Consider the following reaction to generate hydrogen gas: Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) The hydrogen gas is collected by displacement of water at 35.0 °C at a total pressure of 745 torr. The vapor pressure of water at 35.0 °C is 42 torr. Calculate the partial pressure of H2 (in atm) if 353.2 mL of gas is collected over water from this method.

Answer:

See explanation below to full answer

Explanation:

First of all, you are not providing the amounts of acid and hydroxide here, to do the calculations. However, in order to help you, I will use these values that are taken from a similar exercise. Then, replace your data with this procedure and you should get the correct answer.

For this part, I will say that the student weights about 210 mg of oxalic acid, (H2C2O4) and the volume of NaOH used to reach equivalent point was 150 mL in a beaker of 250 mL.

Now the equivalence point is the point where both moles of acid and hydroxide are the same. In other words:

nA = nB

The reaction that it's taking place is the following:

2NaOH + H2C2O4 ----------> Na2C2O4 + 2H2O

This means that 2 moles of NaOH reacts with 1 mole of H2C2O4, therefore the expression in (1) corrected is:

nB = 2 nA

So, we need to calculate first the moles of the acid. To do that we need the molar mass of the acid (the reported is 90.03 g/mol)

nA = 0.210 / 90.03 = 0.0023 moles

We have the moles of acid used, so the moles of the hydroxide is:

nB = 2 * 0.0023 = 0.0046 moles

We have the volume used of hydroxide, which is 150 mL, so finally the concentration is:

MB = 0.0046 / 0.150 = 0.031 M

Now, replace the actual values that you have in here, and you should get an accurate result.

Answer:

The limiting reactant is NaOH (option B)

Explanation:

2S(s)  +  3O₂(g)  +  4NaOH(aq)   →   2Na₂SO₄(aq)  +  2H₂O(l)

The reaction is ballanced. OK

We need to know how many moles do we have from each compound.

Mass / Molar weight = Mol

Molar weight S = 32 g/m

Molar weight O₂ = 32 g/m

Molar weight NaOH = 40 g/m

Mol S: 2g/ 32g/m = 0.0625 mol

Mol O₂: 3g / 32 g/m = 0.09375 mol

Mol NaOH: 4g/ 40g/m = 0.1 mol

Now, we can play with the reactants. The base is: 2 moles of S, react with 3 mol of O₂ and 4 moles of hydroxide to make 2 moles of sulfate and 2 moles of water. Pay attention to the rules of three.

2 moles of S __ react with __ 3 moles of O₂ __ and __ 4 moles of NaOH

0.0625 moles S __________ 0.09375 moles O₂ ___ 0.125 moles NaOH

The limiting reactant is the NaOH. I need to use 0.125 moles and I only have 0.1 moles.

Let's do the same with O₂

3 moles of O₂ __ react with __ 2 moles of S __ and __ 4 moles of NaOH

0.09375 moles of O₂ _______ 0.0625 mol of S _____ 0.125 moles NaOH

Answer:

The new volume of the gas is 1.04L

Explanation:

You have to apply Charles's Law to solve this:

In two different situations, when you have a gas with the same quantity and  pressure, relation between volume and T° must be the same

Volume / Temperature = Constant

Temperature in K

So;

475 mL/248K = Vol₂ / 548K

(475 mL/248K) 548K = Vol₂

1049 mL = Vol₂

The fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C is 92.0%. The van der Waals constant b is 0.0322 L/mol.

The van der Waals constant b is defined as four times the total volume actually occupied by the molecules of a mole of gas 1. To calculate the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C, we can use the following formula:

V_real = V_ideal - nb

where V_real is the real volume of the gas, V_ideal is the ideal volume of the gas, n is the number of moles of the gas, and b is the van der Waals constant. At 0 C, the ideal volume of one mole of any gas is 22.4 L 2. Therefore, the ideal volume of Ar atoms is 22.4 L/mol.

To calculate the real volume of Ar atoms, we need to know the number of moles of Ar atoms present in the container. We can use the ideal gas law to calculate the number of moles of Ar atoms:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Substituting the given values, we get:

n = PV/RT = (230 atm * V)/(0.0821 L atm/mol K * 273 K) = 9.03 V

Substituting the values of n and b into the formula for V_real, we get:

V_real = V_ideal - nb = 22.4 L/mol - 0.0322 L/mol * 4 * 9.03 mol = 20.6 L

Therefore, the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C is:

(V_real/V) * 100% = (20.6 L/V) * 100% = 92.0%

Answer:

C

Explanation:

The answer is shape.

Basically, quantities can either be fundamental or derived. While fundamental are the basic quantities, derived are obtained from combining fundamentals.

To get a shape, we would need the combination of lengths. This makes the shape a derived quantity

Answer:

[Ba^2+] = 0.160 M

Explanation:

First, let's calculate the moles of each reactant with the following expression:

n = M * V

moles of K2CO3 = 0.02 x 0.200 = 0.004 moles

moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles

Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.

Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3

As you can see, 0.04 moles of  K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of

0.012 - 0.004 = 0.008 moles of Ba(NO3)2

These moles are in total volume of 50 mL (30 + 20 = 50)

So finally, the concentration of Ba in solution will be:

[Ba] = 0.008 / 0.050 = 0.160 M

Answer : The quantity of heat released is -11.30 kJ

Explanation :

First we have to calculate the number of moles of water.

[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}[/tex]

Molar mass of water = 18 g/mole

[tex]\text{Moles of water}=\frac{5.00g}{18g/mole}=0.278mole[/tex]

Now we have to calculate the amount of heat released.

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

where,

[tex]\Delta H[/tex] = heat of vaporization = 40.66 kJ/mol

q = heat released = ?

n = number of moles of water = 0.278 mole

[tex]40.66kJ/mol=-\frac{q}{0.278mol}[/tex]

[tex]q=-11.30kJ[/tex]

In vaporization process, the amount of heat is absorbed but in the process of condensation the amount of heat is released.

Therefore, the quantity of heat released is -11.30 kJ

To calculate the quantity of heat absorbed/released when 5.00 g of steam condenses to liquid water at 100°C, use the equation Q = mL. Convert the mass of steam to moles, calculate the heat absorbed or released. Substitute the values and calculate the heat absorbed/released.

To calculate the quantity of heat that is absorbed/released when 5.00 g of steam condenses to liquid water at 100°C, we can use the equation Q = mL, where Q is the heat absorbed or released, m is the mass of the substance, and L is the latent heat of vaporization. First, we need to convert the mass of steam to moles using the molar mass of water. Then, we can calculate the heat absorbed or released using the given latent heat of vaporization.

First, calculate the moles of water vapor:

moles = (mass of water vapor) / (molar mass of water)

Next, calculate the heat absorbed or released using the formula:

Q = (moles of water vapor) * (latent heat of vaporization)

Plugging in the given values, we get:

Q = (5.00 g / (molar mass of water)) * (latent heat of vaporization)

Finally, calculate the molar mass of water using the atomic masses of hydrogen and oxygen:

molar mass of water = (2 * atomic mass of hydrogen) + atomic mass of oxygen

Substitute the molar mass of water and the given latent heat of vaporization into the equation, and calculate the value of Q to find the quantity of heat absorbed/released.

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Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.  

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):  

[tex]V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol[/tex]

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

[tex]n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol[/tex]

[tex]V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21 [/tex]

[tex]P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm[/tex]

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.  

Final answer:

Using the ideal gas law, we can determine the moles of nitrogen in an 'empty' container and calculate the partial pressure of oxygen in air at atmospheric pressure. The moles of nitrogen is 78% of the total moles of air in the container. The partial pressure of oxygen is 21% of the atmospheric pressure.

Explanation:

To calculate the number of moles of nitrogen in a two-liter container at atmospheric pressure and room temperature, we can use the ideal gas law PV = nRT. Given room temperature 25°C (which is 298.15 K), a volume of 2.00 liters (2.00 x 10-3 m3), and atmospheric pressure (1 atm or 101.325 kPa), we can solve for n, the number of moles of nitrogen gas (N2).

The air is approximately 78% nitrogen by moles. Therefore, to find the moles of nitrogen, we first calculate the moles of air using the ideal gas law and then multiply this by 0.78.

For the partial pressure of oxygen, we acknowledge that air is around 21% oxygen by moles. Thus, the partial pressure of oxygen would be 0.21 times the total atmospheric pressure, which results in a partial pressure of 0.21 atm.

Answer:

a. making popcorn in a microwave oven.  Endothermic

b. a burning match.  Exothermic

c. boiling water.  Endothermic

d. burning rocket fuel. Exothermic

e. the reaction inside a heat pack Exothermic

Explanation:

In order to answer, we need to review the definitions of exothermic and endothermic reactions.

Exothermic reactions give out heat. They cause increase in the energy of the system.

Endothermic reactions absorb heat. They cause decrease in the energy of system.

By this definition,

a. making popcorn in a microwave oven. Endothermic as heat energy is provided to the corn which causes it to pop.

b. a burning match. Exothermic as heat energy is given out by a burning match.

c. boiling water. Endothermic as heat energy is provided to the water which causes it to boil.

d. burning rocket fuel. Exothermic as heat energy is given out by burning fuel.

e. the reaction inside a heat pack. Exothermic as reaction which takes place inside heat pack gives out heat. This heat provides comfort to painful joints and muscles.

Energy is absorbed in an endothermic reaction while energy is released in an exothermic reaction.

An exothermic process is a process in which energy is released. This implies that heat is evolved in the process. In an endothermic process, heat is absorbed in the process. We shall now classify the following process as endothermic or exothermic accordingly;

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Raising the control rods out of the reactor

The reactor regulates the number of neutrons that are involved in the chain reaction. This is accomplished by the reactor absorbing some of the neutrons, produced in the splitting of the atoms, in its walls.

Explanation:

If the rods are taken out of the reactor, the rods would heat up very fast and most probably an explosion would occur. This is because most of the neutrons produced in the splitting of the radioactive atoms in the rod would go ahead and bombard other atoms in the rods hence spiking up the chain reaction rate. This would release a lot of energy at a go.

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Answer:

Raising the control rods out of the reactor

Explanation:

Raising the rods will allow the chain reaction to flow more freely, therefore increasing the energy output of the nuclear reactor

Answer:

a) Reduction:

Ag⁺(aq) + e⁻ → Ag(s)

Oxidation:

Pb(s) → Pb⁺²(aq) + 2e⁻

Overall reaction:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺

b) Silver; oxidation;  from the lead electrode to the silver electrode.

Explanation:

a) Ag⁺ had lost 1 electron, so need to gain 1 electron to become Ag(s). Pb needs to lose 2 electrons to become Pb⁺².

Reduction:

Ag⁺(aq) + e⁻ → Ag(s)

Oxidation:

Pb(s) → Pb⁺²(aq) + 2e⁻

Overall reaction:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺ (it will need 2Ag⁺ to gaind the 2 electrons released by Pb)

b) The cation formed in the redox reaction is Pb²⁺, so, to equilibrate the charges, it will flow towards the silver (Ag) electrode.

The lead (Pb) is being oxidized, so oxidation is happening at it.

The electrons flow from the oxidation (anode) to the reduction (cathode), so they flow from the lead electrode to the silver electrode.

In the voltaic cell, the Ag+ is reduced to Ag in the silver half-cell, while Pb is oxidized to Pb2+ in the lead half-cell. The cations flow towards the silver electrode and the electrons flow from the lead to the silver electrode. Hence, the overall reaction is 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq).

In a voltaic cell with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell, the silver half-cell acts as the cathode or reduction half-cell which gains electrons, while the lead half-cell acts as the anode or oxidation half-cell and loses electrons. Therefore, the balanced half-reactions and overall spontaneous reactions are:

(a) Balanced Half-Reactions and Overall Reaction:

Reduction: Ag+(aq) + 1e- -> Ag(s)

Oxidation: Pb(s) -> Pb2+(aq) + 2e-

Overall Reaction: 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq)

(b) The Cation Flow and Electrons Flow:

The cation flow is towards the silver electrode and the electron flow  is from the lead electrode to the silver electrode. In the voltaic cell, the process that occurs at the lead electrode is oxidation.

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Answer:

65.712 grams of oxygen has reacted.

Explanation:

[tex]2 C_2H_2(g) + 5 O_2(g)\rightarrow 4 CO_2(g) + 2 H_2O(g)[/tex]

Mass of acetylene = 21.39 g

Moles of acetylene = [tex]\frac{21.39 g}{26.04 g/mol}=0.8214 mol[/tex]

According to reaction , 2 moles of acetylene reacts with 5 moles of oxygen gas.

Then 0.8214 moles of oxygen gas will react with :

[tex]\frac{5}{2}\times 0.8214 mol=2.0535 mol[/tex] of oxygen gas.

Mass of 2.0535 moles of oxygen gas :

2.0535 mol × 32 g/mol = 65.712 g

65.712 grams of oxygen has reacted.

Final answer:

To find the final solution temperature, we need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other. By plugging in the given values and solving the equation, we find that the final solution temperature will be 24.77 °C.

Explanation:

To find the final solution temperature, we can use the principle that the heat given off by the reaction is equal to that taken in by the solution. We need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other.

First, we calculate the heat exchanged by the reaction using the equation:

q_reaction = C_reaction * ΔT_reaction

where C_reaction is the heat capacity of the reaction solution and ΔT_reaction is the change in temperature of the reaction.

Next, we calculate the heat exchanged by the solution using the equation:

q_solution = m_solution * C_solution * ΔT_solution

where m_solution is the mass of the solution, C_solution is the specific heat of the solution, and ΔT_solution is the change in temperature of the solution.

Now we can set the two heat exchanges equal to each other and solve for the final solution temperature:

q_reaction = q_solution

C_reaction * ΔT_reaction = m_solution * C_solution * ΔT_solution

Plugging in the given values:

C_reaction = C_solution = 3.98 Jg⁻¹°C⁻¹

m_solution = (50.00 mL of NaOH * 1.02 g/mL) + (25.00 mL of HCl * 1.02 g/mL) = 76.50 g

ΔT_reaction = (28.9 °C - 24.66 °C) = 4.24 °C

ΔT_solution = ?

Now we can solve for ΔT_solution:

3.98 Jg⁻¹°C⁻¹ * 4.24 °C = 76.50 g * 3.98 Jg⁻¹°C⁻¹ * ΔT_solution

ΔT_solution = (3.98 Jg⁻¹°C⁻¹ * 4.24 °C) / (76.50 g * 3.98 Jg⁻¹°C⁻¹) = 0.1107 °C

Finally, we calculate the final solution temperature:

Final Temperature = 24.66 °C + 0.1107 °C = 24.77 °C

The final temperature of the solution after the reaction is approximately 33.51°C.

To find the final temperature of the solution after the neutralization reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl), we can follow these steps:

The reaction between NaOH and HCl can be written as:

[tex]\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]

Calculate the moles of NaOH and HCl:

 [tex]\[ \text{Moles of NaOH} = 1.05 \, \text{M} \times 0.05000 \, \text{L} = 0.0525 \, \text{moles} \] \[ \text{Moles of HCl} = 1.88 \, \text{M} \times 0.02500 \, \text{L} = 0.0470 \, \text{moles} \][/tex]

  Since HCl is the limiting reagent (0.0470 moles compared to 0.0525 moles of NaOH), the reaction will produce 0.0470 moles of water.

2.Calculate the heat released during the reaction:

The enthalpy change for the neutralization of strong acid and base (like HCl and NaOH) is typically [tex]\(-57.3 \, \text{kJ/mol}\).[/tex]

The total heat released q can be calculated as:

[tex]\[ q = \text{moles of HCl} \times \Delta H_{\text{neutralization}} \] \[ q = 0.0470 \, \text{moles} \times -57.3 \, \text{kJ/mol} = -2.6931 \, \text{kJ} = -2693.1 \, \text{J} \][/tex]

  (The negative sign indicates that the heat is released, but we will use the magnitude for temperature calculation.)

3. Determine the total mass of the solution:

The total volume of the solution is:

[tex]\[ \text{Volume} = 50.00 \, \text{mL} + 25.00 \, \text{mL} = 75.00 \, \text{mL} \][/tex]

Given the density of the solution is 1.02 g/mL, the total mass (\(m\)) is:

  [tex]\[ m = 75.00 \, \text{mL} \times 1.02 \, \text{g/mL} = 76.50 \, \text{g} \][/tex]

4.Calculate the temperature change:

[tex]\[ \Delta T = \frac{q}{mc} \] \[ \Delta T = \frac{2693.1 \, \text{J}}{76.50 \, \text{g} \times 3.98 \, {J/gC}} = \frac{2693.1}{304.47} \approx 8.85 \°C} \][/tex]

5.Calculate the final temperature:

The initial temperature of both solutions is 24.66°C. Thus, the final temperature [tex](\(T_f\))[/tex] is:

[tex]\[ T_f = 24.66 \, \°C} + 8.85 \, \°C} = 33.51 \, \°C} \][/tex]

So, the final temperature of the solution after the reaction is approximately 33.51°C.

Answer: Electrovalent or Ionic Compounds

Explanation:

Electrovalent Compounds Form bonds that are characterised by transfer of electrons from metallic atoms to non-metal licenses atoms during a chemical reaction.

The metallic atom after donating their valence electrons, become positively charged, while the non-metal license atoms becomes negatively charged after acquiring extra electrons.

A typical example of electrovalent compounds can be found between the association of Group 1(Alkali Metals) elements and the Group 7(Halogen Family) elements.

Answer: The type of compound involves the transfer of electrons is called the ionic compounds.

Explanation: ionic compounds are compounds in which one atom or molecule completely transfers an electron to another.

Ions that have gained an electron are negatively charged and they are called anions while ions that have lost an electron are positively charged and they are called cations.

Answer:

The molal boiling point elevation constant is 1.59 ≈  1.6 [tex]Kkgmol^{-1}[/tex]

Explanation:

To solve this question , we will make use of the equation ,

Δ[tex]T_{b} = i*K_{b} *m[/tex]

where ,

           ∴ [tex]m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]

    4. [tex]K_{b}[/tex] ⇒ ?

∴

[tex]1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]

⇒ [tex]K_{b}[/tex] = [tex]1.59[/tex] ≈ 1.6 [tex]Kkgmol^{-1}[/tex]

The molal boiling point elevation constant (Kb) of substance X is 4.1.

The molal boiling point elevation constant (Kb) can be calculated using the formula: ΔT = Kb × m

Where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.

In this case, the change in boiling point (ΔT) is 1 degree C (124.3 - 123.3), the molality (m) can be calculated by dividing the molal mass of urea by the mass of the solvent water, which gives a value of 0.0246 kg urea / 0.100 kg water = 0.246 mol/kg, and the formula becomes: 1 = Kb × 0.246

Now, rearrange the equation to solve for Kb: Kb = 1 / 0.246 = 4.07

Rounding to 2 significant digits, the molal boiling point elevation constant Kb of substance X is 4.1.

Answer:

[tex]m_{CO2}=260.7 g CO2[/tex]

Explanation:

First of all we need to calculate the energy required:

[tex]KE= 0.5*m*v^2[/tex]

where:

[tex]m=2700kg[/tex]

[tex]v=67 mph=29.95 m/s[/tex]

[tex]KE= 0.5*2700kg*(29.95)^2[/tex]

[tex]KE= 1210953 J=1210.953 kJ[/tex]

Octane's combustion enthalpy: [tex]\Delta H_{comb}=- 5450 kJ/mol[/tex]

The reaction:

[tex]C_8H_{18} + 25/2 O_2 longrightarrow 8 CO_2 +9 H_O[/tex]

Mass of CO2:

[tex]m_{CO2}=\frac{1210.953 kJ}{5450mol}*\frac{1}{0.3}*\frac{8 mol CO2}{1 mol}*\frac{44 g CO2}{mol CO2}[/tex]

[tex]m_{CO2}=260.7 g CO2[/tex]

Answer:

The correct answer is AMP+H2O→ Adenosine + pi

Explanation:

The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.

As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.

Answer:

See explanation below

Explanation:

First, you need to know the density of each compound in order to know this.

The density of 1-chlorobutane is 0.88 g/mL,

The density of water is 1 g/mL

The density of sodium bicarbonate is 2.2 g/cm3.

therefore, the one that has a greater density will always go at the lower phase.

In this case, after the reflux, it will stay in the lower phase, basically because you don't have another solvent with a greater density than the butane.

After adding water, it will be in the upper phase, water has a greater density.

After adding bicarbonate, it will be in the upper phase too.

Answer:

336.8 kilo Joules is the change in internal energy of the system.

Explanation:

The equation for first law of thermodynamics follows:

[tex]\Delta U=Q+W[/tex]

where,

Q = heat added to the system

ΔU = Change in internal energy

W = work done

We have :

Amount of heat given out by the system will be negatuive as heat relased by the system = Q

Q= [tex]-5.00\times 10^2 kJ[/tex]

Work done on the system will positive as work is done on the system:

w = [tex]2.00\times 10^2 kCal=836.8 kJ[/tex]

[tex]\Delta U=-5.00\times 10^2 kJ+836.8 kJ=336.8 kJ[/tex]

336.8 kilo Joules is the change in internal energy of the system.

The net change in internal energy of the system is calculated using the first law of thermodynamics and is found to be 3.368 × 105 Joules.

The change in internal energy (ΔU) of a system can be calculated using the first law of thermodynamics, which states that the change in internal energy is equal to the heat (Q) added to the system minus the work (W) done by the system on its surroundings: ΔU = Q - W.

In your question, work is done on the system (2.00 × 102 kcal), which equates to 2.00 × 105 cal or 8.368 × 105 Joules (since 1 kcal = 4.184 kJ or 4184 Joules). Heat is released by the system (5.00 × 102 kJ), which is already in Joules. Since work is done on the system, it's positive, but heat released by the system is negative for the internal energy calculation. So, ΔU = Q - W = -5.00 × 102 kJ + 8.368 × 105 J.

Here's the calculation:

Therefore, the net change in internal energy of the system is 3.368 × 105 Joules.

Answer:

Ciliary body.

Explanation:

Ciliary body: It is the known for the part of the eye that includes the ciliary muscle, which helps in the control the ciliary epithelium and lens shape, which are helping in the production of aqueous humor.

Through active secretion mechanism helping in to produce eighty percent of aqueous humor, and through the plasma ultra-filtration mechanism twenty percent of aqueous humor is produced.

Ciliary body is the part of the layer which helps to deliver the nutrients, and oxygen to the eye tissues, and this layer is known as uvea.

The aqueous humor, a watery fluid in the anterior cavity of the eye, forms during capillary filtration in the ciliary body.

The aqueous humor is a watery fluid that fills the anterior cavity of the eye, which includes the cornea, iris, ciliary body, and lens. It is produced during a process called capillary filtration.

Capillary filtration occurs when fluid moves from an area of high pressure to an area of lower pressure on the other side of the capillary wall. In the eye, this process takes place in the ciliary body, a part of the eye that has a rich capillary network, and results in the formation of aqueous humor.

The production of aqueous humor is essential for maintaining intraocular pressure and providing nutrients to the cornea and lens, which do not have their own blood supply.

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Answer:

N2

Explanation:

We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get

Given:

P = 2atm

1atm = 101,325pa

2atm = 202,650pa

T = 27 degrees Celsius = 27 + 273.15 = 300.15K

V = 2.2L

R = molar gas constant = 8314.46 L.Pa/molK

PV = nRT

Rearranging n = PV/RT

Substituting these values will yield:

n = (202,650 * 2.2)/(8314.46* 300.15)

n = 0.18 moles

To get the molar mass, we simply divide the mass by the number of moles.

5.1/0.18 = 28.5g/mol

This is the closest to the molar mass of diatomic nitrogen N2.

Hence, the gas is nitrogen gas

The diatomic gas could be identified using the ideal gas law and the given conditions. The calculated molar mass matched with the molar mass of Oxygen, so the diatomic gas is likely Oxygen (O2).

By examining the given conditions and the difference in mass, we can identify the gas X2 using the ideal gas law. The ideal gas law is given by 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 in Kelvin. Here, the pressure P = 2.00 atm, the volume V = 2.2 L, R = 0.0821 L.atm/mol.K and T = 27°C = 300.15 K. Inserting these values gives us the number of moles of gas. The molar mass of the gas can be calculated by dividing the mass of the gas by the number of moles. Using the molar mass and comparing it to the periodic table, the diatomic gas appears to be Oxygen (O2).

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Answer:

The coefficients should be: 2, 5, 2, 5

Explanation:

Given redox reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺+ SO₄²⁻

To balance the given redox reaction in acidic medium, the oxidation and the reduction half-reactions should be balanced first.

Reduction half-reaction: MnO₄⁻ → Mn²⁺

Oxidation state of Mn in MnO₄⁻ is +7 and the oxidation state of Mn in Mn²⁺ is +2. Therefore, Mn accepts 5e⁻ to get reduced from +7 to +2 oxidation state.

⇒ MnO₄⁻ + 5e⁻ → Mn²⁺

Now the total charge on reactant side is (-6) and the total charge on product side is +2. Therefore, to balance the total charge, 8H⁺ must be added to the reactant side.

⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺

To balance the number of hydrogen and oxygen atoms, 4H₂O must be added to the product side.

⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O               .....equation 1

Oxidation half-reaction: SO₃²⁻ → SO₄²⁻

Oxidation state of S in SO₃²⁻ is +4 and the oxidation state of S in SO₄²⁻ is +6. Therefore, S loses 2e⁻ to get oxidized from +4 to +6 oxidation state.

⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻

Now the total charge on reactant side is (-2) and the total charge on product side is (-4). Therefore, to balance the total charge, 2H⁺ must be added to the product side.

⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻ + 2H⁺

To balance the number of hydrogen and oxygen atoms, 1 H₂O must be added to the reactant side.

⇒ SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺                .....equation 2

Now, to cancel the electrons transferred, equation (1) is multiplied by 2 and equation (2) is multiplied by 5.

Balanced Reduction half-reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O ] × 2

⇒ 2MnO₄⁻ + 10e⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O                .....equation 3

Balanced Oxidation half-reaction:

SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ ] × 5

⇒ 5SO₃²⁻ + 5H₂O → 5SO₄²⁻ + 10e⁻ + 10H⁺                  .....equation 4  

Now adding equation 3 and 4, to obtain the overall balanced redox reaction:

2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O

Therefore, the coefficients should be: 2, 5, 2, 5

Answer:

Tend to keep the product concetration low and therefore drive the reaction righward

Explanation:

The fact the products of a reaction are quickly consumed by the next one would tend to keep the product concetration low and therefore drive the reaction righward (to the products).

This happens because the system will not achive equilibrium between the reactants and the product, and will keep producing it util the system achives equilibrium or the reactants dry out.

Answer:53gm

Explanation:

To calculate the density of a substance, divide its mass by its volume. For the mass of a diamond or volume of mercury, multiply or divide, respectively, the given quantity by the substance's density. Densities are significant as they indicate how much matter is contained within a space.

The density of a substance is defined as its mass per unit volume. The formula for density (d) is d = mass (m) / volume (v), where the mass is measured in grams (g) and the volume in cubic centimeters (cm3) for solids and liquids, or in milliliters (mL) as 1 mL equals 1 cm3.

To find the density of a block of marble, we use the formula with the given values: d = 853 g / 310 cm3.

To find the mass of a diamond with a known density, multiply the volume by the density: mass = 0.350 cm3
x 3.26 g/cm3.

The volume of liquid mercury given its mass and density can be calculated by rearranging the formula: volume = 76.2 g / 13.6 g/mL.

Lastly, to find the density of a sample of ore, apply the formula: d = 74.0 g / 20.3 cm3.

Remember that densities can vary greatly among different materials and are particularly high for substances such as gold and mercury.

Answer:

1.94 L

Explanation:

21°C = 21 +273 = 294 K

27°C = 27 + 273 = 300 K

T1/V1 = T2/V2

294 K/1.9 L = 300 K/x L

x = (1.9*300)/294 ≈ 1.94 L

Changes in pressure and temperature can affect a reaction's effectiveness. A drop in temperature or a rise in pressure makes the given reaction more effective, while a rise in temperature or a drop in pressure makes it less effective.

In the given reaction, the relative enthalpy and the entropy can provide insights on how the temperature and pressure changes can affect the reaction.

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More effective reaction conditions for the specified equilibrium include decreasing temperature and increasing pressure, leading to more product formation. Increasing temperature and decreasing pressure are less effective. Effectiveness is based on shifting equilibrium towards the product side.

The reaction in question is SO₂ (g) + 2H₂S (g) ↔ 3S(s) + 2H₂O (g). To determine the effectiveness of the reaction under different temperature and pressure changes, one must consider how each change affects the equilibrium. Here’s the analysis:

The transition elements fill d sublevels, coming after the s sublevel of the same principal energy level. Lanthanides begin filling the 4f sublevel after the 6s, positioned two principal energy levels behind. Many transition element compounds display bright colors from d electron transitions.

The sublevels that are filling across the transition elements are primarily the d sublevels. The electron configurations of these elements have their outermost s sublevel either completely filled or missing one electron. However, the defining characteristic of transition elements is the filling of the inner d sublevel, which typically occurs after the s sublevel of the same principal energy level has been filled. When discussing the f-block elements, specifically the lanthanides, these elements begin filling their 4f sublevels after the 6s sublevel. This is due to the behaviour of electron filling, where the f sublevels are two principal energy levels behind the current one being filled.

Many transition element compounds are known for their brightly colored appearances as a result of inner-level d electron transitions. Unlike the transition elements, the lanthanides are not grouped together in the periodic table and instead are inserted in a separate row, reflecting their unique electron configurations and properties.

Answer:

The most appropriate answer here would be :

Internal energy of a system is the sum of the rotational, vibrational, and translational energies of all of its components

Explanation:

Internal energy of a system is the total energy the system possess. It is represented by U (I'll be referring to internal energy as U now). This option is particularly true for ideal gases. In ideal  monoatomic gases, U is the sum of translational kinetic energies only. In di and polyatomic gases, U is the sum of translational and rotaional kinetic energies. Also, vibrational kinetic energies come into play as we increase the temperature and this also adds to U. But, in real substances such as real gases, solids, liquids, there is also interatomic forces and these accounts for intermolecular potential energies. Intermolecular potential energies also add to U in these type of systems. But even for real gases, under many circumstances, the intermolecular potential energy can be neglected.

So, the most appropriate answer here is: Internal energy of a system is the sum of the rotational, vibrational, and translational energies of all of its components

The internal energy of a system is the total of both kinetic and potential energies of its atoms and molecules. It includes various energy forms such as translational, vibrational, and rotational energy. Thus, option C is correct.

The internal energy of a system is the sum of all the kinetic and potential energies of its component atoms and molecules. This encompasses various forms of energy, including translational, vibrational, and rotational kinetic energy, as well as potential energy from molecular interactions and chemical bonds.

The correct choice is C. It is the sum of the potential and kinetic energies of the components.

Complete Question: -
The internal energy of a system ________.

A. refers only to the energies of the nuclei of the atoms of the component molecules

B. is the sum of the kinetic energy of all of its components

C. is the sum of the potential and kinetic energies of the components

D. is the sum of the rotational, vibrational, and translational energies of all of its components none of the above

Answer:

Density of carbon tetrachloride = 12.8 g/mL

Explanation:

Given :

[tex]m_{flask}=321.9\ g[/tex]

[tex]m_{flask}+m_{CCl_4}=523.6\ g[/tex]

Mass of carbon tetrachloride: -

[tex]m_{flask}+m_{CCl_4}=523.6\ g[/tex]

[tex]m_{CCl_4}=523.6-m_{flask}\ g=523.6-321.9\ g=201.7\ g[/tex]

Mass of carbon tetrachloride = 201.7 g

Given, Volume = 15.7 mL

Considering the expression for density as:

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

So,

[tex]Density=\frac {201.7\ g}{15.7\ mL}[/tex]

Density of carbon tetrachloride = 12.8 g/mL

Answer:Shielding and penetration are essentially the same thing and occurs when one electron is blocked from the full effects of the nuclear charge so that the electron experiences only a part of the nuclear charge

Explanation:

Penetration is how well the outer electrons are shielded from the nucleus by the core electrons. The outer electrons therefore experience less of an attraction to the nucleus.