The principal source of the difference in the normal boiling points of ICI (97°C; molecular mass 162 amu) and Br2(59°C; molecular mass 160 amu) is:a. ICI has greater strength of hydrogen bonding than Br2 b. ICI has stronger dipole-dipole interactions than Br2c. ICI has stronger dispersion forces than Br2 d. The I-CI bond in ICl is stronger than the Br-Br bond in Br2 e. ICI has a greater molecular mass than Br2

Answer:

31.37%

Explanation:

For this case, you should consider the following reaction:

Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O

For which you obtain the precipitate of BaSO₄

In order to obtain the mas of Barium on the precipitate, you may use the following formula:

gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)

Where:

gBa= mass of Barium

M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate

M₍Ba₎= mass of Barium from the original sample

M₍BaSO₄₎= mass of BaSO₄ from the precipitate

gBa= (0.5331)x(137.327/233.39)= 0.3136 g

Then we ontain the percentage of Barium multiplying by 100:

% Ba on the original sample= 31.36%

Final answer:

The percentage of barium in the compound is calculated by finding the moles of BaSO₄, converting it to moles of Ba, and then finding the mass of Ba before expressing it as a percentage of the initial sample mass. The resulting percentage of barium is approximately 43.99%.

Explanation:

To determine the percentage of barium in the compound, we must calculate the moles of BaSO₄ and then use the molar mass of barium to find the mass of barium in the initial sample. We can calculate the moles of BaSO₄ using the following steps:

Moles of BaSO₄ = mass of precipitate / molar mass of BaSO₄

= 0.5331 g / 233.39 g/mol

= 0.002284 mol (rounded to six decimal places)

= 0.002284 mol × 137.327 g/mol

= 0.3137 g (rounded to four decimal places)

Finally, we calculate the percentage of barium in the sample using the initial sample mass (0.713 g).

Percentage of barium = (mass of Ba / initial sample mass) × 100%

= (0.3137 g / 0.713 g) × 100%

= 43.99%

Therefore, the percentage of barium in the compound is approximately 43.99%.

Answer: The atomic number of an element is the same as the number of its protons.

Explanation: For example, the atomic number of oxygen is 8. So, this would mean that the number of protons would be 8.

Hope this helps^

Answer:

1.93

Explanation:

Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH = 38/1000 × 0.50 = 0.019mol

Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol

Final pH = pKa + log([C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/[C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH]

= -log Ka  + log(mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH

= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17

change in pH = final - initial pH

= 4.17 - 2.24

=1.93

pH is the depiction of the acidity and basicity of the solution and can range from 0 to 14 on the pH scale. The change in pH of benzoic acid is 1.93.

pH is the concentration of the hydrogen ion in the solution depicting the acidity and the basicity of the solution.

Calculate moles of Benzoic acid as:

[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{38}{1000}\times 0.50\\\\&= 0.019\;\rm mol\end{aligned}[/tex]

Calculate moles of Sodium benzoate as:

[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{2.64}{144.10}\\\\&= 0.01832\;\rm mol\end{aligned}[/tex]

pH can be calculated as:

[tex]\begin{aligned}\rm pH &= \rm pKa + log\dfrac{([C_{6}H_{5}COONa]}{[C_{6}H_{5}COOH]}\\\\&= \rm -log(6.5 \times 10^{-5}) + log (\dfrac{0.018321}{0.019})\\\\&= 4.17\end{aligned}[/tex]

Change in the pH can be calculated as:

[tex]\begin{aligned} \rm pH &= \rm final - initial \; pH\\\\&= 4.17 - 2.24\\\\&=1.93\end{aligned}[/tex]

Therefore, 1.93 is the change in the pH of benzoic acid.

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Answer:

V= 0.147 L

Explanation:

This is simply the application of combined gas law twice, to find the unknowns.

Combined gas law states that: [tex]PV = nRT[/tex]

P= pressure of air

V= volume of air

n= moles of air

R= Universal gas constant ( 0.08205 L atm mol⁻¹ K⁻¹)

T= Absolute temperature in kelvin.

[tex]n=\frac{P * 0.115}{R * 377}[/tex]

Now, applying the same gas law at 483K and substituting for n

[tex]V = \frac{P * 0.115}{R * 377} * \frac{R* 483}{P}[/tex]

V= [tex]\frac{0.115 * 483}{377}[/tex]

V= 0.147 L

Answer:

The mixture is made up of different atoms and pure substance is made up of same type of atom.

The main difference is that mixture can be separated into its component by physical mean while pure substances can not be separated by physical process

Explanation:

Mixture:

Examples are:

Pure Substance:

Pure substances are those made of same type of atoms all elements and compounds are pure substances.

Examples are:

Answer:

The % of displaced volume of nitrogen is 29.06%.

Explanation:

Volume of nitrogen = 1.2 L = 1200 mL

Density of nitrogen = 0.807 g/ml

Mass of nitrogen = [tex]Density \times Volume= 0.807 \times 1200 = 1044 g[/tex]

Molar mass of nitrogen = 28 g/mol

[tex]Number\,of\,moles\,nitrogen=\frac{Mass}{Molar\,mass}[/tex]

[tex]= \frac{1044}{28}=37.28[/tex]

The ideal gas equation is as follows

[tex]PV = nRT[/tex]

Rearrange the equation is as follows.

[tex]V= \frac{nRT}{P}...............(1)[/tex]

n= Number of moles = 37.28

R = Gas constant = 0.0820

T = Temperature = 22+ 273 = 295

P = Pressure = 1.2 atm

Substitute the all values in equation (1)

[tex]V= \frac{37.28 \times 0.0820 \times 295 }{1.2}= 751.5L= 0.751 \,m^{3}[/tex]

[tex]0.751 \,m^{3}[/tex] of nitrogen will displace same amount of air.

[tex]Volume \,\,of\,\, closet= 1 \times 1.3 \times 2 = 2.6\,m^{3}[/tex]

[tex]%\,displaced\,volume=\frac{0.751}{2.6}=28.8%[/tex]

Therefore, The % of displaced volume of nitrogen is 29.06%.

Final answer:

The percent by volume of air displaced by the evaporation of liquid nitrogen is calculated using the density of liquid nitrogen, the molar mass of nitrogen, and the Ideal Gas Law, accounting for the specific temperature and pressure conditions.

Explanation:

To calculate the percent by volume of air displaced if all the liquid nitrogen evaporated, we need to consider the Ideal Gas Law as well as the density of the liquid nitrogen. Since liquid nitrogen has a density of 0.807 g/mL, we can first find the mass of the nitrogen and then use the molar mass to determine the number of moles of nitrogen gas at the given temperature and pressure when it evaporates.

Using the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature in Kelvin. At standard temperature and pressure, the density of nitrogen gas can be calculated using the given molar mass and standard gas equation factors.

Once the volume of the gaseous nitrogen is determined, that volume will represent the amount of air volume displaced. To find the percent by volume displaced, divide the volume of gaseous nitrogen by the volume of the container and multiply by 100%.

In an ecosystem model of earth , white colored paper represents areas that are covered with snow and reflect sunlight.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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Answer: Option (D) is the correct answer.

Explanation:

A reaction in which heat energy is absorbed by the molecules of a substance is known as an endothermic reaction.

Whereas when energy is released by the molecules of a substance in a reaction then it is known as an exothermic reaction.

Thus, we can conclude that the vaporization of rubbing alcohol is endothermic.

Among the provided options, the vaporization of rubbing alcohol is the endothermic process, as it involves the absorption of heat from its surroundings.

In the context of chemistry, an endothermic process is one in which heat is absorbed from the surroundings. This results in an overall increase in the internal energy of the system. Given the options, the vaporization of rubbing alcohol is the process that is endothermic. This is because when rubbing alcohol vaporizes, it absorbs heat from its surroundings, causing the surrounding environment to feel cooler. Other examples might be the melting of ice or the evaporation of water, both of which require an external heat source to facilitate the process.

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Answer:

H⁺/H₂//Ni/Ni⁺²; H⁺/H₂//Cd/Cd⁺²; Ni⁺²/Ni//Cd/Cd⁺²

Explanation:

The redox reactions are spontaneously when the E° of the substance that is reduced is higher than the E° of the substance that is oxidized. The substance is reduced when gaining electrons and is oxidized when loses electrons.

So, the reactions that occur spontaneously are:

2H⁺ + Ni → H₂ + Ni⁺² (H⁺ is reduced and Ni is oxidized)

2H⁺ + Cd → H₂ + Cd⁺² (H⁺ is reduced and Cd is oxidized)

Ni⁺² + Cd → Ni + Cd⁺² (Ni⁺² is reduced and Cd is oxidized)

Pairs: H⁺/H₂//Ni/Ni⁺²; H⁺/H₂//Cd/Cd⁺²; Ni⁺²/Ni//Cd/Cd⁺²

The enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.

Thermochemical equation is defined as a chemical equations that are properly balanced and take into account the physical conditions of all reactants, products, and energy change. A chemical equation is an equation that shows the beginning molecule, reactants, and final products separated by arrows, but a thermochemical equation is a balanced stoichiometric chemical process that also includes the enthalpy change.

Given heat capacity = 227.14 J/°C

Temperature = 9.82 °C

Number of moles of HCl and NaOH is 0.04 moles

q (solution) = -C (solution) x ΔT

= 227.14 J/°C x 9.82 °C

= 2.2305 kJ

For enthalpy of reaction we have to divide it by mole of reactant

= 2.2305 / 0.04

= 55.4 kJ/ mole

Thus, the enthalpy of reaction for the following thermochemical equation, when the heat capacity of the solution and calorimeter is 227.4 J/°C is 55.7 kJ /mole.

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To determine the enthalpy of reaction for HCl reacting with NaOH, the heat released by the reaction is calculated using the mass of the solution, the specific heat capacity, and the change in temperature. The result is then divided by the number of moles reacted to express the enthalpy change in kJ per mole.

The problem here concerns the measurement of an enthalpy change during the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O). To find the enthalpy of reaction for the provided thermochemical equation, we use the temperature change and the heat capacity of the solution to calculate the total heat released by the reaction, and then express this heat change in kJ per mole of reactants.

First, calculate the total heat (q) released using the formula q = mc∆T, where m is the mass of the solution, c is the specific heat capacity, and ∆T is the change in temperature. As the reaction takes place in 50.0 mL of water, and assuming the density of water is 1.00 g/mL, the mass of the solution is 50.0 g. Therefore, q = 50.0 g × 227.4 J/°C × 9.82°C.

After calculating q, divide this value by the number of moles of HCl reacted to obtain the enthalpy of reaction per mole. Remember to convert from joules to kilojoules since enthalpy is usually expressed in kJ/mol. Please note that the reaction is exothermic, so the enthalpy change should be a negative value, indicating heat is released.

Answer:

Explanation:

It wouldn't be able to separate negative gram from positive. it would not wash away some of the crystal violet stain effectively. Gram - Cells will appear to be the same hue. Water can also dilute cells or wash them off the slide

Answer:

Explanation:

Would not be able to differentiate gram - from + it would not efficiently wash some of the crystal violet stain away. gram - cells will still remain the same color. also water may dilute or wash cells off the slide

Answer:

B: Discovering the internal structure of the atom.

Explanation:

Dalton’s experiments with gases marked the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton’s atomic theory is based can be

summarized as follows:

Dalton made no attempt to describe the structure or composition  of atom —he had no idea what an atom is really like. But he did realize that the  different properties shown by elements such as hydrogen and oxygen can be explained  by assuming that hydrogen atoms are not the same as oxygen atoms.

Answer:

so total pressure is = 261.92 kPa

partial pressure of Ar gas=  33.50 Kpa

partial pressure of N2 gas = 227  kPa

Explanation:

PV = nRT  

n = PV / RT

as we know that 1dm3 = 1 Liter  

smaller flask before mixing

ideal gas constant

n = (803 kPa) x (4L) / ((8.3144621 L kPa/K mol) x (25 + 273 K))  

n = (3212 kPa L) / ((8.3144621 L kPa/K mol) x (298 K))  

n = (3212 kPa L) /2477.7

n=  1.29 mol

In the larger flask:

n = (47.2 kPa) x (10 L) / ((8.3144621 L kPa/K mol) x (25 + 273 K)) =  

n = (472 kpaL/ 2477.7

n= 0.19 mol  

PV = nRT  

P = nRT / V  

After mixing:

P = (1.29 mol + 0.19 mol) x (8.3144621 L kPa/K mol) x (25+ 273 K) / (4 L + 10 L)  

    = (1.48) x (2477.7 / (14 L)    

    = 261.92 kPa total pressure  

so total pressure is = 261.92 kPa

a)

(261 kPa) x (0.19 mol Ar) / (1.29 + 0.19mol) =    

= 49.59/1.48 = 33.50 Kpa for Arg

b)

(261 kPa total) - (33.50 kPa Ar) = 227  kPa  for N2

partial pressure of Ar gas=  33.50 Kpa

partial pressure of N2 gas = 227  kPa

Answer: The enthalpy of formation of accetylene is, 226.2 kJ/mol

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The chemical equation for the combustion of acetylene follows:

(1) [tex]C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)[/tex]

[tex]\Delta H^o_{rxn}=-1299kJ/mol[/tex]

The formation of [tex]CO_2[/tex] will be,

(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]    [tex]\Delta H_f_{(CO_2)}=-393.5kJ/mol[/tex]

The formation of [tex]H_2O[/tex] will be,

(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex]    [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]

The formation of [tex]C_2H_2[/tex] will be,

(4) [tex]2C(s)+H_2(g)\rightarrow C_2H_2(g)[/tex]    [tex]\Delta H_f_{(C_2H_2)}=?[/tex]

Now we are reversing equation 1, multiplying equation 2 by 2 and then adding equation 1, 2 and 3, we get:

Reaction (1) :

[tex]2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)[/tex]

[tex]\Delta H^o_{rxn}=1299kJ/mol[/tex]

Reaction (2) :

[tex]2C(s)+2O_2(g)\rightarrow 2CO_2(g)[/tex]    [tex]\Delta H_f_{(CO_2)}=2\times -393.5kJ/mol=-787kJ/mol[/tex]

Reaction (3) :

[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex]    [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]

[tex]\Delta H_f_{(C_2H_2)}=\Delta H^o_{rxn}+\Delta H_f_{(CO_2)}+\Delta H_f_{(H_2O)}[/tex]

[tex]\Delta H_f_{(C_2H_2)}=(1299kJ/mol)+(-787kJ/mol)+(-285.8kJ/mol)[/tex]

[tex]\Delta H_f_{(C_2H_2)}=226.2kJ/mol[/tex]

Therefore, the enthalpy of formation of accetylene is, 226.2 kJ/mol

To calculate the enthalpy of formation of acetylene, we can use the enthalpy change of the combustion reaction and the enthalpies of formation of CO2 and H2O. By rearranging the equation and applying Hess's law, we can determine the enthalpy of formation of acetylene.

The enthalpy of formation of a compound can be calculated using Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken. To calculate the enthalpy of formation of acetylene (C2H2), we can use the equation:

C2H2 (g) + (5/2)O2 (g) → CO2 (g) + H2O (l)

The enthalpy change of this reaction is -1299 kJ/mol. By rearranging the equation, we can compare it to the standard formation equations for CO2 (g) and H2O (l) to determine the enthalpy of formation of acetylene.

The enthalpy of formation of acetylene is:

ΔHf(C2H2) = -1299 kJ/mol - [(-393.5 kJ/mol) + (-285.8 kJ/mol)]

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Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

[tex]ka = \frac{[H_3O^+][X^-]}{[HX]}[/tex]

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

[tex][H_3O^+]} = 10^{-2,40}[/tex]

[H₃O⁺] = 0,004M

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = 0,085M

[tex]ka = \frac{[0,004][0,004]}{[0,085]}[/tex]

ka = 1,88x10⁻⁴

And pka = 3,73

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

pkb = 10,27

I hope it helps!

The correct answers are:

a. The pKa of X-281 is 2.40.

b. The pKb of the conjugate base of X-281 is 11.60.

a. To find the pKa of X-281, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the concentration of the acid and its conjugate base. For a monoprotic weak acid HA, the equation is:

[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]

Given that the pH of the solution is 2.40 and the concentration of X-281 (HA) is 0.089 M, we can assume that the concentration of the conjugate base A^- is negligible compared to the concentration of the acid HA because the pH is close to the pKa. This means that the ratio[tex]\([\text{A}^-] / [\text{HA}]\)[/tex]is approximately zero, and thus the log term is approximately zero. Therefore, the pH is approximately equal to the pKa of the acid:

[tex]\[ \text{pH} \approx \text{pKa} \] \[ \text{pKa} \approx 2.40 \][/tex]

b. At 25°C, the sum of pKa and pKb for any conjugate acid-base pair is 14.00, which is the pKw of water:

[tex]\[ \text{pKa} + \text{pKb} = 14.00 \][/tex]

We have already determined that the pKa of X-281 is 2.40. Therefore, we can solve for the pKb of the conjugate base of X-281:

[tex]\[ 2.40 + \text{pKb} = 14.00 \] \[ \text{pKb} = 14.00 - 2.40 \] \[ \text{pKb} = 11.60 \][/tex]

Thus, the pKb of the conjugate base of X-281 is 11.60.

Answer:

0.094 mols of CO2 are present in 2.09 L.

0.132 mols of NH3 are present in 3.08 L

The moles of ammonium chloride produced are 0.094

Explanation:

First of all, think the reaction

NH3 + 2H2O + NaCl + CO2 → NH4Cl + NaOH + H2CO3

To get the moles with volume, you must need density

Density = mass / volume

Density CO2 = 0,001976 g/ml

Density NH3 =  0,00073 g/ml

Volume of CO2 = 2.09L

2.09 L . 1000 = 2090 mL

0,001976 g/ml = mass / 2090 mL

4.13 g = mass

Molar weight = 44 g/m

mass / molar weight = 4.13 g / 44 g/m = 0.094 moles

Volume of NH3 = 3.08 L

3.08 L . 1000 = 3080 mL

0,00073 g/ml = mass/ 3080 mL

2.25 g = mass

Molar weight = 17 g/m

mass / molar weight = mols → 2.25 g / 17g/m = 0.132 mols

In my reagents, the least amount I have is CO2. This is my limiting reactant. Take account the reaction.

1 mol of NH3 reacts with 1 mol of CO2, so as I have 0.132 moles of NH3 I need 0.132 moles of CO2; I only have 0.094.

Relation between CO2 and NH4Cl is 1:1 so, 1 mol of CO2 is needed to produce 1 mol e chloride, so 0.094 mols are needed to produce the same amount of chloride.

Answer:

Explanation:

The balanced chemical reaction is:

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_1+p_2[/tex]

As the hydrogen is collected over water, the total pressure will be sum of pressure of water and pressure of dry hydrogen.

[tex]p_{total}=p_{H_2}+p_{H_2O}[/tex]

[tex]p_{total}[/tex] = 742.5 mm Hg

[tex]p_{H_2}[/tex] = ?

[tex]p_{H_2O}[/tex] = 23.8 mm Hg

Putting in the values:

[tex]742.5=p_{H_2}+23.8[/tex]

[tex]742.5-23.8=p_{H_2}[/tex]

[tex]p_{H_2}=718.7mmHg[/tex]

Thus the pressure of dry hydrogen gas is 718.7 mm Hg

Answer:

Nitrogen triiodide is NI3.

Explanation:

Answer:

there are three Fluorine atom. This is because Nitrogen give 3 electron away but fluorine can only take 1. So nitrogen gives 1 neutron to 3 fluorine each which makes it a stable compound

Answer : The number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.

Explanation :

Molecule : It is defined as the smallest particle in an element or a compound which has chemical properties of that element or a compound.

Molecules are made up of atoms and the atoms are bonded by the sharing of electrons.

The given molecules is, [tex]NF_3[/tex]

The nitrogen trifluoride molecule is made up of two elements which are nitrogen and fluorine.

In the given molecules [tex]NF_3[/tex], there are 1 nitrogen atom and there are 3 fluorine atoms.

Hence, the number of atoms of fluoride present in one molecule of [tex]NF_3[/tex] are, 3.

Answer:

sulfur oxides

Explanation:

Answer:

a) 189.7 kPa

b) 45.4 kPa

c) 235 kPa

Explanation:

After the mixture, the temperature must remain the same, because both gases are at 17°C. Besides, the gases will not react because both nitrogen and argon are non-reactive. So, the partial pressure can be calculated by Boyle's Law:

P1*V1 = P2*V2

Where P is the pressure, V is the volume, 1 is the initial state, and 2 the final state.

When the valve is opened, the gases mix and occupy all the two vases. So, V2 = 13 L.

a) For nitrogen, P1 = 822 kPa, V1 = 3L, V2 = 13 L

822*3 = P2*13

P2 = 189.7 kPa

b) For argon, P1 = 59 kPa, V1 = 10 L, V2 = 13 L

59*10 = P2*13

P2 = 45.4 kPa

c) By Dalton's law, the total pressure of a gas mixture is the sum of the partial pressures of the components, so:

P = 189.7 + 45.4

P = 235 kPa

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the energy available to do work, of a system for a constant-temperature process is:

[tex]ΔG = ΔH - TΔS[/tex]

spontaneous in the opposite direction.

If both ΔH and ΔS are positive, then ΔG will be negative only when the TΔS  term is greater in magnitude than ΔH. This condition is met when T is large.

Answer:

c) albumin inside

Explanation:

Let us see look closer at each answer and see why it is wrong/correct.

a) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside

b) There cannot be water inside only because it would mean starch and albumin diffused out and they cannot pass through the bag

c) This is correct. The bag isn't permeable to albumin so it stays inside

d) Albumin cannot be outside the bag because the bag isn't permeable to albumin. The albumin stays inside

e) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside

After dialysis, albumin - a colloid, will remain inside the dialyzing bag because the dialysis membrane is designed to let only small molecules and ions pass through, not large colloidal particles. Starch, another colloid, will remain inside for the same reason. Water will reach equilibrium, moving in and out of the membrane.

The question relates to the process of dialysis, which is used to separate different substances in a mixture based upon their size. An aquatic mixture containing both colloids like starch and albumin and small molecules like NaCl (sodium chloride) and glucose is placed in a dialyzing bag to see where the substances will be after the process is complete. A characteristic of dialysis is that the dialyzing membrane allows small molecules and ions to pass through, while retaining larger colloidal particles.

Referring to the given options, the correct descriptions after dialysis are as follows:

Therefore, the accurate statement concerning the location of the substances post-dialysis is

c) albumin inside

Answer:

F centripetal force (tension) = 275.9  N

Explanation:

Given data:

Mass = 1.50 kg

Radius = 0.520 m

Velocity of ball = 9.78 m/s

Tension = ?

Solution:

F centripetal force (tension) =  m.v² / R

F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m

F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m

F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m

F centripetal force (tension) = 275.9  N

Answer:

1)20. ML at 100.°C

Explanation:

Average kinetic energy is not related with volume.  so increase in temperature will also increase kinetic energy. so this increase means the highest temperature will be a right answer.

Final answer:

The water sample with the highest average kinetic energy is the 20 mL of water at 100.0°C, as average kinetic energy is directly proportional to temperature.

Explanation:

The molecules with the highest average kinetic energy are in the sample of water at the highest temperature. Kinetic energy is directly proportional to the absolute temperature of the substance, meaning that as temperature increases, kinetic energy also increases. Therefore, ignoring the volume of the water samples, the rank from highest kinetic energy to lowest based on temperature would be:

The sample at 100.0°C holds the highest average kinetic energy because it has the highest temperature.

Photophosphorylation of sugar moiety produces ATP  (adenosine triphosphate) which stores chemical energy that can be used for the Calvin cycle.

The Calvin cycle, often known as the Calvin-Benson cycle after its discoverers, is a series of chemical processes that collect carbon dioxide.

It is also referred to as the C3 cycle alone. C3 plants are defined as those that only fix carbon through the Calvin cycle. In the stroma of chloroplasts, carbon dioxide diffuses and mixes with the five-carbon sugar ribulose 1, 5-biphosphate (RuBP).

Rubisco, a large molecule that may be the most prevalent organic compound on Earth, is the name of the enzyme that catalyzes this reaction.

The three-carbon chemical 3-PGA is changed into another three-carbon compound termed G3P by ATP and NADPH using their stored energy. A reduction reaction is the name for this kind of reaction.

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To find the final pressure of the gas mixture after transferring the gases, we can use the ideal gas law equation. The final pressure is calculated based on the total number of moles of gas and the final volume of the container. The final pressure of the gas mixture is 0.229 atm.

To find the final pressure of the gas mixture after the transfer, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We can calculate the number of moles of N2 gas using the information given:

n = PV/RT = (5.23 atm)(1.20 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 0.287 mol

Similarly, we can calculate the number of moles of O2 gas:

n = PV/RT = (5.21 atm)(5.10 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 1.05 mol

To find the final pressure, we need to find the total number of moles of gas and the final volume of the container:

Total moles of gas = moles of N2 + moles of O2 = 0.287 mol + 1.05 mol = 1.34 mol

Final volume of the container = sum of the initial volumes = 1.20 L + 14.5 L = 15.7 L

Now we can use the ideal gas law again to calculate the final pressure:

P = nRT/V = (1.34 mol)(0.0821 atm·L/mol·K)(20 + 273 K)/(15.7 L) = 0.229 atm

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Answer:

The false statement regarding an exothermic reaction is: A. the products have a higher enthalpy than reactants

Explanation:

An exothermic reaction is a type of chemical reaction that involves the release of energy from the system to the surroundings. Thus increasing the temperature of the surroundings.

In this reaction, the enthalpy or energy of the reactants is greater than the enthalpy or energy of the products. ([tex]\Delta H_{f} (Products) < \Delta H_{f} (reactants)[/tex])

As the enthalpy change of a reaction: [tex]\Delta H_{r} = \sum \Delta H_{f} (Products) - \sum \Delta H_{f} (reactants)[/tex]

Therefore, the enthalpy change for an exothermic reaction is negative ([tex]\Delta H_{r} < 0[/tex])

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

[tex]\frac{1}{2}[/tex]Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is [tex]\frac{1134.9}{4}[/tex] = 283.725 kJ ⋅ mol − 1

The  average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule is 283.725 kJ ⋅ mol − 1

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3)

⇒ 2Br2(g) + C(s)

⇒ 4 Br(g) + C(g) , Δ H ∘

= 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4)

⇒ 4 Br(g) + C(g)

⇒ CBr4(g) , Δ H ∘

= -1134.9 kJ ⋅ mol − 1

so,

average bond enthalpy is  = 283.725 kJ ⋅ mol − 1

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