Look up the density of n-butyl chloride (1-chlorobutane). Assume that this alkyl halide was prepared instead of the bromide. Decide whether the alkyl chloride would appear as the upper or the lower phase at each stage of the separation procedure: after the reflux, after the addition of water, and after the addition of sodium bicarbonate.

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%.

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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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:

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:

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.

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:

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:

The concentration of acetic acid in the vinegar is 7,324 (%V/V)

Explanation:

The titration equation of acetic acid with NaOH is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

The moles required were:

1,024M×0,02500L = 0,02560 moles NaOH. These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:

0,02560 moles CH₃COOH×[tex]\frac{60,052g}{1mol}[/tex]= 1,537g of CH₃COOH

As density is 1,01g/mL:

1,537g CH₃COOH×[tex]\frac{1mL}{1,01g}[/tex]= 1,522mL of CH₃COOH

As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:

[tex]\frac{1,522mLCH_{3}COOH}{20,78mL}[/tex]×100= 7,324 (%V/V)

I hope it helps!

Considering the definition, molarity, density and volume percentage, the concentration of acetic acid in the vinegar is 7.32% (v/v).

The balanced reaction is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

Molarity is the number of moles of solute that are dissolved in a given volume.  Molarity is determined by:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

In this case you know for NaOH:

So, the moles required are calculated as:

[tex]1.024 M=\frac{number of moles of solute}{0.025 L}[/tex]

Solving:

1.024 M× 0.025 L= number of moles of solute

0.0256 moles= number of moles of solute

By observing the balanced reaction, you can see that, since the ratio of CH₃COOH to NaOH is 1: 1, these moles are equivalent to the number of moles of CH₃COOH.

On the other hand, since the molar mass of CH₃COOH is 60.052 g/mol, this is the mass that contains one mole of the compound, the mass in 0.0256 moles of CH₃COOH is:

[tex]0.0256 molesx\frac{60.052 grams}{1 mole} =[/tex] 1.537 grams

Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

[tex]density=\frac{mass}{volume}[/tex]

In this case, being the density 1.01 [tex]\frac{g}{mL}[/tex], the volume can be calculated as:

[tex]1.01 \frac{g}{mL} =\frac{1.537 grams}{volume}[/tex]

Solving:

1.01 [tex]\frac{g}{mL}[/tex]× volume= 1.537 grams

[tex]volume=\frac{1.537 grams}{1.01 \frac{g}{mL} }[/tex]

volume= 1.522 mL of CH₃COOH

Volume Percentage (%v/v) is a measure of concentration that indicates the volume of solute per 100 volume units of the solution. In other words, the volume percent of a component in the solution is defined as the ratio of the volume of the component to the volume of the solution, expressed as a percentage.

The volume percentage of a solution is determined by the following expression:

[tex]volume percentage=\frac{volume of solute}{volume of solution}x100[/tex]

Then, as volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:

[tex]volume percentage=\frac{1.522 mL}{20.78 mL}x100[/tex]

Solving:

volume percentage= 7.32%

Finally, the concentration of acetic acid in the vinegar is 7.32% (v/v).

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Answer:

Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg

Explanation:

The vapor pressure can be calculated by using Clausius‐Clapeyron equation.

ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)

Where

p₁ is the vapor pressure at T₁ (Initial Temperature)

p₂ is the vapor pressure at T₂ (final Temperature)

ΔHvap is molar heat of vaporization of the substance

R is the real gas constant = 8.314 x 10⁻³ kJ/mol.K

Data Given:

p₁ = ?

p₂ = 394.98 mm Hg

T₁ = 1.84×10³ K

T₂ = 1.81×10³ K

ΔHvap = 115 kJ/mol

Put the values in the Clausius‐Clapeyron equation

                             ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)

ln(p₁/394.98 mm Hg) = (-115 kJ/mol / 8.314 x 10⁻³ kJ/mol.K)(1/1.84×10³ K- 1/1.81×10³ K)

ln(p₁ /394.98 mm Hg) = (- 13.8321 x 10³)(-0.5519)

ln(p₁ /394.98 mm Hg) = 7633.936

ln cancel out by E, e is raise to a power x

So,

p₁/394.98 mm Hg = e^7633.936

p₁/ 394.98 mm Hg =    20.75 x 10³

p₁ = 20.75 x 10³ x 394.98 mm Hg

p₁ = 8.19 x 10⁴ mm Hg

Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg

Answer:

338 mm Hg

Explanation:

[tex]ln(\frac{p2}{400mmHg} )=-115*\frac{1}{0.008314} *(\frac{1}{1.8*10^3} -\frac{1}{1.84*10^3}[/tex])

ln(p2/400)=-0.16705

p2/400= e^-0.16705=0.84615

p2=0.84615*400=338

Note that T2 is lower than T1 and that the vapor pressure decreases as the temperature decreases.

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:

Nitrogen triiodide is NI3.

Explanation:

Answer:

K2Cr2O7 + HI + HClO4  → Cr(ClO4)3  +  KClO4  + I2 + H2O

                                          ↓

8HClO4 + K2Cr2O7 + 6HI  →  3I₂ +  2Cr(ClO4)3 + 2KClO4  +  7H2O

Explanation:

I put and example of what you said.

Potassium dichromate and iodide acid react with perchloric acid to generate chromium perchlorate, potassium chlorate, iodine and water.

First of all think all the oxidation number of each element. One ox. number will increase and the other decrease, so those will be our half reactions (one of reduction, the other of oxidation).

In the case above, I in HI acts with -1 and I2 has 0 (all elements in ground state has 0 as oxidation number). (Increase) - Oxidation

Cr in K2Cr2O7 acts with +6, in Cr(ClO4)3 is 3+ (Decrease) -  Reduction

So, the first half reaction is:

2I⁻  →  I₂ + 2e⁻           (OXIDATION)

I have to put 2 iodides to ballance, so the total charge is 2-. It has to release 2 electrons.

Cr2O7²⁻  →  2Cr³⁺

In products side, I have to add 2 chromes but we don't have the charges ballanced. At the main equation, we have acids so this redox occurs in an acidic medium. In the acidic medium we add water, the same as oxygen we have, so:

Cr2O7²⁻  →  2Cr³⁺  +  7H2O

and in reactant side, we add protons the same as hydrogen, we have, in this case like this

14H⁺  +  Cr2O7²⁻  →  2Cr³⁺  +  7H2O

finally, we add the electrons. Chrome to decrease +6 to +3 had to lose 3 electrons but, we have 2 Cr, so 6 in total. These are final the 2 half reaction

14H⁺  +  Cr2O7²⁻  + 6e-  →  2Cr³⁺  +  7H2O  (REDUCTION)

2I⁻  →  I₂ + 2e⁻ (OXIDATION)

Electrons are not ballanced, we have to multiply by a minimum common multiple. For 2 and 6, this number is 12 so:

(14H⁺  +  Cr2O7²⁻  + 6e-  →  2Cr³⁺  +  7H2O) .2

(2I⁻  →  I₂ + 2e⁻) .6

Afterwards, we can sum the reactions:

28H⁺ + 2Cr2O7²⁻ + 12e- + 12I⁻  →  6I₂ + 12e⁻ + 4Cr³⁺  +  14H2O

As we have 12e- in both sides, we cancel them

28H⁺ + 2Cr2O7²⁻ + 12I⁻  →  6I₂ +  4Cr³⁺  +  14H2O (still balanced)

Look that all the stoichiometry is even, so we can /2.

14H⁺ + Cr2O7²⁻ + 6I⁻  →  3I₂ +  2Cr³⁺  +  7H2O

So the final reaction is:

8HClO4 + K2Cr2O7 + 6HI  →  3I₂ +  2Cr(ClO4)3 + 2KClO4  +  7H2O

We have in total 14H+, so 6 protons are for HI and 8 for the HClO4.

Answer:

a) FeCl2, FeCl3, MgCl2, KClO2.

KClO2 --> K+ + ClO2-; ClO2- will hydrolyse to form HClO +OH-

Mg+2, Fe+2 and Fe+3 ions will form acidic solutions, since theyfom slightly amount of

Mg+2 + 2H2O <-> Mg(OH)2 + 2H+

Fe+2 + 2H2O <-> Fe(OH)2 + 2H+

Fe+3 + 3H2O <-> Fe(OH)3 + 3H+

Therefore;

decreasing pH is high pH to low pH:

KClO2 > MgCl2 > FeCl2 > FeCl3

b) NH4Br, NaBrO2, NaBr, NaClO2.

NH4Br is acidic, forms NH4+ and NH4+ dnates H+ to form NH3 andH+

NaBrO2 is basic, forms Na+ + BrO2- then H2O + BrO2- HBrO2

NABr is neutral, NaClO2 is basics, forms Na+ + ClO2-then H2O + ClO2- HClO2

decreasig pH:

NaClO2 > NaBrO2 > NaBr >NH4Br

Note that HClO2 is stronger acid than HBrO2, therefore, expectmore HBrO2 formation

NaBrO2 > NaClO2 > NaBr >NH4Br

The order in (a) is; FeCl2, FeCl3, MgCl2, KClO2. The order in (b) is; NaBrO2 > NaClO2 > NaBr >NH4Br

The term pH refers to the degree of acidity or alkalinity of a solution. We must recall that salts are solvated in solution. The pH of the solution after solvation depends on the ions produced by the salt in solution.

Since FeCl2  yields a basic solution, then it has the highest pH. Similarly, KClO2 yields an acid solution hence it has the lowest pH. The order in (a) is; FeCl2, FeCl3, MgCl2, KClO2. The order in (b) is; NaBrO2 > NaClO2 > NaBr >NH4Br

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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:

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:

sulfur oxides

Explanation:

Answer:

rate = kxyz

Explanation:

It is worth knowing that the rate low can only be determined by experimentation only not by just balancing equations. So here we are told that all the reactants x , y and z are all first order. This is important because we use this as exponents. That is why the exponents of all the reactants will be 1.

rate = kxyz

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:

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:

2

Explanation:

There are four quantum numbers:

Principal quantum number (n)

Azimuthal quantum number (l)

Magnetic quantum number (ml)

Principal quantum number (n)

It tell about the energy levels.  It is designated by n.

For example,

If n =2

It means there are two energy level present.

Azimuthal quantum number (l)

The azimuthal quantum number describe the shape of orbitals. Its value for s, p, d, f... are 0, 1, 2, 3. For l=3

(n-1)

4-1 = 3

it means principle quantum is 4 and electron is present in f subshell.

Magnetic quantum number (ml)

It describe the orientation of orbitals. Its values are -l to +l. For l=3  the ml will be -3 -2 -1 0 +1 +2 +3.

Spin quantum number (ms)

The spin quantum  number tells the spin of electron either its clock wise (+1/2) or anti clock wise (-1/2).

If the electron is added in full empty orbital its spin will be +1/2 because it occupy full empty. If electron is already present and another electron is added then its spin will be -1/2.

The reaction of 30 g of C(s) in the given equation absorbs approximately 119.95 kJ of heat.

In chemistry, the amount of heat absorbed or released during a chemical reaction is given by the product of the mole ratio and the enthalpy change (ΔH°). Looking at the balanced equation: 5 C(s) + 2 SO2(g) → CS2(l) + 4 CO(g) ΔH° = +239.9 kJ, the reaction of 5 moles of C(s) absorbs 239.9 kJ of heat. However, we only have 30 g of C(s) - or 30/12.01 (since the molar mass of carbon is 12.01 g/mol) equals roughly 2.5 mol of C(s).

To find out how much heat 2.5 mol of C(s) absorbs, we use a proportional relationship: (2.5 moles C/5 moles C) * 239.9 kJ = 119.95 kJ. Therefore, when 30 g of C(s) reacts in the presence of excess SO2(g) to produce CS2(l) and CO(g), it absorbs approximately 119.95 kJ of heat.

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The heat absorbed when 30.00 g of C(s) reacts in the given chemical reaction is approximately 119.95 kJ. This value is determined based on the moles of carbon involved and the enthalpy change given for the reaction.

First, we need to find the number of moles of carbon (C) involved in the reaction:

The balanced chemical equation shows that 5 moles of C are needed for one reaction cycle, which absorbs 239.9 kJ of heat.

To find the heat absorbed for 2.50 moles of C, we use the proportion:

Heat absorbed = (2.50 moles / 5 moles) × 239.9 kJ ≈ 119.95 kJ

Therefore, the heat absorbed when 30.00 g of C(s) reacts is ≈ 119.95 kJ.

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:

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

Molarity and mole fraction of acetone in solution can be calculated by using the densities and molality. Molarity is derived by dividing number of moles of acetone by volume of solution and mole fraction is derived by dividing moles of acetone by total moles in solution.

To calculate the molarity and the mole fraction of acetone in a 1.17 m solution of acetone (CH3COCH3) in ethanol (C2H5OH), we need to first convert the molal concentration (moles of solute per kilogram of solvent) to molar concentration (moles of solute per liter of solution).

Using given data, density of acetone is 0.788 g/cm3 and ethanol 0.789 g/cm3. As volumes are additive, volume of solution = volume of solute + volume of solvent. With this data, we can find number of moles of acetone and ethanol respectively.

For molarity (M), we divide number of moles of solute (acetone) by volume of solution in liters. For mole fraction (χ), we divide number of moles of solute (acetone) by total number of moles in solution (acetone + ethanol moles).

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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:

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.

Learn more about pH here:

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Answer:

Hybridization

Explanation: -

Hybridization occurs when atomic orbitals mix to form a new atomic orbital

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:

24 Lt/mol

Explanation:

Though we have many data here (such as molar mass of Mg, water vapor pressure, etc), we need to focus on data for H₂, which will help us to obtain the molar volume of this gas

Statement refers the following data for H₂:

V = 82.1 ml = 0.082 Lt

T = 22°C = 295 K

P atm = 766.7 mm Hg = 1.0089 atm (which is, the pressure for H₂ before being collected in water)

If we consider H₂ behaves as an ideal gas:

PV = nRT

Then we can move some terms from this ecuation :

V/n = RT/P so we can obtain the relation between V (volume) and n (N° of moles) for H₂, which is the experimental valur for the molar volume

Considering R = 0.082 Lt*atm/K*mol:

V/n = [(0.082 Lt*atm/K*mol)x295 K]/1.0089 atm

V/n = 23.97 Lt/mol (molar volume at this experiment conditions)