue as an indicator, if no CO2 is present, the solution will be blue. If a medium amount of CO2 is present, the solution turns green. A large amount of CO2 would turn the solution yellow due to the acidic conditions created by carbonic acid.

The colors in the test tube depend on the identification or indicator test you are performing. However, if we were referring to the bicarbonate buffer system and the formation of carbonic acid, we could consider the pH indicators' reaction to acidity caused by carbon dioxide concentration. One common pH indicator is bromothymol blue. ~ This works as follows:

It's important to remember, this answer assumes the use of bromothymol blue, a common pH indicator. Different indicators may produce different color changes.

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Answer:

C₃H₄N₂ is the molecular formula

Explanation:

The molecular ion peak is obtained at 68 units, which is the molecular mass of the compound. The even number of nitrogen atoms gives even mass of the molecule. In this problem, the mass is even so there is an even number of nitrogen atoms. It can be either two or four, from which it must be two nitrogen, as four nitrogen atoms will give higher mass. Now, we know that 28 out of 68 units belongs to nitrogen

The number of carbon atoms can also be determined in the same manner. Only one or two carbon atoms will leave much of the remaining units (28 or 16 units respectively) to hydrogen atoms. This leaves the option of three carbon atoms as any number higher than it will give molecular mass more than 68 units.

For hydrogen atoms, only 4 units are remaining this means four hydrogen atoms are present in the molecule. This is possible, if we consider that the compound is cyclic and contains two double bonds.

PS: If we consider two carbon atoms then there will be 16 units left which means sixteen hydrogen atoms. The valency of two carbon and two nitrogen atoms will only allow a maximum of eight hydrogen atoms in the molecule (giving only 60 units).

Imidazole, with a molecular ion of 68, has the molecular formula C3H4N2, determined through understanding the application of the nitrogen rule in mass spectrometry.

The molecular formula for the compound imidazole, which contains carbon (C), hydrogen (H), and nitrogen (N), and has a molecular ion of 68, is C3H4N2. According to the nitrogen rule in mass spectrometry, nitrogen contributes either +1 or -1 to the mass, depending on whether the number of nitrogens in the compound is odd or even. Imidazole has an even number of nitrogens, hence the even-mass molecular ion of 68. The makeup of the remaining 68 mass units results from the carbon and hydrogen in the molecule: 3 carbons (each with a mass of 12, for a total of 36) and 4 hydrogens (each with a mass of 1, for a total of 4), and 2 nitrogens (each with a mass of 14, for a total of 28). So, the total mass (36 + 4 + 28) equals 68, aligning with the M+ value.

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To move a solid compound to the bottom of a melting point capillary tube, one should tap the open end on a hard surface. A needle can be used to adjust the position of the compound if necessary, without compacting it too much.

To force a solid compound to the bottom of a melting point capillary tube, you should tap the open end of the capillary tube on a hard surface, such as a bench top. This action utilizes gravity to help settle the compound into the closed end of the tube. If the compound does not move down to the bottom, you can use a long, thin object like a needle to carefully push the compound down without compacting it too tightly, which could affect its melting behavior. Alternatively, tapping the closed end might cause some compacting and is not usually recommended for settling the compound.

While using the capillary tube, it's important to understand capillary action and the characteristics of liquid-glass interactions. For example, if a capillary tube is placed into a beaker of ethylene glycol, the ethylene glycol will rise into the tube by capillary action due to the strong adhesive forces between the polar Si-OH groups on the surface of glass and the molecules of the ethylene glycol, creating a concave meniscus. This is opposite to what happens with a nonpolar liquid like SAE 20 motor oil, which cannot form strong interactions with the polar Si-OH groups, resulting in the oil having a convex meniscus and a reduction in the capillary action.

Answer:

Please refer to the attachment below for answer and explanation.

Explanation:

Please refer to the attachment below for answer and explanation.

The object's density is calculated by converting the mass into grams (12.4 grams) and the volume into cubic centimeters (1.893 cm3), and then dividing mass by volume. The resulting density is approximately 6.56 grams per cubic centimeter.

The density of an object is calculated by dividing its mass by its volume. We also need to convert the units to be compatible. Since your mass is in kilograms we will convert it to grams, 0.0124 kg is the same as 12.4 grams. Your volume is in cubic millimeters (mm3), and we need it in cubic centimeters (cm3). 1 cm3 is equivalent to 1,000 mm3. Therefore, 1893mm3 is equivalent to 1.893 cm3. Now we simply calculate the density: 12.4g / 1.893cm3 = approximately 6.56 grams per cubic centimeter.

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Final answer:

To find the pH of the final solution after reacting Ca(OH)2 and HNO3, calculate moles of reactants, determine moles needed for the reaction, and use the formula for a strong base to find the final pH.

Explanation:

Step 1: Calculate the moles of HNO3 used: 0.045 L × 1.00 mol/L = 0.045 mol.

Step 2: Determine the moles of Ca(OH)2 needed to react with the HNO3: 2 moles of HNO3 react with 1 mole of Ca(OH)2, so 0.045 mol HNO3 requires 0.0225 mol Ca(OH)2.

Step 3: Use the volume and molarity of the Ca(OH)2 to find the pH using the formula for a strong base: pH = 14 - pOH. From the given data, we can calculate pOH = -log(0.672) = 0.173. Therefore, pH = 14 - 0.173 = 13.827.

The pH of the final solution is 0.324.

To determine the pH of the final solution, follow these steps:

Calculate moles of Ca(OH)₂:

Molar mass of Ca(OH)₂ is 74.10 g/mol. Moles of Ca(OH)₂:

(0.350 g) / (74.10 g/mol) = 0.00472 mol

Calculate moles of HNO₃:

Molarity (M) = moles/volume in L. For HNO₃, moles:

(1.00 M) x (0.0450 L) = 0.0450 mol

Determine reaction and limiting reagent:

The balanced equation for the reaction is:

Ca(OH)₂ + 2 HNO₃ ➞ Ca(NO3)₂ + 2 H₂O

Moles of HNO₃ required to react with Ca(OH)₂:

0.00472 mol Ca(OH)₂ x 2 = 0.00944 mol HNO₃

Excess HNO₃:

0.0450 mol - 0.00944 mol = 0.03556 mol

Final volume of solution = 75.0 mL = 0.0750 L. H⁺ concentration:

Given the balanced equation 2A + 3B = 3C, with 6 moles of A and 2 moles of B, A is the limiting reactant. The maximum moles of C formed is 9 (Option D).

First, let's identify the limiting reactant by comparing the mole ratios of A and B in the balanced equation:

[tex]\[2A + 3B \rightarrow 3C\][/tex]

The mole ratio between A and B is 2:3. If you have 6 moles of A and 2 moles of B, you can compare the actual ratio (6 moles A to 2 moles B) with the ratio in the balanced equation (2 moles A to 3 moles B).

The actual ratio is 6:2, which simplifies to 3:1. The balanced ratio is 2:3. The limiting reactant will be the one with the smaller ratio, so in this case, A is the limiting reactant.

Now, let's determine the maximum number of moles of C that can be formed using the stoichiometry of the balanced equation:

For every 2 moles of A, 3 moles of C are formed.

Since you have 6 moles of A (the limiting reactant), you can use the ratio to find the moles of C:

[tex]\[\text{Moles of C} = \left( \frac{3 \text{ moles C}}{2 \text{ moles A}} \right) \times 6 \text{ moles A} = 9 \text{ moles C}\][/tex]

Therefore, the correct answer is:

D. 9 moles

The probable question may be:

Consider the chemical equation: 2A+3B=3C. If 6 moles of A and 2 moles of  B are reacted, what is the maximum number of moles of C that can be formed?

A. 4 moles

B. 2 moles

C. 3 moles

D. 9 moles

Answer:

88.8 minutes

Explanation:

Graham's law of diffusion relates rate of difusion by the following formula

Rate1 / rate 2 = √( Mass of argon / Mass of Neon)

Where rate = volume divided by time

Rate 1 = 10 ml / t1

Rate 2 = 10 ml / t2

Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1

t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)

125 / t1 = 1.4026

t1 = 125 / 1.4026 = 88.8 minutes

It will take approximately 95.3 seconds for 10.0 mL of Ne gas to effuse through the porous barrier.

To calculate the time it will take for 10.0 mL of Ne gas to effuse through a porous barrier, we can use the effusion rate ratio. In Example 9.21, it is stated that it takes 243 seconds for 4.46 × 10-5 mol Xe to effuse through a tiny hole. Using the effusion rate ratio, we can calculate that it will take 95.3 seconds for 4.46 × 10-5 mol Ne to effuse. Since Ne is lighter than Xe, its effusion rate will be larger, resulting in a smaller effusion time.

Explanation:

It is known that entropy is the measure of degree of randomness present in a substance due to the movement of its molecules.

More randomly the atoms are moving from one place to another more will be the entropy of system.

In order to calculate the entropy, we divide the amount of heat transferred by the temperature at which heat transfer occurs.

As the given temperature is 296.95 K and heat of vaporization is 24.8 kJ/mol. Therefore, calculate the molar entropy of vaporization as follows.

Molar entropy of vaporization = [tex]\frac{24.8 kJ/mol}{296.95 K}[/tex]

                                                  = 0.08351 kJ/mol K

Thus, we can conclude that molar entropy of vaporization of [tex]CC_{3}F[/tex] is 0.08351 kJ/mol K.

CFCs, including Trichlorofluoromethane, have been banned due to their destructive effect on the ozone layer. UV light causes them to release chlorine atoms, which react with and deplete ozone. The harmful side effect of this is increased solar radiation reaching Earth.

Chlorofluorocarbons, or CFCs, like Trichlorofluoromethane, were previously widely used in a variety of industries, such as refrigeration and aerosols. However, the stable nature of these compounds, combined with their ability to break down and release chlorine atoms under ultraviolet light, led to significant depletion of the ozone layer in the stratosphere. This process happens because as CFCs are broken down by UV light, they produce chlorine atoms. These chlorine atoms then react with ozone molecules, leading to a net decrease in stratospheric ozone. This ozone depletion had severe implications, such as increased risks from solar radiation, which prompted the worldwide effort to phase out the use of CFCs under the Montreal Protocol.

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Complete Question:

A small sphere of initial volume V is filled with n moles of helium at initial pressure and temperature and T. Which of the following statements is true?

a) The volume decreases to V/2, and the pressure increases to 4P when the temperature is T/2

b) n/2 moles of gas are removed, the volume is decreased to V/2, and the pressure decreases to P/4 with a drop in temperature of T/2

c) n moles of gas are added, the total sample is heated to 2T, and the pressure drops to P/2 when the volume increases to 8V

d) The amount of gas is doubled to 2n, the pressure is doubled to 2P, and the volume is doubled to 2V, with a corresponding temperature drop to T/2

Answer:

c

Explanation:

Let's consider the helium as an ideal gas, so it can be studied by the ideal gas law, which states:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature. Because R is constant:

PV/nT = R. Thus the initial state must be equal to the final state.

So, let's check the statements:

a) Let's indicate the final state as P₂, V₂, n₂ and T₂. So, if T₂ = T/2:

PV/nT = P₂V₂/n₂T₂

PV/nT = P₂V₂/n₂(T/2)

PV/nT = 2P₂V₂/n₂T

So, if V₂ = V/2 and P₂ = 4P:

PV/nT = 2*(V/2 * 4P)/n2T

PV/nT = 4VP/n2T

Which is not correct!

b) Now, if T₂ = T/2:

PV/nT = 2P₂V₂/n₂T

If n/2 is removes, n₂ = n/2. And, V₂ = V/2 and P₂ = P/4:

PV/nT = 2*(V/2 * P/4)/(n/2)*T

PV/nT = 4*(V/2 *P/4)/nT

PV/nT = PV/2nT

Which is not corret!

c) Now, if V₂ = 8V:

PV/nT = P₂*8V/n₂T₂

And n₂ = n +n = 2n, T₂ = 2T and P₂ = P/2:

PV/nT = (P/2)*8V/2n*2T

PV/nT = 8*(PV)/2*2n*2T

PV/nT = 8*(PV)/8*(nT)

PV/nT = PV/nT

So, it's correct!

d) Now, T₂ = T/2, n₂ = 2n, P₂ = 2P, and V₂ = 2V:

PV/nT = 2P*2V/2n*(T/2)

PV/nT = 4PV/nT

Which is not correct!

Answer:

The extinction coefficient is 1.15 x 10⁴ M⁻¹.cm⁻¹ (value is not rounded off)

Explanation:

According to Lambert-Beer law

[tex]A = ε.b.C[/tex]

Here, A is absorbance, ε is extinction coefficient, b is the length of the cuvette and C is the molar concentration of substance X.

This equation is used for the relation between concentration and absorbance of electromagnetic radiation absorbing species. It is a linear equation and can be used for making a calibration curve, which is used for the analysis of an unknown concentration solution. The slope of this curve according to the equation is the product of extinction coefficient (M⁻¹.cm⁻¹) and the length of the cuvette in cm.

In this problem, the slope is provided, which can be mathematically represented as:

[tex]slope =  ε.b[/tex]

[tex]2.3 X 10^{3}M^{-1} = ε.(0.2 cm)[/tex]

[tex]ε = 1.15 X 10^{4} M^{-1}.cm^{-1}[/tex] (not rounded off)

[tex]ε = 1 X 10^{4} M^{-1}.cm^{-1}[/tex] (rounded off)

The given question is incomplete. The complete question is as follows.

The solubility product of calcium fluoride () is  at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?

Explanation:

Reaction equation for the given chemical reaction is as follows.

       [tex]CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}[/tex]

Therefore, expression for [tex]K_{sp}[/tex] will be as follows.

        [tex]K_{sp} = [Ca^{2+}][F^{-}]^{2}[/tex]

                     =

Also, moles of  per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]

                = [tex]\frac{1.0 \times 10^{-3}}{19.0}[/tex]

               = [tex]5.263 \times 10^{-5} mol[/tex]

Hence,    [tex][F^{-}] = \frac{\text{moles of F^{-}}{volume}[/tex]

                       = [tex]\frac{5.263 \times 10^{-5}}{1}[/tex]

                      =  M

Now, moles of  per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]

            = [tex]\frac{200 \times 10^{-3}}{40.1}[/tex]

           =  M

Also,   [tex][Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}[/tex]

                      = [tex]\frac{4.988 \times 10^{-3}}{1}[/tex]

                     =  M

Hence, ionic product =

                 = [tex](4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}[/tex]

                = [tex]1.38 \times 10^{-11}[/tex]

As, the ionic product is less than the [tex]K_{sp}[/tex], this means that the fluoride will be soluble in water containing the calcium.

To balance the equation Ca(OH)2(aq) + H3PO4(aq) → Ca3(PO4)2(s) + H2O(ℓ), you need to place a coefficient of 3 in front of Ca(OH)2, 6 in front of H3PO4, and 3 in front of H2O.

To balance the equation Ca(OH)2(aq) + H3PO4(aq) → Ca3(PO4)2(s) + H2O(ℓ), we can start by balancing the calcium (Ca) atoms. There are three calcium atoms on the right side (Ca3(PO4)2) and only one on the left side (Ca(OH)2). To balance the calcium atoms, we need to place a coefficient of 3 in front of Ca(OH)2.

Next, we can balance the hydrogen (H) atoms. There are 6 hydrogen atoms on the left side (2 from Ca(OH)2 and 4 from H3PO4), so we need to place a coefficient of 6 in front of H3PO4.

Finally, we can balance the oxygen (O) atoms. There are 4 oxygen atoms on the left side (from Ca(OH)2) and 6 oxygen atoms on the right side (3 from Ca3(PO4)2 and 3 from H2O). To balance the oxygen atoms, we need to place a coefficient of 3 in front of H2O.

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Answer:

                   The answer is 2,3-dimethylbutan-2-ol and the structure is attached below.

Explanation:

                     Although we are not provided with ¹H-NMR spectrum and IR spectrum but still we can elucidate the ¹³C-NMR data and finalize a plausible structure.

                      First of all we look at the molecular formula, we can conclude from the formula that the structure given is saturated in nature because the hydrogen deficiency index of this formula is zero. Hence, we can say that there is no double bond either between Carbon atoms or between carbon and oxygen atom. This can also be proved by the absence of peaks in downfield as unsaturated compounds and carbonyl compounds give value above 100 and 200 ppm respectively.

                      Secondly, we can also conclude that among the six carbon atom two pairs of them are having same electronic environment because we are having only 4 signals hence we can conclude that two pairs have same chemical shift values.

                       Also, after making every possinble isomer of given molecular formula the structure of 2,3-dimethylbutan-2-ol was found to be the most accurate structure.

Despite the absence of certain spectra, analysis of the ¹³C-NMR data suggests a saturated structure. The most fitting isomer is identified as 2,3-dimethylbutan-2-ol based on chemical shift values.

While lacking specific data from the ¹H-NMR and IR spectra, an insightful elucidation of the ¹³C-NMR data allows for a plausible structural determination.

Examining the molecular formula, the absence of a hydrogen deficiency index suggests a saturated structure, ruling out double bonds or carbonyl groups. The lack of peaks in downfield further supports this, as unsaturated and carbonyl compounds typically exhibit values above 100 and 200 ppm, respectively.

Considering the six carbon atoms, the presence of only four signals indicates two pairs with identical chemical shift values, implying equivalent electronic environments.

Upon exploring various possible isomers for the given molecular formula, the most fitting structure emerged as 2,3-dimethylbutan-2-ol. This conclusion aligns with the observed NMR data and satisfies the saturation criteria.

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When presented with a concentration of 1 mg/mL, one can express it through molarity (if the molar mass of the solute is known) or mass-volume percent concentration. In this case, the solution's mass-volume percent concentration would be 0.1%.

The two concentration expressions that can be used to describe a solution with a concentration of 1 mg/mL are molarity and mass-volume percent concentration.

Molarity is defined as the amount of solute in moles divided by the volume of the solution in liters. Assuming the molar mass of the solute is known, the molarity can be calculated using the provided concentration.

Mass-volume percent concentration (% m/v) is another applicable concept. It is defined as the mass of solute in grams divided by the volume of the solution in milliliters, multiplied by 100%. In your case, as the solution is 1 mg/mL, it translates to a mass-volume percent concentration of 0.1%.

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Final answer:

To identify a concentration expression compatible with 1 mg/mL for a solution with density 1.00 g/mL, molarity can be used if the molar mass of the solute is known. Molality is not directly applicable without knowing the weight of the solvent separately.

Explanation:

When considering a solution with a concentration of 1 mg/mL of solute and a density of 1.00 g/mL, it's important to note that molality cannot be used directly, because it requires the weight of the solvent (not the solution), and it's expressed in kilograms. The question pertains to a solution that has its density equivalent to that of water, thus allowing for certain concentration expressions like molarity to provide a good approximation to molality for dilute aqueous solutions.

To convert from mg/mL to molarity (M), we need the molar mass of the solute. For example, if we have potassium bromide (KBr), which has a molar mass of 119.0 g/mole, a 1 mg/mL solution would be 1 g/L or 1000 mg/L. To find the molarity, divide the concentration in grams per liter (1 g/L) by the molar mass of KBr, yielding approximately 0.0084 M. This would be similar to the molality for a dilute KBr solution because the molality and molarity converge when the density is close to that of water (1.00 g/mL).

Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure.

0.1 M [tex]NH_4NO_3(aq)[/tex] , 0.1 M [tex]NaF(aq)[/tex],  0.1 M [tex]LiNO_3(aq)[/tex],  0.1 M [tex]Na_3PO_4(aq)[/tex] and  0.1 M [tex]HC_2H_3O_2(aq)[/tex]

Answer: 0.1 M [tex]Na_3PO_4[/tex]

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]

where,

[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure

i = Van'T Hoff factor  

[tex]x_2[/tex] = mole fraction of solute  

1. For 0.1 M [tex]NH_4NO_3[/tex]

[tex]NH_4NO_3\rightarrow NH_4^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]

2. For 0.1 M [tex]NaF[/tex]

[tex]NaF\rightarrow Na^{+}+F^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]

3. For 0.1 M [tex]LiNO_3[/tex]

[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]

4. For 0.1 M [tex]Na_3PO_4[/tex]

[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]  

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]3\times 0.1+1\times 0.1=0.4[/tex]

5. For 0.1 M [tex]HC_2H_3O_2(aq)[/tex]

[tex]HC_2H_3O_2(aq)\rightarrow CH_3COO^{-}+H^{+}[/tex] [/tex]

, i= 2 as it is a electrolyte and dissociate to give two ions, concentration of ions will be [tex]2\times 0.1=0.2[/tex]

Thus as concentration of solute is highest for 0.1 M [tex]Na_3PO_4[/tex] , the vapor pressure will be lowest.

Answer: the empirical formula is CH2O

Explanation:Please see attachment for explanation

In a tetrahedral molecular geometry, a central atom is located at the center with four substituents that are located at the corners of a tetrahedron.

The number of bonds in the structure is as follows:-

The angle between the elements is 109.5.

The geometry is how the bonds and the lone pairs of electrons are distributed around the atom, and the angles formed by them occur to minimize the repulsive forces of the bonds and the lone pairs.  

For example, the molecule of CO₂ is linear, because the central atom C has no lone pairs (the valence shell of it has 4 electrons, so it shares all of them to be stable), and does only two bonds. But the molecule of H₂O is angular, because the central atom O has two lone pairs (its valence shell has 6 electrons, so it only shares 2 electrons to be stable), and the repulsive forces are minimized with an angle of 104.45°.

To the geometry be tetrahedral, the atom must have 4 bonds and no lone pairs, and it only happens at the two C in the molecule given.

H only does one bond, and O does 2 bonds and has 2 lone pairs.

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In the molecule CH3OCH3, the carbon (C) atoms will display a tetrahedral electron-pair geometry due to the sp³ hybridization commonly found in organic compounds.

In the molecule CH3OCH3, the atoms that will display tetrahedral electron-pair geometry are the carbon (C) atoms. This is because carbon atoms, particularly when involved in organic compounds like this one, often form bonds using sp³ hybridization, which results in a tetrahedral shape. The tetrahedral geometry of carbon occurs when it forms four bonds, often with hydrogen and other carbon atoms.

This conformation is one of the most common in organic chemistry and is noteworthy for the fact that all the bonds are 109.5° apart - the ideal bond angle in a true tetrahedral geometry.

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Answer:

The reactions are:

H₂SO₃  +  H₂ O  ⇄  HSO₃⁻  +  H₃O⁺

HSO₃⁻  +  H₂O  ⇄  SO₃⁻²  +  H₃O⁺

Explanation:

This is the sulfurous acid → H₂SO₃.

It is a weak diprotic acid and it is diprotic because it can release 2 protons to increase H₃O⁺ from water. As every weak acid, the reactions are in equilibrium, that's why it has two Ka. As it can release protons, it is an acid from Bronsted Lowry and an acid from Arrhenius.

H₂SO₃  +  H₂ O  ⇄  HSO₃⁻  +  H₃O⁺   Ka₁

 HSO₃⁻  → this is called acid sulfite

HSO₃⁻  +  H₂O  ⇄  SO₃⁻²  +  H₃O⁺     Ka₂

2H₂O  ⇄   H₃O⁺  +  OH⁻    Kw

(Don't forget to put water equation, always)

Sulfurous acid, a diprotic acid, ionizes in two stages when dissolved in water. The first stage produces the hydrogen sulfite ion, HSO3¯, while the second stage further ionizes this ion to form the sulfite ion, SO3²-. The first ionization is more prominent than the second, reflecting the moderate strength of sulfurous acid.

Sulfurous acid, being a polyprotic acid, ionizes in two stages when dissolved in water. In the first step, sulfurous acid gives up one proton (H⁺) to produce the hydrogen sulfite ion, HSO3¯:

H2SO3(aq) ⟶ H⁺(aq) + HSO3¯(aq)

In the second stage, the hydrogen sulfite ion can further ionize to give up another proton to form the sulfite ion, SO3²-:

HSO3¯(aq) ⟶ H⁺(aq) + SO3²-(aq)

It's important to note that the ionization is much more significant in the first stage than the second, due to sulfurous acid being a moderately strong acid.

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Answer:

Partial pressure O₂  = 2.284 atm

Explanation:

This is problem with unit conversion. Let's convert everything to atm

760 mmHg ___ 1 atm

540 mmHg ___ (540 / 760) = 0.710 atm

101325 Pa ____ 1 atm

203303 Pa ____ (203303 / 101325) = 2.006 atm

Total pressure of the mixture = Sum of partial pressure of each gas

0.710 atm (He) + 2.006 (N₂) + Partial Pressure O₂ = 5 atm

Partial pressure O₂ = 5 atm - 2.006 atm - 0.710 atm = 2.284 atm

Answer:

164 g/mol

Explanation:

According to Graham's law, the rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M).

rH₂/rX = √[M(X)/ M(H₂)]

(rH₂/rX)² = M(X)/ M(H₂)

M(X) = (rH₂/rX)² × M(H₂)

M(X) = (9)² × 2.02 g/mol

M(X) = 164 g/mol

The molar mass of the unknown gas is 164 g/mol.

The molar mass of the unknown gas is 64 g/mol.

The rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M).

It is given by:

[tex]rH_2/rX = \sqrt{[M(X)/ M(H_2)]} \\\\(rH_2/rX)^2 = M(X)/ M(H_2)\\\\M(X) = (rH_2/rX)^2 * M(H₂)\\\\M(X) = (9)^2 * 2.02 g/mol\\\\M(X) = 164 g/mol[/tex]

Thus, the molar mass of the unknown gas is 164 g/mol.

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Answer:

a) pH = 2.573

b) pH = 4.347

Explanation:

a) weak acid: CH3COOH

∴ Ka = 1.8 E-5 = [H3O+][CH3COO-] / [CH3COOH]

∴ C CH3COOH = 0.40 M

mass balance:

⇒ 0.40 M = [CH3COO-] + [CH3COOH].........(1)

charge balance:

⇒ [H3O+] = [CH3COO-].........(2)

(2) in (1):

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ Ka = 1.8 E-5 = [H3O+]² / ( 0.40 - [H3O+] )

⇒ [H3O+]² = 7.2 E-6 - 1.8 E-5[H3O+]

⇒ [H3O+]² + 1.8 E-5[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 2.6743 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.573

b) balanced reations:

∴ C CH3COOH = 0.40 M

∴ C CH3COONa = 0.20 M

mass balanced:

⇒ C CH3COOH + C CH3COONa = [CH3COO-] + [CH3COOH]

⇒ 0.60 = [CH3COO-] + [CH3COOH]......(1)

charge balanced:

⇒ [H3O+] + [Na+] = [CH3COO-]

∴ [Na+] = 0.20 M

⇒ [H3O+] + 0.20 M = [CH3COO-]........(2)

(2) in (1):

⇒ 0.60 M = ( [H3O+] + 0.20 ) + [CH3COOH]

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ 1.8 E-5 = ([H3O+])([H3O+] + 0.20) / (0.40 - [H3O+])

⇒ 7.2 E-6  - 1.8 E-5[H3O+] = [H3O+]² + 0.20[H3O+]

⇒ [H3O+]² + 0.20[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 4.499 E-5 M

⇒ pH = 4.347

The pH of the solution in (a) is 2.57 The pH of the solution in (b) is 4.4.

We have to set up the ICE table for the reaction;

           CH3CO2H + H2O ⇄ H3O^+   +   CH3CO2^-

I            0.40                            0                  0

C          -x                                  +x                 +x

E       0.40 - x                            x                   x

The pKa of  CH3CO2H  is 1.8 x 10-5

Hence,

1.8 x 10-5 = x^2/0.40 - x  

1.8 x 10-5 (0.40 - x ) = x^2

7.2  x 10-6 - 1.8 x 10-5x =  x^2

x^2 +  1.8 x 10^-5x - 7.2  x 10^-6 = 0

x = 0.00267 M

Hence;

pH = -log [0.00267 M] = 2.57

Using the Henderson Hasselbaclch equation;

pH = pKa + log [A-]/[HA]

pKa = - log Ka = -log[1.8 x 10^-5] = 4.7

Hence;

pH = 4.7 + log [0.20 M]/[0.40 M ]

pH = 4.4

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