Which of the following processes have a ΔS > 0? A. N2(g) + 3 H2(g) → 2 NH3(g) B. Na2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s) C. CH3OH(l) → CH3OH(s) D. All of these processes have a ΔS > 0. E. CH4(g) + H2O (g) → CO(g) + 3 H2(g)

Answer : The molecular formula of a caffeine is, [tex]C_8H_{10}N_4O_2[/tex]

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles[/tex]

Moles of N = [tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{4.12}{1.03}=4[/tex]

For H = [tex]\frac{5.19}{1.03}=5.03\approx 5[/tex]

For N = [tex]\frac{2.06}{1.03}=2[/tex]

For O = [tex]\frac{1.03}{1.03}=1[/tex]

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_4H_5N_2O_1=C_4H_5N_2O[/tex]

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]

[tex]n=\frac{194.19}{97}=2[/tex]

Molecular formula = [tex](C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2[/tex]

Therefore, the molecular of the caffeine is, [tex]C_8H_{10}N_4O_2[/tex]

Answer:

The solution in the buret, during a titration is called titrant.

Explanation:

A titration is a useful process, that makes you know the concentration of a solution.  A titrant solution (burette) is evaluated against a titrand to control the pH changes against the volume aggregate. Only a strong acid with a strong base, a strong base with a strong acid, a weak acid with a strong base and a weak base with strong acid are valued.

When the pH reaches the equivalence point, it is said that the normality of the acid by the milliequivalents, is equal to the basic normality by its milliequivalents. In conclusion, the entire base / acid became its conjugate pair.

To check this sudden change in pH, a substance is used, called Indicator that changes the color of the titrand (analyte).

In a titration analysis, the substance in the buret is called the 'titrant'. It is used to react with the analyte, the sample solution whose concentration we're measuring. The goal is to reach the endpoint, the point when a distinct visual change indicates that the titrant has completely reacted with the analyte.

In a titration analysis, the solution in the buret is called the titrant. This solution contains a known concentration of a substance. During a titration, this titrant is added incrementally to a sample solution, called the analyte, which contains the substance whose concentration is to be measured. The titrant and analyte undergo a chemical reaction of known stoichiometry.

By measuring the volume of the titrant solution needed to completely react with the analyte, scientists can calculate the concentration of the analyte. This point where the titrant has completely reacted with the analyte is termed the equivalence point of the titration. The process of adding the titrant is halted when a distinct change is visually detected in the solution - this could be a color change, for example. This is known as the end point.

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Answer:

it is easier to rotate and single bond rather than a double bond made of a sigma bond and pi bond

Explanation:

Rotation around a single bond happens easily but it is very limited around a double bond because of the overlapping electron cloud above and below the imaginary axis between the two atoms.

It is easier to rotate around a single sigma (σ) bond than around a double bond that includes both a sigma and a pi bond. This is because single bonds are not dependent on the orientation of the atom's orbitals, unlike double bonds. In the latter, any rotation can break the bond.

The matter at hand relates to the ease of rotation around different types of bonds - a single sigma (σ) bond versus a double bond made of one sigma (σ) and one pi (π) bond. Simplifying the science, it is easier to rotate around a single sigma (σ) bond than it is around a double bond that includes both a sigma and a pi bond. This is because the orbital overlap, which forms these bonds among atoms, doesn't rely on the relative orientation of the orbitals on each atom for a single bond. So, twisting or rotating doesn't affect the bonding. Whereas, with a double bond, the σ and π bonds form from different types of orbital overlaps. With this type of bond, the stable configuration is planar (flat) as seen in ethene molecules. Any rotation would misalign their overlapping, unstable, and effectively break the pi bond.

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Answer:

ΔH = -44,13 kJ/mol

Explanation:

For the following reaction:

NaOH(s) → Na⁺(aq) + OH⁻(aq)

It is possible to obtain the heat produced using:

Q = 4,18J/g°C×mass×ΔT

Where the mass is 35,1g + 199g = 234,1g

And ΔT is FinalT - InitialT → 62,6°C - 23,0°C = 39,6°C

Replacing:

Q = 38750 kJ = 38,75kJ

The enthalpy change is obtained using:

ΔH = -Q/moles of NaOH

Moles of NaOH are:

35,1g×[tex]\frac{1mol}{40g}[/tex]= 0,878 moles

Thus:

ΔH = -38,75 kJ/0,878mol = -44,13 kJ/mol

I hope it helps!

The solution contains 2.41 m (molality) and 2.14 M (molarity) of ethanol. A volume of 58.5 mL of this solution would contain 0.125 mol of ethanol.

(a) Let's calculate the molality of this solution. Mass of ethanol in 1000g of solution is 100g (10% of 1000g). The molar mass of ethanol (C2H5OH) is (2*12.01 + 6*1.008 + 1*16.00 + 1*1.008) = 46.07 g/mol. Hence, the number of moles = 100g / 46.07 g/mol = 2.17 mol. Since these moles are in 900g of water (1000g - 100g), molality = 2.17 mol / 0.900 kg = 2.41 m.

(b) To calculate molarity, let's find the volume of 1000g of solution: Volume = mass/density = 1000g / 0.984 g/mL = 1016.26 mL = 1.016 L. Therefore, molarity = 2.17 mol/1.016 L = 2.14 M.

(c) To find the volume of solution that would contain 0.125 mol ethanol, we'll use molarity: Volume = moles/Molarity = 0.125 mol / 2.14 M = 0.0585 L or 58.5 mL.

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The molality of the solution is 2.41 mol/kg, the molarity is 2.135 M, and to contain 0.125 mole of ethanol, 58.57 mL of the solution is required.

(a) Molality is defined as the number of moles of solute per kilogram of solvent. Given the density of the solution, we can determine the mass of 100.0 mL of the solution:

Density = 0.984 g/mL

Mass of 100 mL of solution = 100 mL * 0.984 g/mL = 98.4 g

Mass of ethanol (10% by mass) = 0.10 * 98.4 g = 9.84 g

Mass of water = 98.4 g - 9.84 g = 88.56 g

To convert grams of ethanol to moles:

Molar mass of ethanol (C2H5OH) = 46.07 g/mol

Moles of ethanol = 9.84 g / 46.07 g/mol = 0.2135 mol

To convert grams of water to kilograms:

88.56 g = 0.08856 kg

Therefore, the molality is:
Molality = 0.2135 mol / 0.08856 kg = 2.41 mol/kg

(b) Molarity is defined as the number of moles of solute per liter of solution. We already calculated the moles of ethanol as 0.2135 mol in 100 mL of solution.

Volume of solution = 0.100 L (since 100 mL = 0.1 L)

Therefore, the molarity is:
Molarity = 0.2135 mol / 0.100 L = 2.135 M

(c) We need to find the volume of solution that contains 0.125 mole of ethanol. Using the molarity calculated:

Molarity = 2.135 M = moles of solute / volume of solution (L)

Volume of solution = moles of solute / M

Volume of solution = 0.125 mol / 2.135 M = 0.05857 L = 58.57 mL

Therefore, the volume of the solution that would contain 0.125 mole of ethanol is 58.57 mL.

The work done by a gas on its surrounds is positive when the volume of the gas increases because the gas is doing work to push against the external pressure.

The work done by a gas on its surroundings is defined as positive when the gas expands, meaning when the volume of the gas increases. This is because the gas is doing work to push against the external pressure and expand its volume. So, the correct answer to your question 'When the volume of a gas increases, is the work done by the gas on its surroundings positive, negative, or zero?' would be (c) positive.

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When the volume of a gas increases, the work done by the gas on its surroundings is positive. This occurs because the gas is doing work on its surroundings, resulting in energy loss for the gas. The key equation is w = -Pext ΔV.

So the answer is C)Positive

To determine this, we need to understand the concept of work in thermodynamics. Specifically, the work done by a gas during expansion is given by the equation:

w = -[tex]P_{ext[/tex] ΔV,

where:

When the volume of the gas increases, ΔV is positive because the final volume ([tex]V_{final[/tex]) is greater than the initial volume ([tex]V_{initial[/tex]). According to the equation, since ΔV is positive and there is a negative sign in front, the work (w) done by the gas is negative.

This negative sign indicates that the system (the gas) is doing work on its surroundings, thus losing energy. Therefore, when the volume of a gas increases, the work done by the gas on its surroundings is positive.

Answer:

The heat capacity of the bomb calorimeter is 7.58 J/°C.

Explanation:

[tex]C_6H_6(I) + \frac{15}{2} O_2 (g) \rightarrow 6CO_2 (g) + 3H_2O (g) ,\Delta H^o = -3267.5 kJ[/tex]

First, we will calculate energy released on combustion:

[tex]\Delta H[/tex] = enthalpy change =  -3267.5 kJ/mol

q = heat energy released

n = number of moles benzene= [tex]\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{2.15 g g}{78 g /mol}=0.02756 mol[/tex]

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

[tex]q=\Delta H\times n =-3267.5 kJ/mol\times 0.02756 mol=-90.0657 kJ[/tex]

q = -90.0657 kJ = -90,065.7 J

Now we  calculate the heat gained by the calorimeter let it be Q.

Q = -q= -(-90,065.7 J) = 90,065.7 J (conservation of energy)

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

where,

Q = heat gained by calorimeter

c = specific heat capacity of calorimeter =?

[tex]T_{final}[/tex] = final temperature = [tex]34.34^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]22.46^oC[/tex]

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

[tex]90,065.7 J=c\times (34.34-22.46)^oC[/tex]

[tex]c=\frac{90,065.7 J}{(34.34-22.46)^oC}=7.58 J/^oC[/tex]

The heat capacity of the bomb calorimeter is 7.58 J/°C.

Answer:

a.

Explanation:

By the Linus Pauling's diagram, we know that the order of the energy of the subshell at the same atom, from the less to the higher, is:

1s < 2s < 2p < 3s < 3p < 4s < 4d ...

As larger is the atom, the attraction energy between the electron and the nuclei is weak. The size of the atom increase from right to left in the periods, and from the top to the bottom in the families. The cations are smaller than the neutral atoms (the attraction is more effective with fewer electrons), and the anions are larger than the neutral atoms. So, in order of size, and also the energy of the similar subshells:

He⁺ < He < H

How higher the energy, less negative it is, so less stable is the orbital.

a. As shown above, the energy of the orbital at H is higher than the energy at the orbital at He⁺, so the 1s orbital is more stable at He⁺, and the sentence is incorrect.

b. As shown above, the energy of the orbital at He is higher than the energy of the orbital He⁺, so the 2s orbital at He is less stable than the 2s orbital at He⁺, and the sentence is correct.

c. From the Linus Pauling's diagram, the energy of 2p is higher than the energy of the 2s, so the 2s subshell is more stable, and the sentence is correct.

Explanation:

The number of electron density regions that encircle an atom is calculated by the steric number, which also defines the hybridisation of the central atom. In that scenario, there are six hybridised orbitals in the central atom, leading to a [tex]\displaystyle sp3d^2[/tex]hybridization.

I am joyous to assist you anytime.

Answer:  d: They are poor conductors of heat and electricity .

Explanation:

Although is common knowledge that metals are characterized by high electrical conductivity and heat conductivity, these properties can be justified by looking at the metallic bonding.

In a metal, the bonding electrons are delocalized over the entire crystal. In fact, metal  atoms in a crystal can be imagined as an array of positive ions immersed in a sea of delocalized valence electrons.The mobility of the delocalized electrons makes metals good conductors of heat and electricity.

Answer:

25 coolers are need to be produce and sell in order to minimize the cost.

Explanation:

[tex]C = 0.005x^2-0.25x+12[/tex] ..[1]

Differentiating the given expression with respect to dx.

[tex]\frac{dC}{dx}=\frac{d(0.005x^2-0.25x+12)}{dx}[/tex]

[tex]\frac{dC}{dx}=0.01x-0.25+0[/tex]

Putting ,[tex]\frac{dC}{dx}=0[/tex]

[tex]0=0.01x-0.25+0[/tex]

[tex]0.01x=0.25[/tex]

x = 25

Taking second derivative of expression [1]

[tex]\frac{d^2C}{dx^2}=\frac{d(0.01x-0.25)}{dx}=0.01[/tex]

[tex]\frac{d^2C}{dx^2}>0[/tex] (minima)

25 coolers are need to be produce and sell in order to minimize the cost.

Explanation:

No, an exothermic reaction will not be  favored or less favored if it were carried out within a super heated chamber. In the super heated chamber the heat will flow back to the reaction( considering heat emitted by the reaction is less than the heat in a super heated chamber). So, the reaction will shift backward instead, the product will again change into reactant.

Answer:

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

Explanation:

More the pressure of the gas, more will be its solubility.

So, for two different pressure, the relation between them is shown below as:-

[tex]\frac {P_1}{C_1}=\frac {P_2}{C_2}[/tex]

Given ,  

P₁ = 1 atm

P₂ = 0.76 atm

C₁ = 6.8 × 10⁻⁴ mol/L

C₂ = ?

Using above equation as:

[tex]\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}[/tex]

[tex]C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L[/tex]

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

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Answer:

A) 9 moles of reactants chemically change into 11 moles of product.

Explanation:

C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)

The first step is to check if the equation is indeed a balanced one before proceeding to compare the moles.

This is done by comparing the number of atoms on the reactant side with that in the product side, if equal the reaction is indeed a balanced one.

Number of moles of Carbon atoms

Reactant = 5

Product = 5

Number of moles of Hydrogen atoms

Reactant = 12

Product = 12

Number of moles of Oxygen atoms

Reactant = 16

Product = 10 + 6 = 16

Since the equation is balanced, we can now compare the moles.

From the reactant side we have a total number of 9 (1 + 8 ) moles reacting to yield a total of 11 ( 5 + 6 ) moles of product.

This alone leads us  to take option A as our answer.

Let us see how other options are wrong however;

Option B) It cannot be grams as what chemical equations shows us the molar (number of moles) relationship between elememts/molecules/compounds/ions etc. The gram relationship can be calculated however if the molar mass of the compounds is taken into consideration.

Option C) This is wrong again due to same reasons mentioned for option B.

Option C) In the compound C5H12 alone, there are 17 (5 C and 12 H) atoms. Hence this option is also wrong.

The heat capacity of the bomb calorimeter is approximately 31.391 kJ/°C, and the energy of combustion of acetylene is approximately 1096.5 kJ/mol.

The heat capacity of a bomb calorimeter can be determined using the formula: q = CΔT, where q represents the energy absorbed or released, C is the heat capacity, and ΔT is the change in temperature.

First, we will determine the amount of energy absorbed by the bomb calorimeter from the combustion of methane. Given that the molar mass of methane is 16.04 g/mol, 6.79 g of methane corresponds to 0.423 mol. Therefore, the amount of energy absorbed is 0.423 mol x 802 kJ/mol = 339.046 kJ. Subsequently, we can calculate the heat capacity of the bomb calorimeter by rearranging the initial formula to: C = q/ΔT. So, C = 339.046 kJ/10.8 °C, which gives a heat capacity of approximately 31.391 kJ/°C.

Secondly, the energy of combustion of acetylene is given by the formula: q = CΔT. We already know the heat capacity of the bomb calorimeter (31.391 kJ/°C) and the change in temperature (16.9 °C). Therefore, the energy of combustion of acetylene is 31.391 kJ/°C x 16.9 °C = 530.7 kJ. As the molar mass of acetylene (C2H2) is 26.04 g/mol, 12.6 g corresponds to 0.484 mol. Thus, the energy of combustion of acetylene per mole is given by: 530.7 kJ/0.484 mol, which is approximately 1096.5 kJ/mol.

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The heat capacity of the bomb calorimeter is about 50229 kJ/°C. The energy of combustion of acetylene is found to be approximately 84710 kJ, which can then be converted back to kJ/mol from g.

The subject question refers to a bomb calorimetry experiment in which different substances, namely methane and acetylene, are being combusted. The heat capacity of the bomb calorimeter can be calculated using the formula q = CΔT, where q represents heat, C is heat capacity and ΔT is the change in temperature. The energy of combustion for the second substance can be calculated using the same formula rearranged as q/ΔT = C.

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Answer: Polyethylene

Explanation:

The waxy solid had been formed during high-pressure experiments conducted in 1933 by ICI scientists Reginald Gibson and Eric Fawcett. They had heated a mixture of ethylene and benzaldehyde to 170°C (338°F), using apparatus that could submit materials to a pressure of 1,900 atmospheres (1,925 bars). But the reactions were explosive and safety concerns prompted the now defunct ICI, which merged into Dutch-based Akzo Nobel, to halt the research.

Answer:

4C3H5N3O9 (l)  ---------> 12CO2 (g)  +  H20 (g)  + 6N2 (g) + 6O2 (g)

Explanation:

Nitroglcerin is a drug basically used to treat chest pain. It is a dense, colourless and explo9sive liquid. Its molecular formula is C3H5N3O9.

It decomposes to gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon di oxide

The equation for its decomposition is shown below;

4C3H5N3O9 (l)  ---------> 12CO2 (g)  +  H20 (g)  + 6N2 (g) + 6O2 (g)

Answer:

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)+3Br^{-}(aq.)\rightarrow Al^{3+}(aq.)+3Br^{-}(aq.)+3H_{2}O(l)[/tex]

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)\rightarrow Al^{3+}(aq.)+3H_{2}O(l)[/tex]

Explanation:

Aluminium hydroxide ([tex]Al(OH)_{3}[/tex]) is a base and hydrobromic acid (HBr) is a strong acid.

Hence an acid-base reaction occurs between [tex]Al(OH)_{3}[/tex] and HBr

Balanced molecular equation:

[tex]Al(OH)_{3}(s)+3HBr(aq.)\rightarrow AlBr_{3}(aq.)+3H_{2}O(l)[/tex]

Balanced total ionic equation:

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)+3Br^{-}(aq.)\rightarrow Al^{3+}(aq.)+3Br^{-}(aq.)+3H_{2}O(l)[/tex]

Balanced net ionic equation:

[tex]Al(OH)_{3}(s)+3H^{+}(aq.)\rightarrow Al^{3+}(aq.)+3H_{2}O(l)[/tex]

(net ionic equation is written by removing common ions present in both side of total ionic equation)

To synthesize 3-bromo-5-isopropyltoluene from m-isopropyltoluene, brominate using Br2/FeBr3, then rearrange with NaOH. Product: 3-bromo-5-isopropyltoluene with bromine at meta position and isopropyl group at para position.

To synthesize 3-bromo-5-isopropyltoluene starting with m-isopropyltoluene, we need to perform two main reactions: bromination and rearrangement. Here are the steps and reagents needed:

1. **Bromination**: We need to brominate the m-isopropyltoluene at the meta position. We can achieve this using bromine (Br2) in the presence of a Lewis acid catalyst like iron (III) bromide (FeBr3).

[tex]\[ \text{m-isopropyltoluene} + \text{Br}_2 + \text{FeBr}_3 \rightarrow \text{3-bromomethyl-isopropylbenzene} \][/tex]

2. **Rearrangement**: Next, we need to rearrange the 3-bromomethyl-isopropylbenzene to form the desired product, 3-bromo-5-isopropyltoluene. This rearrangement can be achieved using a strong base like sodium hydroxide (NaOH).

[tex]\[ \text{3-bromomethyl-isopropylbenzene} + \text{NaOH} \rightarrow \text{3-bromo-5-isopropyltoluene} \][/tex]

Now, let's draw the resulting product, 3-bromo-5-isopropyltoluene:

In this structure, the bromine atom (Br) is attached to the third carbon atom of the benzene ring, and the isopropyl group (-CH(CH3)2) is attached to the fifth carbon atom of the benzene ring.

Answer:

3.6

Explanation:

The pH of a solution can be defined as the negative logarithm to base 10 of the concentration of hydrogen ions in the solution. In the equation form, it can be written as :

pH = -Log [H+]

In the question, we have been given the concentration of the hydrogen ions but we do not know the value of the pH. We then insert the value of the concentration into the equation.

This means [H+] = 2.5 * 10^-4

pH = -Log [ 2.5 * 10^-4] = 3.6

Answer:

neutron

Explanation:

which affects mass number

Answer: The mass of magnesium in given amount of alloy is 0.396 g

Explanation:

We are given:

Percent mass of manganese = 4.5 %

Percent mass of copper = 3.5 %

Percent mass of magnesium = 0.45 %

Percent mass of aluminium = (100 - 4.5 - 3.5 - 0.45) = 91.55 %

Mass of alloy = 88 g

We need to calculate the mass of magnesium in 88 g of alloy, we get:

[tex]\Rightarrow 88\times \frac{0.45}{100}=0.396g[/tex]

Hence, the mass of magnesium in given amount of alloy is 0.396 g

Final answer:

The final equilibrium temperature for both the aluminum and the water is approximately 23.39°C.

Explanation:

To find the final equilibrium temperature for both the aluminum and the water, we can use the principle of thermal equilibrium and the conservation of energy.

First, let's calculate the heat gained or lost by the aluminum and the water:

The heat gained or lost by a substance can be calculated using the formula:

Q = mcΔT

where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the aluminum:

Qalu = malu * calu * ΔTalu = (8.6 g) * (0.900 J/g°C) * (T - 100°C)

For the water:

Qwater = mwater * cwater * ΔTwater = (402.4 g) * (4.18 J/g°C) * (T - 25°C)

Since the total heat gained by the aluminum is equal to the total heat lost by the water:

Qalu = -Qwater

Substituting the values and solving for T:

(8.6 g) * (0.900 J/g°C) * (T - 100°C) = -(402.4 g) * (4.18 J/g°C) * (T - 25°C)

Simplifying the equation:

(7.74 T - 774) = -(1681.832 T - 42045)

1765.832 T = 41271

T ≈ 23.39°C

The final equilibrium temperature for both the aluminum and the water is approximately 23.39°C.

Answer:Reliability

Explanation:

Reliability of a test refers to how consistently a test measures a characteristic under the same conditions.

Reliability can be defined as the degree of consistency of which a chemical test gives a similar result. measure. A test is said to be highly reliable when it gives the same repeated result under the same conditions of measure.

But when a test gives different results under the same condition of measure it has a low reliability.

Hence, If a test yields consistent results every time it is used, it has a high degree of reliability.

The rate law for the reaction can be determined by analyzing the effects of changing reactant concentrations. Based on the given observations, the rate law can be written as rate = k[A]/[NaSH] and rate = [tex]k[A][NaSH]^2[/tex] for the given conditions. Additionally, the rate constant, k, is temperature-dependent.

The rate law for the reaction can be determined by examining the effects of changing the concentrations of reactants on the rate of the reaction. Based on the given observations, we can conclude the following:

(a) The rate triples when the concentration of [A] is tripled and the concentration of [NaSH] is held constant. This suggests that the reaction rate is directly proportional to the concentration of A, so the rate law can be expressed as rate = k[A]

(b) The rate is decreased when the concentration of [A] is doubled and the concentration of [NaSH] is cut by a factor of 3. This implies that the rate is inversely proportional to the concentration of [NaSH], so the rate law can be written as rate = [tex]k[A]/[NaSH][/tex]

(c) The rate doubles when the concentration of [A] is cut in half and the concentration of [NaSH] is quadrupled. This indicates that the rate is quadratically proportional to the concentration of [NaSH], so the rate law can be expressed as rate = [tex]k[A][NaSH]^2[/tex]

(d) The rate increases with an increase in temperature. This suggests that the rate constant, k, in the rate law equation is temperature-dependent.

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Answer:

The answer to your questions is below

Explanation:

1.- Gas  matter with no definite shape or volume

2.- Liquid matter with no definite shape but with definite volume

3.- Sublimation the process of a solid changing directly to a gas without forming a liquid.

4.- Plasma matter in a high-energy state in which electrons are separated from their nuclei

5.- Solid matter with definite shape and volume

6.- Intermolecular bonds weak electrostatic bonds that form between particles of a substance

Final answer:

The matchings are 1-a, 2-c, 3-f, 4-d, 5-e, and 6-b, identifying the physical state or process corresponding to each description given in Part A using basic knowledge of the properties of the different states of matter.

Explanation:

To match the given descriptions in Part A with the correct terms in Part B, we need to understand the basic properties of the different states of matter and recognize the processes they undergo:

Gas: Matter with no definite shape or volume.

Answer:

1 electron

Explanation:

Fluorine is a nonmetal with a high electronegativity and electron affinity so it tends to gain electrons to reach stability. According to the octet rule, fluorine will gain enough electrons so as to complete its valence shell with 8 electrons. Fluorine is in the Group 17 in the Periodic Table, which means it has 7 electrons in its valence shell. Therefore, it will gain 1 electron to complete its octet and form the anion F⁻.

Answer:

No

Explanation:

If you heat the air inside a rigid, sealed container, then you are keeping it's mass and volume constant.

Then, Pressure of the container is directly proportional to the temperature.

Gay-Lussac's Law: This law states that the pressure of a given amount of gas when at constant volume is directly proportional to gas's absolute temperature in Kelvin.

So, if Kelvin temperature doubles, the air pressure in the container will also double.

The relationship between Celsius temperature and Kelvin temperature is not linear.

[tex]T_{K} = T_{C} + 273.15[/tex]

If [tex]T_{K}[/tex] becomes [tex]2*T_{K}[/tex], then [tex]T_{C}[/tex]  will become [tex](2 * T_{K})- 273.15[/tex] .

i.e, it is increasing for sure, but doesn't double.

Answer:

D is the correct option.

Explanation:

The carbon atoms are bonded together by a double bond.

The carbon atoms are bonded together by single bonds.

Lets analyse each option:

A) Its false. Lets look at the obvious: Ethylene is a gas and polyethylene is a solid. Monomers and polymers have very different properties.

B) Its false. Polyethylene doesnt have double bonds.

C) False as well. Polyethylene doesnt have double bonds.

D) True. It is evident from the structure of the the monomer and polymer that I've shown above. I'll also provide neat structures in the attachments.

Answer:

26.99 % of air that will be displaced

Explanation:

Step 1: Data given

A 1.2-L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m

Temperature = 23.5 °C

Atmospheric pressure = 1.2 atm

Liquid nitrogen has a density of 0.807 g/mL

Molar mass of N2 = 28 g/mol

Step 2: Calculate mass of nitrogen

Mass of nitrogen = density * volume

Mass of nitrogen = 0.807 g/mL * 1200 mL

Mass of nitrogen = 968.4 grams

Step 3: Calculate moles of N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 968.4 grams /28 g/mol

Moles N2 = 34.586 moles

Step 4: Calculate volume

p*V = n*R*T

⇒ p = the the pressure = 1.2 atm

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = the number of moles = 34.586 moles

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

⇒ T = the temperature = 23.5 °C = 296.65

V = (n*R*T)/p

V = (34.586 * 0.08206 * 296.65)/1.2

V = 701.61 L

Step 5: Calculate the total volume of the chamber

1.0 m * 1.3 m * 2 m = 2.6 m³ = 2600 L

Step 6: Calculate the percent volume displaced

(701.61 L  / 2600 L) * 100% = 26.99%

26.99 % of air that will be displaced

The percent (by volume) of air that would be displaced is 26.99 %.

Temperature = 23.5 °C

Atmospheric pressure = 1.2 atm

Liquid nitrogen has a density of 0.807 g/mL

Molar mass of N2 = 28 g/mol

Mass of nitrogen = density  × volume

[tex]= 0.807 g/mL \times 1200 mL[/tex]

= 968.4 grams

Now

Moles N2 = mass N2 ÷  molar mass N2

[tex]= 968.4 grams \div 28 g/mol[/tex]

= 34.586 moles

Now volume

[tex]p\times V = n \times R \times T[/tex]

Here p = the the pressure = 1.2 atm

n = the number of moles = 34.586 moles

R = the gas constant = [tex]0.08206 L\times atm/K\times mol[/tex]

T = the temperature = 23.5 °C = 296.65

[tex]V = (n\times R\times T)\div p\\\\= (34.586 \times 0.08206 \times 296.65)\div 1.2[/tex]

= 701.61 L

Now the total volume of the chamber is

[tex]= 1.0 m \times 1.3 m \times 2 m \\\\= 2.6 m^3[/tex]

= 2600 L

Now finally the percent volume displaced

[tex]= (701.61 L \div 2600 L) \times 100\%[/tex]

= 26.99%

Find out more information about the  Temperature here:https://brainly.com/question/7510619?referrer=searchResults