The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K. Calculate the standard change in enthalpy for this reaction (assuming ∆H° is temperature independent).

Answer:

744.2 g of C2H6 must burn to raise the temperature of 39.0 L of water by 58.0°C

Explanation:

This excersise is about calorimetry.

Q = m . C . ΔT

For water, 58°C is the ΔT, and the specific heat is 4.18 kJ/kg°C. We are missing the mass, but with the density data, we can know it.

Water density = water mass / water volume

1 g/ml = water mass / 39000 mL

(Notice we had to convert 39 L in mL (.1000))

Water mass = 39000 g

But this is in grams, and we have to make again a conversion, to kg because the units of specific heat.

Q = 39 kg . 4.18 kJ/ kg.°C . 58°C

Q = 9455.16 kJ

This is the heat required to change water temperature with that water mass, and the heat released for one mol of C2H6 is 382kJ.

How many mol of C2H6, for the heat required to change water, need the chemist?. The rule of three will be:

382 kJ ____ 1 mol of C2H6

9455.16 kJ _____  (9455.16 / 382) = 24.7 moles of C2H6

For mass, just work with the molar weight.

Mol . molar weight = mass

24.7 mol . 30.07g/m =744.2 g

To raise the temperature of 39.0 L of water by 58.0°C, you will need to burn 756.504 grams of C2H6.

Given: 39.0 L of water; Density = 1.00 g/cm3

Mass = Volume x Density

Mass = 39.0 L x 1.00 g/cm3

Mass = 39.0 kg

Heat = Mass x Specific Heat x Temperature Change

Heat = 39.0 kg x 4.184 J/g °C x 58.0 °C

Heat = 9619.4 kJ

Energy Released = 382 kJ

According to the balanced chemical equation, 1 mole of C2H6 releases 382 kJ

Therefore, we need to calculate the number of moles of C2H6 present in 382 kJ using the molar enthalpy change.

1 mole of C2H6 = 382 kJ

x moles of C2H6 = 9619.4 kJ

x = 9619.4 kJ * (1 mole of C2H6/382 kJ)

x = 25.2 moles of C2H6

Molar mass of C2H6 = 30.07 g/mol

Mass of C2H6 = Moles x Molar Mass

Mass of C2H6 = 25.2 moles x 30.07 g/mol

Mass of C2H6 = 756.504 g

Therefore, 756.504 grams of C2H6 must burn to raise the temperature of 39.0 L of water by 58.0°C.

https://brainly.com/question/30218216

#SPJ3

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.

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

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:

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:

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:

Osmotic pressure(π) = 0.661 Torr.

Depression in freezing point =  6.64 * 10⁻⁵ °C.

Explanation:

To calculate depression in freezing point and osmotic pressure, let's start by calculating Molarity of the solution.

Also, protein undergoes no dissociation or association when in solution.

Molarity = [tex]\frac{Moles of solute}{liters of solution}[/tex]

Molarity = [tex]\frac{3.2 g/L}{9.0 * 10^{4}g/mol }[/tex]

Molarity= 3.56 * 10⁻⁵ mol/L

Temperature = 25 +273 = 298 K

Osmotic pressure(π) = M R T

= (3.56 * 10⁻⁵ mol/L)* (0.08206 L atm/ mol K) * (298 K)

= 87.055 *  10⁻⁵ atm

But 1 atm= 760 Torr

So, Osmotic pressure(π) = (87.055 *  10⁻⁵ atm) * ( 760 torr/ 1atm)

= 0.661 Torr.

The depression in freezing point Δ[tex]T_{f} =K_{f} * m[/tex]

[tex]K_{f}[/tex]= molal freezing point depression constant of the solvent (1.86 °C/m for water).

m= molality or molal concentration of the solution.

m= moles of solute in 1kg of solvent.

Density of solution = [tex]1.0 \frac{g}{cm^{3} }[/tex]

Mass of 1L(1000 cm³) of this solution is= density * volume of solution

= 1000g

Molarity means 3.56 * 10⁻⁵ mol of protein in 1L of solution

Mass of protein=  3.56 * 10⁻⁵ * 9.0 * 10⁴ = 3.2 g of protein

1000g of solution- 3.2 g of protein = 996.8 g of solvent

Molality =  [tex]\frac{3.56 * 10⁻⁵ mol}{0.9968 kg}[/tex]

=3.57 *  10⁻⁵ m

depression in freezing point Δ[tex]T_{f} =K_{f} * m[/tex]

= 1.86 * 3.57 *  10⁻⁵ = 6.64 * 10⁻⁵ °C.

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

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:

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:

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;

Learn more: https://brainly.com/question/4612545

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.

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.

Learn More:

For more on nuclear reactors check out;

https://brainly.com/question/7588011

https://brainly.com/question/1253457

#LearnWithBrainly

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:

[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 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:

True

Explanation:

The majority of the reactions happened with a flow of heat. When there's no heat, the reaction is adiabatic.

For no adiabatic reactions, the heat can be released (evolution) by the system, so the reaction will be exothermic, or absorbed by the system (absorption), then the reaction is called endothermic.

Answer: The specific heat of the metal is 0.277 J/g.ºC.

Explanation:

The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. Its units are J/g.ºC.

If we know the specific heat and the amount of a substance, then the change in  the sample’s temperature (ΔT) will tell us the amount of heat (q) that has been absorbed  or released in a particular process. The equation for calculating the heat change is  given by:

[tex]: q=m.s.ΔT[/tex]      

Where ΔT is the temperature change: [tex]ΔT= tfinal - tinitial[/tex], m the mass and s the specific heat.

If no energy is lost to the sorroundings, then all the heat lost by the metal will be absorbed by the water. Therefore, the heat change of the system (qsystem) must be zero and we can write:

[tex]qsystem = qwater + qmetal[/tex]

[tex]qwater = -qmetal[/tex]

Replacing each term with the equation for calculating heat change:

[tex]mw.sw.ΔTw = -[mm.sm.ΔTm ][/tex]

A recommendation is to carry the units through the entire sequence of calculations. Therefore, if the equation is set up correctly, then all the units will cancel except the desired one.

[tex]53.2 g . 4.184 J/g°C . (26.1 - 24.0)ºC = -[23.0 g . sm . (26.1 - 99.0)°C][/tex]

[tex]464.436 J = -[23.0 g . sm . (-72.9)°C][/tex]

[tex]sm = 464.436 J/ -[-1676.7 g°C][/tex]

[tex]sm = 0.277 J/g.°C[/tex]

Thus, the specific heat of the metal is 0.277 J/g.ºC.

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.

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:

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:

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:

1.04 mM

Explanation:

The conversion reaction given is reversible, and for reversible reactions, the free-energy can be calculated by:

ΔG = -RTlnK

Where R is the constant of the gases(8.3145 J/mol.K), T is the temperature( 25°C + 273 = 298 K), and K is the equilibrium constant.

K = [F6P]/[glucose-6-phosphate]

Because T = 25ºC, ΔG = ΔG°' = 1670 J/mol

1670 = -8.3145*298*ln[F6P]/2.05

-2477.721*ln[F6P]/2.05 = 1670

ln[F6P]/2.05 = -0.6740

[F6P]/2.05 = 0.50966

[F6P] = 1.04 mM

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:

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.

https://brainly.com/question/34762749

#SPJ12

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.

https://brainly.com/question/35904400

#SPJ12

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:

1 Lower

2 wider

Explanation:

It is lower and wider in range because impurities affects the crystalline lattice of sample structure theory causing a deviation from real melting point of pure sample.

Final answer:

Impure samples display a wider and lower melting point range compared to a pure sample due to melting point depression caused by impurities.

Explanation:

Impure samples have melting point ranges that are both wider and lower compared to a pure sample. This is due to the presence of impurities which cause a phenomenon known as melting point depression. When assessing the purity of a substance, melting point determination is crucial as a pure sample typically has a very narrow melting point range of 1 - 2 0C. In contrast, an impure sample will start melting at a lower temperature and continue to melt over a broader range, with the extent of this range depending on the amount and type of impurity present.

For example, if we examine the melting points of samples of benzoic acid contaminated with acetanilide, as the quantity of impurity increases, the onset of melting begins at a progressively lower temperature, and the breadth of the melting range expands. This makes the melting point range a valuable tool for a rough assessment of a sample's purity.

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:

The correct answer is meters/second squared m/s2

Explanation:

Acceleration corresponds to a magnitude of vector type, is the relationship between a delta velocity and a delta time. The speed has units of m / second, km / second, km / minute for example, and time in seconds, minutes, etc.

Acceleration is measured in meters per second squared (m/s2), indicating the amount an object's speed changes every second.

The correct unit for acceleration is meters/second squared (m/s2). Acceleration is the rate at which an object changes its velocity. This means it's measuring how quickly an object's speed or direction of motion changes in a given period of time. In physics, the primary units are typically expressed using meters for distance, seconds for time; thus, it's in terms of how much the speed (meters / second) changes every second, leading us to meters/second squared.

https://brainly.com/question/11789833

#SPJ6