At 25 ∘C the reaction CaCrO4(s)←→Ca2+(aq)+CrO2−4(aq) has an equilibrium constant Kc=7.1×10−4. What is the equilibrium concentration of Ca2+ in a saturated solution of CaCrO4?

The question is incomplete, here is the complete question:

An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?

Answer: The mass of oxalic acid in final solution is 0.0234 grams

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]     ......(1)

Given mass of oxalic acid = 0.5842 g

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M[/tex]

To calculate the molarity of the diluted solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated oxalic acid solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted oxalic acid solution

We are given:

[tex]M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL[/tex]

Putting values in above equation, we get:

[tex]0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M[/tex]

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of oxalic acid solution = 0.0026 M

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

[tex]0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g[/tex]

Hence, the mass of oxalic acid in final solution is 0.0234 grams

The question involves dilution of an oxalic acid solution and subsequent calculation of the molarity for titration, which is a typical procedure in a college-level chemistry course.

The student's question pertains to the dilution of an aqueous solution of oxalic acid and the calculation of molarity after dilution and titration. Starting with a mass of 0.5842g of oxalic acid dissolved to make a 100 ml solution, a 10 ml aliquot is further diluted to 250 ml. The goal is to understand the change in concentration as a result of the dilution process and to apply this understanding to various problems presented, such as titration against potassium permanganate (KMnO4).

The pressure P2 at the temperature which CO2 gets solidifies is 0.857 atm.

Explanation:

The relation between temperature T and the pressure P is that it is proportional to each other.

                                      T ∝ P

As the temperature decreases, the pressure also decreases which is given by

                             T1 / T2 = P1 / P2

At the triple point of water, the temperature equals 273 K.

Consider T1 = 273 K,  P1 = 1.20 atm

The temperature T2 of the CO2 solidifies equals 195 K

                            (273 / 195) = (1.20 / P2)

                              P2 = (195 x 1.20) / 273

                             P2 = 0.857 atm.

The pressure P2 at the temperature which CO2 gets solidifies is 0.857 atm.

Answer:

The answers I think you are looking for is Gasoline or Fuel (chemical energy)

Explanation:

Energy transformation involves change or conversion of energy from one form to another. Having two examples of the same form of chemical energy, convert into two or more types of energy, the best example would be gasoline.

GASOLINE/FUEL → CAR BATTERY + LIGHT BULBS (TRAFFICATOR LIGHT)

(chemical)                     (electrical)               (radiant)

GASOLINE/FUEL → INTERNAL COMBUSTION ENGINE

(chemical)                   (mechanical)

Answer:

C - result in radiant energy

Explanation:

Answer : The value of new volume is, 50.0 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]

or,

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure at STP = 1 atm

[tex]P_2[/tex] = final pressure =  4.00  atm

[tex]V_1[/tex] = initial volume at STP = 200.0 mL

[tex]V_2[/tex] = final volume = ?

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

[tex]1atm\times 200.0mL=4.00atm\times V_2[/tex]

[tex]V_2=50.0mL[/tex]

Therefore, the value of new volume is, 50.0 mL

Answer:

4.332 millimoles of NaOH

Explanation:

Molarity (M) = moles per liter (mole/L); liters x moles/liter = moles

0.04332 L x 0.1 mole/liter = 0.004332 moles of NaOH (4.332 millimoles)

millimoles / milliliters = moles/liter = molarity (M).

4.332 millimoles / 50 milliliters = 0.08664 M HCl

In the Volumetric titration at the end point, moles of NaOH = moles of HCl

NaOH + HCl --> H2O + NaCl

Answer:

B

Explanation:

The interaction between magnetism and electricity is called electromagnetism. The movement of a magnet can generate electricity. The flow of electricity can generate a magnetic field.

Answer: B: electromagnetism

Explanation:

A changing magnetic field produces an electric current in a wire or conductor.

Answer:

minerals

Explanation:

A potted geranium is plant that is capable of manufacturing complex carbohydrates from sugar synthesized through the process of photosynthesis.

The process of photosynthesis primarily requires water and carbohydrate together with resources such as essential minerals. Hence, when a plant food is added to a plant, the most plausible term this refers to is minerals. This is because food contains minerals necessary for various metabolic activities of living organisms.

Answer:

nitrogen monoxide: NO

nitrogen dioxide: NO₂

Explanation:

Nitrogen monoxide is composed by 1 atom of O (prefix "mono-") and 1 atom of N. Nitrogen dioxide is composed by 2 atoms of O (prefix "di-") and 1 atom of N. As the oxigen atom in oxides has the valency -2 (it shares 2 electrons), the nitrogen has valency +2 in NO and +4 in NO₂.

Answer:

42 liters

Explanation:

Let the volume of 2% sulfuric acid required be X while that of 12% sulfuric acid required be Y.

According to the concentration: 0.02X + 0.12Y = 0.05(X + Y).......equation 1

According to the volume: X + Y = 60...............equation 2

Substitute equation 2 into equation 1.

0.02X + 0.12(Y) = 0.05(60)

2X + 12Y = 300

X + 6Y = 150..............................equation 3

From equation 2, Y = 60 -X...................equation 4

Substitute equation 4 into equation 3.

X + 6(60-X) = 150

X + 360 - 6X = 150

-5X = -210

X = 42

Hence, the volume of 2% sulfuric acid required is 42 liters.

Answer:

0.3 Coulomb charge flows through the toy car.

0.9 Joules of energy is lost by the battery

Explanation:

[tex]Current(I)=\frac{charge(Q)}{Time (T)}[/tex]

a)

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

milli Ampere = 0.001 Ampere

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

1 min = 60 seconds

Charge flows through the toy car be Q

[tex]0.50\times 0.001 A=\frac{Q}{600 s}[/tex]

[tex]Q=0.50 \times 0.001 A\times 600 s=0.3 C[/tex]

0.3 Coulomb charge flows through the toy car.

b)

[tex]Heat(H)=Voltage(V)\times Current(I)\times Time(T)[/tex]

Voltage of the battery = V = 3.0 V

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

[tex]H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s[/tex]

H = 0.9 Joules

0.9 Joules of energy is lost by the battery

Answer : The final temperature of the aluminum is, [tex]155.9^oC[/tex]

Explanation :

Formula used :

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

where,

q = heat = 3600 cal = 15062.4 J    (1 cal = 4.184 J)

m = mass of aluminum = 125 g

c = specific heat of aluminum = [tex]0.90J/g^oC[/tex]

[tex]T_{final}[/tex] = final temperature = ?

[tex]T_{initial}[/tex] = initial temperature = [tex]22^oC[/tex]

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

[tex]15062.4J=125g\times 0.90J/g^oC\times (T_{final}-22)^oC[/tex]

[tex]T_{final}=155.9^oC[/tex]

Thus, the final temperature of the aluminum is, [tex]155.9^oC[/tex]

Answer:

= 238.6J

Explanation:

According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another.

Endothermic reactions are those in which heat is absorbed by the system and thus the energy of products is higher than the energy of reactants.

For the given reaction:

A + B ⇄ C + D

Energy of A = 85.1 J

Energy of B = 87.9 J

Product C contains 38.7 J

Energy balance:  

∑ enthalpy of the reactants + energy added = ∑ enthalpy of the products + energy released.

∑ enthalpy of the reactants = 85.1 J + 87.9 J = 173 J

energy added = 104.3 J

∑ enthalpy of the products = 38.7 J + D

energy released = 0

Equation:

173J + 104.3J = 38.7 + D + 0  

⇒ D = 173J + 104.3J - 38.7J

= 238.6J

which is the chemical energy of the product D

Product D must contain 238.6 J of chemical energy based on the principles of energy conservation in a chemical reaction.

In a chemical reaction, the law of conservation of energy states that energy cannot be created or destroyed; it can only change forms. This principle applies to chemical energy as well. Therefore, the total energy in the reactants must equal the total energy in the products.

Let's analyze the given information and calculate the chemical energy in product D:

Reactant A contains 85.1 J of chemical energy.

Reactant B contains 87.9 J of chemical energy.

Product C contains 38.7 J of chemical energy.

The total initial chemical energy in the reactants (A and B) is:

Initial Energy (A + B) = 85.1 J + 87.9 J = 173.0 J

The reaction absorbs 104.3 J of chemical energy as it proceeds. This means that the total energy in the products (C + D) must be equal to the initial energy plus the absorbed energy:

Total Energy in Products (C + D) = Initial Energy (A + B) + Absorbed Energy

Total Energy in Products (C + D) = 173.0 J + 104.3 J = 277.3 J

Now, we know the total energy in the products (C + D) is 277.3 J, and we already know the energy in product C is 38.7 J. To find the energy in product D, we can subtract the energy in C from the total energy in the products:

Energy in Product D = Total Energy in Products - Energy in Product C

Energy in Product D = 277.3 J - 38.7 J = 238.6 J

Therefore, product D must contain 238.6 J of chemical energy.

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Answer : The value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]

Explanation :

For the given chemical reaction:

[tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=(P_{NH_3})^2\times P_{CO_2}[/tex]        ........(1)

                      [tex](NH_4)(NH_2CO_2)(s)\rightleftharpoons 2NH_3(g)+CO_2(g)[/tex]

Initial:                                                      0               0

At eqm:                                                   2x              x

As we are given that:

Total pressure of gas at equilibrium = 0.116 atm

2x + x = 0.116 atm

3x = 0.116 atm

x = 0.0387 atm

Putting values in expression 1, we get:

[tex]K_p=(2x)^2\times (x)[/tex]

[tex]K_p=(2\times 0.0387)^2\times (0.0387)[/tex]

[tex]K_p=2.32\times 10^{-4}[/tex]

Thus, the value of [tex]K_p[/tex] is, [tex]2.32\times 10^{-4}[/tex]

The number of liters of oxygen formed at STP when 20.8 g of potassium chlorate decomposes is 5.70 L.

Potassium chlorate decomposes according to the following balanced equation:

2KClO3 → 2KCl + 3O2

The molar mass of KClO3 is 122.55 g/mol.

To calculate the number of moles of KClO3, we divide the given mass (20.8 g) by the molar mass:

20.8 g / 122.55 g/mol = 0.1697 mol

According to the balanced equation, for every 2 moles of KClO3, 3 moles of O2 are produced.

Therefore, the number of moles of O2 produced is:

0.1697 mol × (3/2) = 0.2546 mol

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies approximately 22.4 liters.

Therefore, the number of liters of O2 produced at STP is:

0.2546 mol × 22.4 L/mol = 5.70 L

Answer:

12 molecules of carbon dioxide

24 molecules of water

Explanation:

Given parameters:

Number of molecules of methane = 12 molecules

Unknown:

Number of molecules of carbon dioxide = ?

Number of molecules of water = ?

Solution:

The given amount of methane is the limiting reagent in this reaction. By this, we can ascertain the amount of products that will be formed and the extent of the reaction.

We first write the balanced chemical equation for this reaction;

             CH₄     +    2O₂    →     CO₂   +     2H₂O

From balanced equation;

                 1 mole of methane produced 1 mole of carbon dioxide

  12 molecules of methane will produce 12 molecules of carbon dioxide

Also;

             1 mole of methane produced 2 moles of water;

12 molecules of methane will produce (12 x 2)molecules = 24 molecules of water

When 12 molecules of methane react with excess oxygen, 12 molecules of carbon dioxide and 24 molecules of water are produced, following a 1:2:1:2 ratio of methane to oxygen to carbon dioxide to water.

If 12 molecules of methane (CH4) react with plenty of oxygen (O2), the balanced chemical equation for the combustion of methane shows a 1:2:1:2 ratio. This means for every one methane molecule, two oxygen molecules are consumed, and one carbon dioxide (CO2) molecule and two water (H2O) molecules are produced. Therefore, 12 molecules of methane reacting with excess oxygen will produce 12 molecules of carbon dioxide and 24 molecules of water.

This equation states that for every 1 molecule of methane and 2 molecules of oxygen that react, we will produce 1 molecule of carbon dioxide and 2 molecules of water.

Final answer:

The equilibrium concentration of [[tex]S_2[/tex]] is [tex]1.12 \times 10^{-3} M[/tex].

Explanation:

To find the equilibrium concentration of [[tex]S_2[/tex]], we need to use the equilibrium constant expression [tex]\( K_c = \frac{{[H_2]^2[S_2]}}{{[H_2S]^2}} \)[/tex]. Given the initial concentrations of [tex]H_{2}S[/tex] and [tex]H_2[/tex], and assuming x moles of [tex]H_{2}S[/tex] decompose, we can set up an ICE table and solve for the equilibrium concentrations. Substituting the equilibrium concentrations into the equilibrium constant expression and solving for [[tex]S_{2[/tex]], we find the equilibrium concentration to be [tex]1.12 \times 10^{-3} M[/tex]. This indicates the concentration of [tex]S_2[/tex] at equilibrium after the reaction has reached its dynamic equilibrium state at 800°C.

Answer:

The answer to the question is;

The inter-molecular forces of water are stronger than those of hydrogen.

Explanation:

Ammonia is a compressible gas at room temperature with molecules free to move about and so fill up the volume of the container in which it is placed due to the weaker inter-molecular Van der Waals forces such as Keesom, Debye and London dispersion forces holding the particles of ammonia together in a given volume of the compound.

The inter-molecular forces between water molecules is hydrogen binding and dipole moments due to the strongly electronegative oxygen and hydrogen which tends to move the electrons towards the oxygen creating a charge imbalance that causes the hydrogen surrounding the water molecule to aggregate to neutralize the the charge imbalance  forming the bases for the strong hydrogen bonds.

Therefore water is a liquid at room temperature while ammonia is a gas due to the difference in strength of their inter-molecular forces.

The intermolecular forces in water are stronger than the intermolecular forces in ammonia.

The intermolecular forces hold substances together in a particular state of matter. There are three states of matter;

The strongest degree of intermolecular interaction occurs between matter in the solid state. weaker intermolecular interactions occur in the liquid state and the weakest intermolecular interaction occurs in the gaseous state.

Since ammonia is a gas at room temperature, it has weaker intermolecular interaction between its molecules compared to molecules of water at room temperature:

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Answer: The percent change in volume will be 25 %

Explanation:

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

Mathematically,

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where,

[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.

[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.

We are given:

[tex]V_1=2L\\T_1=T_1\\V_2=?\\T_2=75\% \text{ of }T_1=0.75\times T_1[/tex]

Putting values in above equation, we get:

[tex]\frac{2L}{T_1}=\frac{V_2}{0.75\times T_1}\\\\V_2=\frac{2\times 0.75\times T_1}{T_1}=1.5L[/tex]

Percent change of volume = [tex]\frac{\text{Change in volume}}{\text{Initial volume}}\times 100[/tex]

Percent change of volume = [tex]\frac{(2-1.5)}{2}\times 100=25\%[/tex]

Hence, the percent change in volume will be 25 %

To find the amount of NaCl needed to increase the boiling temperature of water by 1.500°C, one must understand molal boiling point elevation and van't Hoff factor. Using the formula and given constants, we calculate the grams of NaCl required through the steps of determining molality, calculating the moles of NaCl, and then converting to grams.

To calculate the amount of NaCl needed to raise the boiling temperature of water by 1.500°C, we first understand the concept of molal boiling point elevation and van't Hoff factor. Given that the Kb for water is 0.5100°C/m, we use the formula ΔTb = i*Kb*m, where ΔTb is the boiling point elevation, i is the van't Hoff factor for NaCl (which is 2 because NaCl disassociates into Na+ and Cl- ions), Kb is the ebullioscopic constant for the solvent (water), and m is the molality of the solution.

To find the molality (m), we rearrange the equation as m = ΔTb / (i*Kb). Substituting the known values, we get m = 1.500 / (2*0.5100) = 1.4706 mol/kg. Knowing the molality and the mass of the solvent (water), we can then calculate the moles of NaCl required, which is molality * mass of solvent in kg. Finally, converting moles of NaCl to grams using its molar mass (58.44 g/mol), gives us the total grams of NaCl needed. This step-by-step process provides a clear link between the theoretical concepts and their practical application.

Approximately 85.78 grams of [tex]NaCl[/tex] would need to be added to 1001 grams of water to increase the boiling temperature of the solution by [tex]1.500 \°C[/tex].

To solve this problem, we will use the concept of boiling point elevation, which states that the boiling point of a solvent is increased by an amount proportional to the molal concentration of the solute. The proportionality constant is known as the ebullioscopic constant (Kb). The formula to calculate the boiling point elevation is:

[tex]\[ \Delta T_b = i \cdot K_b \cdot m \][/tex]

 where:

- [tex]\( \Delta T_b \)[/tex] is the increase in boiling point temperature,

- [tex]\( i \)[/tex] is the van 't Hoff factor (the number of moles of particles in solution per mole of solute; for [tex]NaCl[/tex], [tex]\( i = 2 \)[/tex] because it dissociates into two ions, [tex]Na^+ and Cl^-)[/tex],

- [tex]\( K_b \)[/tex] is the ebullioscopic constant for the solvent (given as [tex]0.5100 \°C[/tex]/m for water),

- [tex]\( m \)[/tex] is the molality of the solution (moles of solute per kilogram of solvent).

Given:

- [tex]\( \Delta T_b = 1.5000 °C \)[/tex]

- [tex]\( K_b = 0.5100 °C/m \)[/tex]

- [tex]\( i = 2 \)[/tex] for [tex]NaCl[/tex]

- Mass of water (solvent) = 1001 g (which is approximately 1 kg, since 1000 g = 1 kg)

First, we will rearrange the formula to solve for the molality (m):

[tex]\[ m = \frac{\Delta T_b}{i \cdot K_b} \][/tex]

Substitute the given values:

[tex]\[ m = \frac{1.5000 °C}{2 \cdot 0.5100 °C/m} \][/tex]

[tex]\[ m = \frac{1.5000 °C}{1.0200 °C/m} \][/tex]

[tex]\[ m \approx 1.4695 m \][/tex]

Now that we have the molality, we can calculate the number of moles of [tex]NaCl[/tex] needed to achieve this molality in 1 kg of water:

[tex]\[ \text{moles of NaCl} = m \cdot \text{mass of solvent in kg} \][/tex]

[tex]\[ \text{moles of NaCl} = 1.4695 \text{ m} \cdot 1 \text{ kg} \][/tex]

[tex]\[ \text{moles of NaCl} \approx 1.4695 \][/tex]

The molar mass of [tex]NaCl[/tex] is approximately 58.44 g/mol. To find the mass of [tex]NaCl[/tex] needed, we multiply the number of moles by the molar mass:

[tex]\[ \text{mass of NaCl} = \text{moles of NaCl} \cdot \text{molar mass of NaCl} \][/tex]

[tex]\[ \text{mass of NaCl} \approx 1.4695 \text{ mol} \cdot 58.44 \text{ g/mol} \][/tex]

[tex]\[ \text{mass of NaCl} \approx 85.7797 \text{ g} \][/tex]

Explanation:

The relation between [tex]K_a\&K_b[/tex] is given by :

[tex]K_w=K_a\times K_b[/tex]

Where :

[tex]K_w=1\times 10^{-14}[/tex] = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = [tex]K_{a1}=1.4\times 10^{-2}[/tex]

The value of [tex]K_{b1}[/tex]:

[tex]1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}[/tex]

[tex]K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}[/tex]

The value of the second ionization constant of sodium sulfite = [tex]K_{a2}=6.3\times 10^{-8}[/tex]

The value of [tex]K_{b2}[/tex]:

[tex]1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}[/tex]

[tex]K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}[/tex]

Final answer:

To find the equilibrium constant for the ionization of HSO4−, we use the equilibrium concentrations of H3O+, HSO4−, and SO42− in the given reaction formula to compute Ka.

Explanation:

To compute the equilibrium constant for the ionization of the HSO4− ion, we need to use the expression for the equilibrium constant (Ka) which is based on the concentrations of products over reactants, excluding water because its concentration is considered constant in dilute aqueous solutions. The given equilibrium is HSO4−(aq) + H2O(l) ⇒ H3O+(aq) + SO42−(aq). Given equilibrium concentrations are [H3O+] = 0.027 M, [HSO4−] = 0.29 M, and [SO42−] = 0.13 M. Hence, the equilibrium constant (Ka) is calculated as Ka = [H3O+][SO42-−]/[HSO4−] = (0.027)(0.13)/(0.29).

Answer:

secrete cytotoxic substance which triggers apoptosis of target cell.

Explanation:

Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.

After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.

Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.

Answer:

the value of Kp at 25°C is 2.37 × 10⁻⁴atm³

Explanation:

Given that Kp = 0.116atm

NH₄(NH₂CO₂)(s) ⇄ 2NH₃(g) + CO₂(g)

                               2x atm         xatm

Pt                     =      PNH₃         PCO₂

0.116                =      2x          +        x

0.116                =      3x

x = 0.116/3

x = 0.039 atm

PNH₃ = 2x =2(0.039) = 0.078 atm

PCO₂ = x = 0.039 atm

Now,

KP = PNH₃² × PCO₂

     = 0.078² × 0.039 atm³

     = 2.37  × 10⁻⁴ atm³

the value of Kp at 25°C is 2.37 × 10⁻⁴atm³

Answer:

The mass of 1.00 mol of allicin, rounded to the nearest integer, is 162 g

Explanation:

Allicin is a compound that derivates from the alliin which is produced by the catalyzis of an enzime.

The molecular formula is: C₆H₁₀OS₂.

Let's determine the molar mass which is the mass that corresponds to 1 mol

Molar mass C . 6 + Molar mass H . 10 + Molar mass O + Molar mass S . 2 =

12 g/mol . 6 + 1 g/mol . 10 + 16 g/mol + 32.06 g/mol . 2 =  162.1 g/mol

The mass of 1 mole of allicin has been 162 g.

Allicin has been the organic compound of carbon, hydrogen, and sulfur. It has a molecular formula, [tex]\rm C_6H_10OS_2[/tex].

For the calculation of mass in a mole compound, the molecular mass of the compound has been given as:

[tex]mwt=M_C\;+\;M_H\;+\;M_O\;+\;M_S[/tex]

Where the mass of carbon, [tex]M_C=12\;\rm g/mol[/tex]

The mass of hydrogen, [tex]M_H=1\;\rm g/mol[/tex]

The mass of oxygen, [tex]M_O=16\;\rm g/mol[/tex]

The mass of sulfur, [tex]M_S=32\;\rm g/mol[/tex]

Substituting the values for molar mass of allicin as:

[tex]mwt=(6\;\times\;12)\;+\;(10\;\times\;1)\;+\;(1\;\times\;16)\;+\;(2\;\times\;32)\\mwt=72\;+\;10\;+\;16\;+\;64\;\text{g/mol}\\mwt=162\;\text{g/mol}[/tex]

The molar mass of allicin has been 162 g/mol.

The mass of 1 mole sample has been given as:

[tex]\rm Mass=moles\;\times\;\textit {mwt}[/tex]

Substituting the values for the mass of allicin as:

[tex]\rm Mass=1\;mol\;\times\;162\;g/mol\\Mass=162\;g[/tex]

The mass of 1 mole of allicin has been 162 g.

For more information about moles, refer to the link:

https://brainly.com/question/20674302

Final answer:

NaCN is likely to increase the solubility of ZnSe in water due to complexation with Zn2+, which drives the dissolution equilibrium forward.

Explanation:

The question asks which compound is most likely to increase the solubility of ZnSe when added to water. Considering the principles of solubility and the common ion effect, the best compound to increase ZnSe solubility would not be one that provides a common ion, as that would decrease solubility. Among the options provided, NaCN seems to be the most suitable choice.

Cyanide ions from NaCN can form a complex with Zn2+, effectively removing Zn2+ from the solution and driving the dissolution equilibrium of ZnSe forward. This complexation increases the solubility of ZnSe due to Le Chatelier's principle, where the reaction shifts to alleviate the stress of the added CN- ions by dissolving more ZnSe.

Answer:

ground water / leaking storage tanks

Answer:

1.2,dibromoethane is the sha'awa .

The central atom in C2H4Br2 is a carbon (C) atom, with each carbon atom being central to its respective part of the molecule, exhibiting a trigonal planar shape and bond angles of approximately 120°.

The central atom in C2H4Br2 is a C (carbon) atom, with other C, H (hydrogen), and Br (bromine) atoms as surrounding atoms. In this molecule, there are typically two central C atoms which are connected by a double bond, and each C atom has attached hydrogen and bromine atoms. Commonly in molecules, the central atom is the least electronegative element that is not hydrogen or a halogen, as these are usually terminal atoms. In C2H4Br2, each carbon atom is the central atom of its respective part of the molecule, with a trigonal planar shape due to the double bond and the single bonds to hydrogen and bromine. This arrangement leads to bond angles of about 120° around each central C atom.

Answer:

There are 1.29 moles of toluene in the solution.

Explanation:

m = ∆Tb/Kb

m is molality of the solution

∆Tb is the change in boiling point of cyclohexane = 90.3 ° C - 80.9 °C = 9.4 °C

Kb is the boiling point elevation constant of cyclohexane = 2.92 °C/m

m = 9.4/2.92 = 3.22 mol/kg

Number of moles of toluene = molality × mass of cyclohexane in kilogram = 3.22 × 401/1000 = 1.29 moles

Answer:

The limiting reactant can be described as the substance that is depleted first and stops

Explanation:

Imagine you have hydrogen and oxygen to produce water.

The reaction is:  2H₂(g) + O₂(g) → 2H₂O(g)

You have 1 mol of each reactant. As you see ratio is 2:1, so the limiting reactant is the hydrogen.

You know by stoichiometry, that 2 moles of H₂ need 1 mol of O₂ to react

If I have 1 mol of H₂, I will need the half of moles of O₂, so 0.5 moles. It is ok  because I have 1 mole, as I need the half, then half a mole will remain unreacted. This is what is called excess reagent,

If I make to react 1 mol of oxgen I need 2 moles of H₂. As I have 1 mol, of course I will need 2 moles but the thing is I have 1 mol.

This is the limiting reactant. I do not have enough of reactant so the reaction will happen until I complete to use it, that's why we can say that is depleted first and stops.

In a chemical reaction, when you have data of both reactants you can determine the limiting. Otherwise the excersise must tell that one ractant is in excess, to work with the limiting. Limiting reactant is the first step to work with the reaction, all the operations must be done by it. You do not use the reagent in excess

Answer: Volume of the balloon at this altitude is 46.9 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 737 mm Hg

[tex]P_2[/tex] = final pressure of gas =  385 mm Hg

[tex]V_1[/tex] = initial volume of gas = 28.6 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]26.8^oC=273+26.8=299.8K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-16.3^oC=273-16.3=256.7K[/tex]

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

[tex]\frac{737\times 28.6}{299.8}=\frac{385\times V_2}{256.7}[/tex]

[tex]V_2=46.9L[/tex]

Thus the volume of the balloon at this altitude is 46.9 L

Answer:

a) pPCl5 = 0.856 atm

b)pPCl5 = 0.0557 atm

pCl2 = pCl3 = 0.800 atm

c)  Ptotal = 1.66 atm

d) 93.5

Explanation:

Step 1: Data given

Temperature = 600 K

Kp = 11.5

Mass of PCl5 = 2.010 grams

Volume of the bulb = 555 mL = 0.555 L

The bulb is heated to 600 K

Step 2: The balanced equation

PCl5(g) ⇄ PCl3(g) + Cl2(g)

Step 3:

a) pv = nrt

⇒with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 0.555 L

⇒ with n =the number of moles PCl5 = 2.010 grams / 208.24 g/mol = 0.00965 moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 600K

p = (0.00965 *0.08206*600)/0.555

pPCl5 = 0.856 atm

b)

The initial pressures

pPCl5 = 0.856 atm

pCl2 = pCl3 = 0 atm

For 1 mol PCl5 we'll have PCl3 and 1 mol Cl2

The pressure at the equilibrium

pPCl5 = (0.856 -x) atm

pCl2 = pCl3 = x atm

Kp = pPCl3 * pCl2/pPCl5  

11.5 = x*x / (0.856 - x)

11.5 = x²/(0.856- x)

x = 0.8003

pPCl5 = (0.856 -x) atm = 0.0557 atm

pCl2 = pCl3 = x atm = 0.800 atm

c) Since x = 0.8003 and PCl3 and PCl2 are x  

Ptotal = 0.8003 + 0.8003 +0.0557 = 1.66 atm

d)

The degree of dissociation = (x / initial pressure PCl5)

(0.8003/0.856) * 100 = 93.5