When combining with nonmetallic atoms, metallic atoms generally will(a) lose electrons and form negative ions(b) lose electrons and from positive ions(c) gain electrons and from negative ions(d) gain electrons and form positive ions

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: C10H8 + 12O2 ----> 10CO2 + 4H2O

Explanation:

To balance this, we have to make sure that the same number of atoms exist at both sides (Conservation of energy)

Note:The reaction is exothermic, giving off heat energy.

Left hand side: C= 10,

H=8,

O= 2×12=24

Right hand side: C= 10×1=10,

H= 4×2=8,

O=10×2=20 (for 10CO2) and 4×1=4 (All equal to 24)

With the above analysis, it is clear that numbers had to be added to molecules to get an equal number of atoms for both sides.

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:

Steep slope with high mountain, as these are formed by a composite volcano

Explanation:

Answer:

Steep slope with high mountain, as these are formed by a composite volcano

Explanation:

Answer:

6.935g

Explanation:

From the question above, we can see that 1 mole of magnesium hydroxide neutralized 2 moles of hydrochloric acid.

Now, let's calculate the actual number of moles of magnesium hydroxide reacted. We can do this by dividing the mass of the magnesium hydroxide by the molar mass. Molar mass of the magnesium hydroxide is 24 + 2(17) = 58g/mol

The number of moles thus produced is 5.5/58 = 0.095moles

From the first relation we established that 1 mole of magnesium hydroxide reacted with 2 moles of hydrochloric acid. Hence, 0.095moles of magnesium hydroxide will yield 2 × 0.095 moles of hydrochloric acid = 0.190 moles

We then calculate the mass of HCl that be neutralized by multiplying the number of moles by the molar mass. The molar mass of HCl is 1 + 35.5 = 36.5g/mol.

The mass of HCl neutralized = 36.5 × 0.190 = 6.935g

Answer: The mass of [tex]HCl[/tex] neutralized can be, 6.88 grams.

Explanation : Given,

Mass of [tex]Mg(OH)_2[/tex] = 5.50 g

Molar mass of [tex]Mg(OH)_2[/tex] = 58.3 g/mol

Molar mass of [tex]HCl[/tex] = 36.5 g/mol

First we have to calculate the moles of [tex]Mg(OH)_2[/tex]

[tex]\text{Moles of }Mg(OH)_2=\frac{\text{Given mass }Mg(OH)_2}{\text{Molar mass }Mg(OH)_2}[/tex]

[tex]\text{Moles of }Mg(OH)_2=\frac{5.50g}{58.3g/mol}=0.0943mol[/tex]

Now we have to calculate the moles of [tex]HCl[/tex]

The balanced chemical equation is:

[tex]Mg(OH)_2(aq)+2HCl(aq)\rightarrow 2H_2O(l)+MgCl_2(aq)[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]Mg(OH)_2[/tex] react with 2 mole of [tex]HCl[/tex]

So, 0.0943 moles of [tex]Mg(OH)_2[/tex] react with [tex]0.0943\times 2=0.1886[/tex] moles of [tex]HCl[/tex]

Now we have to calculate the mass of [tex]HCl[/tex]

[tex]\text{ Mass of }HCl=\text{ Moles of }HCl\times \text{ Molar mass of }HCl[/tex]

[tex]\text{ Mass of }HCl=(0.1886moles)\times (36.5g/mole)=6.88g[/tex]

Therefore, the mass of [tex]HCl[/tex] neutralized can be, 6.88 grams.

Answer:

[CH3OH(g)] = 1.7 M

Explanation:

∴ Kc(25°C) = 2.3 E4 = [CH3OH(g)] / [CO(g)]×[H2(g)]²

⇒ [CH3OH(g)] = Kc.[CO(g)][H2(g)]²

∴ [CO(g)] = 0.02 M

∴ [H2(g)] = 0.06 M

⇒ [CH3OH(g)] = (2.3 E4)(0.02)(0.06)²

⇒ [CH3OH(g)] = 1.7 M

Final answer:

To find the equilibrium concentration of methanol, the equilibrium-constant expression is rearranged and the given concentrations of reactants are plugged in, yielding an equilibrium concentration of methanol equal to 1.7 M.

Explanation:

To calculate the equilibrium concentration of methanol ([tex]CH_3OH[/tex]) using the equilibrium constant (Kc), we start by writing the equilibrium-constant expression for the reaction:

Kc = [[tex]CH_3OH[/tex]] / ([[tex]CO][H_2]_2[/tex])

Given that Kc = 2.3 × 104, [CO] = 0.02 M, and [[tex]H_2[/tex]] = 0.06 M, we can rearrange the expression to solve for [[tex]CH_3OH[/tex]]:

[[tex]CH_3OH[/tex]] = Kc × [CO] × [tex][H_2]_2[/tex]

Plug in the values:

[[tex]CH_3OH[/tex]] = 2.3 × 104 × 0.02 M × (0.06 M)2

Calculating the above expression gives us:

[[tex]CH_3OH[/tex]] = 2.3 × 104 × 0.02 × 0.0036 = 1.656 M

Rounding to one decimal place, the equilibrium concentration of methanol is 1.7 M.

The molecular formula of the gas compound given is approximately CCl3. The geometry surrounding each carbon atom would be trigonal planar with each carbon being surrounded by three chlorine atoms. This can be determined using the given mass, ideal gas laws, and the percent composition of the constituents.

The data given can be used alongside the Ideal Gas Law (PV=nRT) to find the molar mass of the gas. This molar mass, along with the percent composition of carbon (C) and chlorine (Cl), can be used to deduce the molecular formula of the compound.

The molar mass of chlorine is around 35.45 g/mol and for carbon, it's approximately 12.01 g/mol. With 25.305% C and 74.695%, the molecular formula becomes approximately CCl3, such that each carbon atom in the structure is surrounded by 3 chlorine atoms.

This geometry surrounding the carbon atom is trigonal planar based on VESPR theory. The Lewis structure would illustrate this with a carbon in the center surrounded evenly by three chlorine atoms.

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Answer:

its B

Explanation:

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.

Final answer:

For a molecule with the formula AB2, if there are no lone pairs on the central atom, the molecular shape is linear, as in BeH2 or CO2. If one lone pair exists, it creates a bent or V-shaped structure, seen in molecules like SO2. Other geometries like trigonal planar or T-shaped are not possible for AB2.

Explanation:

For a molecule with the formula AB2, there are potentially different shapes that the molecule can have, depending on the presence and arrangement of lone pairs of electrons on the central atom. If there are no lone pairs on the central atom, the molecule would have a linear shape, with the two B atoms and the A atom arranged in a straight line. This can be seen in examples like BeH2 and CO2, where the central atom contains only two electron groups, and they orient themselves as far apart as possible—180° apart.

However, if there is one lone pair on the central atom, the shape will be bent, or V-shaped. This is because the molecule can be thought of as a trigonal planar structure with one vertex missing due to the lone pair. The lone pair takes up additional space, causing the bond angle to be less than 120°, as seen in molecules like SO2.

Lastly, with additional lone pairs, other geometries like trigonal planar can be eliminated. The molecule AB2 could also not exhibit a T-shaped geometry as that would require more than three electron groups on the central atom.

Answer:

B. At the end of the reaction, both product and reactants are of a constant concentration.

Explanation:

[tex]^{*}[/tex]first half time is the when concentration of reactant reduces to 50% of initial concentration which you can nearly assume on or before the point of intersection of both the concentration graphs.

[tex]^{**}[/tex]end of reaction is the time when the reaction completes which is theoretically infinite but generally we take end of the reaction as the time when the slope of concentration curve becomes nearly zero or the time when change in concentration of reactant and product is negligible.

Answer:

B. At the end of the reaction, both product and reactants are of a constant concentration.

Explanation:

Got it right on the test!

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:

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:

Lithium loses one electron to fluorine and forms ionic bond, having formula [tex]LiF[/tex].

Explanation:

Lithium is the element of the group 1 and period 2 which means that the valence electronic configuration is [tex][He]2s^1[/tex].

Fluorine is the element of the group 17 and period 2 which means that the valence electronic configuration is [tex][He]2s^22p^5[/tex].

Thus, lithium loses 1 electron and become positively charged. Fluorine on the other hand accepts this electron and become negatively charged. This is done in order that the octet of the atoms are complete.  These both ions then form ionic bond as their will be electrostatic interaction between the two oppositely charged ions.

Thus, the formula of calcium chloride is [tex]LiF[/tex].

Ionic bonding between Lithium and Fluorine involves transferring 1 electron from Lithium to Fluorine, resulting in Lithium becoming a positive ion and Fluorine becoming a negative ion. This creates an electrostatic force that forms the bond. Both elements reach a stable electron configuration resembling those of noble gases.

The process of ionic bonding between Lithium (Li) and Fluorine (F) involves the transfer of electrons. In the noble gas configuration, Lithium has 1 electron in its outer shell and Fluorine has 7. For both elements to reach a stable state (similar to that of noble gases), Lithium needs to lose 1 electron and Fluorine needs to gain 1.

When these elements come together, Lithium donates its 1 electron to Fluorine. Now, Lithium has no electrons in its outer shell and is left with the inner shell that resembles the electron configuration of Helium (He), a noble gas. Fluorine, on receiving the electron from Lithium, now has 8 electrons in its outer shell and resembles the electron configuration of Neon (Ne), another noble gas. So, both have achieved a stable state.

On losing 1 electron, Lithium becomes a positively charged ion (Li+) and Fluorine, on gaining 1 electron, becomes negatively charged (F-). This creates an electrostatic force between the two ions and they stick together, resulting in an ionic bond.

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Metallic and nonmetallic mineral resources are considered nonrenewable because natural processes make minerals much more slowly than we can mine them. Therefore, the correct option is option A.

Natural resources that cannot be easily replenished by natural processes at a rate rapid enough even to keep up with use are considered non-renewable resources. Fossil fuels made of carbon are one instance. With the use of temperature and pressure, the original biological substance transforms into a fuel like gas or oil.

On the other hand, resources like lumber and wind are seen as renewable resources, primarily since their localized replenishment may take place during times that are also significant to people. Metallic and nonmetallic mineral resources are considered nonrenewable because natural processes make minerals much more slowly than we can mine them.

Therefore, the correct option is option A.

To learn more about non renewable resources, here:

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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.

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:

.B.When oxygen is removed from a combustion reaction, the reaction slows down or stops.

Explanation:

Le Chatelier's principle -

The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,

It states that ,

When a reaction is at equilibrium , Changing the concentration , pressure,  temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the change.

The combustion reaction of carbon dioxide is as follows -

C + O₂ → CO₂

Hence ,

Removing O₂ from the system , i.e. decreasing the concentration of O₂ ,  according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced O₂ .

Hence, reaction will go in backward direction.

The statement that correctly describes the possible effects of changing conditions on a chemical reaction is (B) When oxygen is removed from a combustion reaction, the reaction slows down or stops.

Le Chatelier's  principle states that when a system at equilibrium experiences a change in concentration, temperature, or pressure, the equilibrium will shift to counteract the imposed change and restore a new equilibrium. If we consider a combustion reaction at equilibrium, we can apply Le Chatelier's principle to predict the effects of changing conditions:

Overall, according to Le Chatelier's principle, the system will adjust to a change by shifting the equilibrium position to either increase the concentration of reactants or products, depending on the direction of the change.

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.

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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:

The heat absorbed by the metal is 91.27 kJ.

Explanation:

This is a sort of excersise about calorimetry which formula is:

Q = m . C . ΔT

where Q = heat

where m = mass

where C  = Specific heat

ΔT = T° final - T° initial

You can find specific heat in tables, if you don't know it. It is generally given in the statement unless, you have to find it out.

Specific heat for copper is 0.390 kJ/kg.°K

Notice that units in specific heat are in °K, but it is the same K or C.

The difference doesn't change the sense of units. We can use °C.

Q = 2.21 kg . 0.390 kJ/kg.°K (126.4°C - 20.5°C)

Q = 91.27 kJ

Final answer:

The copper metal absorbed approximately 898 kJ of heat when its temperature was increased from 20.5°C to 126.4°C.

Explanation:

To calculate the heat absorbed by a substance when its temperature changes, we can use the formula:

Q = mcΔT

where ΔT is the temperature change, m is the substance's mass, c is its specific heat capacity, and Q is the heat absorbed.

The specific heat capacity of copper (c) is approximately 0.385 J/g°C. We need to use the mass (m) in grams and the temperature change (ΔT) in Celsius.

First, we convert the mass from kg to g:

2.21 kg = 2210 g

Then calculate the temperature change (ΔT):

ΔT = Final temperature - Initial temperature

ΔT = 126.4°C - 20.5°C = 105.9°C

Now, we can calculate Q:

Q = (2210 g)(0.385 J/g°C)(105.9°C)

Q = 897993.35 J

Q = 897.99335 kJ

Q ≈ 898 kJ (rounded to three significant figures)

So, the copper metal absorbed approximately 898 kJ of heat when its temperature was increased from 20.5°C to 126.4°C.

Answer:

The answer to your question is CuSO₄

Explanation:

To answer your question just remember the following information

- A chemical reaction is divided into sections

 reactants and products

 reactants on the left side of the reaction

products on the right side of the reaction

- All the symbols have a meaning

 (s) means that that compound is in solid phase

 (aq) means that that compound is dissolved in solution.

Then, the answer is CuSO₄

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.

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Answer:

Deforestation can directly affect the nitrogen cycle within a forest ecosystem. It is because the nitrogen is used by the micro-organisms such as nitrogen fixing bacteria that helps in converting the atmospheric nitrogen into useful ammonia that are taken up by the plants, in order to carry out the process of photosynthesis.

By cutting down the trees and plants, these cycling of nitrogen will be disturbed and also these nitrogen containing articles will be eroded and eventually will mix up with the rivers and stream affecting the aquatic ecosystem.

Thus, by cutting down the trees, the nitrogen cycle will be disrupted in a forest ecosystem.

Answer:

The element A is S (sulfur)

Explanation:

The elements for the 3erd period in the periodic table are Na, Mg, Al, Si, P, S, Cl and Ar.

The one that has 6 e⁻ in its valence shell is the S, because it is missing 2 e⁻ to reach the octet rule. 2 e⁻ to has the most stable noble gas conformation.

The IE of S = 3360 kJ/mol

It is a little lower than Cl because the electron is so far from the nucleus, that's why we have to apply a very low ionization energy to rip the electron off.

The element 'A' in question, which belongs to period 3 and has 6 valence electrons, is most likely aluminum (Al).

Based on the given information, the element 'A' belongs to period 3 and has 6 valence electrons. To identify the element, we need to find the ionization energy values that match the given pattern. Looking at the successive ionization energies provided, we can see that the jump in ionization energy occurs after the third ionization. Since 'A' has 6 valence electrons, it is likely to be aluminum (Al), which fits the pattern.

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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.

Final answer:

The interhalogen with the lowest boiling point is bromine chloride (BrCl) because both bromine and chlorine are smaller atoms compared to iodine.

Explanation:

The boiling point of a compound is determined by the strength of the intermolecular forces between its molecules. In the case of the interhalogens, the boiling point generally increases as the size and atomic mass of the halogen atoms increase. Therefore, the interhalogen with the lowest boiling point would be the one with the smallest halogen atom.

Among the given examples, bromine chloride (BrCl) has the lowest boiling point because both bromine and chlorine are smaller atoms compared to iodine. Iodine bromide (IBr) would have a higher boiling point since iodine is larger than both bromine and chlorine.

Similarly, bromine fluoride (BrF) would have a higher boiling point compared to bromine chloride due to the presence of a larger fluorine atom. Lastly, chlorine fluoride (ClF) would have the highest boiling point because both chlorine and fluorine are smaller atoms compared to bromine.

Answer:

The given statement is true.

Explanation:

Chlorine gas is prepared by heating manganese dioxide with hydrochloric acid.T he reaction takes place in two steps

1) Reaction between manganese dioxide and HCl gives manganese(II) chloride along with water and nascent oxygen.

[tex]MnO_2+2HCl\rightarrow MnCl_2+H_2O+O[/tex]...[1]

2) Then this nascent oxygen formed in above reaction oxidizes HCl into water and chlorine gas.

[tex]2HCl+O\rightarrow Cl_2+H_2O[/tex]..[2]

So , the net chemical reaction comes out to be:

[tex]MnO_2+4HCl\rightarrow MnCl_2+H_2O+Cl_2[/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.