a buffer is made by combining 20.0 ml 0.250 m nh4cl with 30.0 ml 0.250 m nh3. a. calculate the ph of the buffer. 14 b. calculate the ph of the buffer after 1.00 ml of 6.00 m hcl is added and equilibrium is re- established. c. calculate the ph of the buffer after 1.00 ml of 6.00 m naoh is added to a fresh sample of the buffer and equilibrium is re-established.

Charles's Law-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

Where:-

As per question, we are given that -

Now that we are given all the required values, so we can put them into the formula and solve for V₂ :-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= \dfrac{V_1}{T_1}\times T_2\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= \dfrac{5.31}{200}\times 300\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2 = 0.02655\times 300\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf V_2=7.965 \\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf\underline{ V_2 = 7.965\: L}\\[/tex]

Therefore, the final volume of 5.31L of an ideal gas when heated from 200 K to 300 K at constant pressure will be 7.965 L.

The pH of the new solution after adding the HCl is 2.82.

How to calculate pH after adding the HCl ?

To calculate the new pH of the buffer solution after adding the HCl, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and base.

First, we need to find the concentrations of the acid and its conjugate base in the buffer solution. We know that the total volume of the buffer solution is 1.00 L, and the number of moles of HX and NaX are equal, so each one has a concentration of:

C = n/V = 0.50 mol / 1.00 L = 0.50 M

Since HX is an acid, it will donate a proton (H+) to form its conjugate base (X-). Therefore, the acid and base concentrations in the buffer solution are:

[HX] = 0.50 M

[X-] = 0.50 M

The pKa of HX can be calculated from the pH and the acid dissociation constant (Ka):

pH = pKa + log([X-]/[HX])

4.925 = pKa + log(0.50/0.50)

4.925 = pKa

Now we can use the Henderson-Hasselbalch equation to find the pH of the buffer solution after adding the HCl:

pH = pKa + log([X-]/[HX])

pH = 4.925 + log(0.50/0.50)

pH = 4.925

This means that the pH of the original buffer solution is unchanged by the addition of the HCl.

However, if we want to calculate the pH of the new solution after adding the HCl, we need to use the stoichiometry of the reaction:

HX + HCl -> H2O + XCl

We know that 25.0 mL of 2.0 M HCl were added to the buffer solution, so the number of moles of HCl added is:

n(HCl) = C x V = 2.0 mol/L x 0.025 L = 0.050 mol

Since HX and HCl react in a 1:1 ratio, the number of moles of HX consumed is also 0.050 mol. This leaves us with:

n(HX) = 0.50 mol - 0.050 mol = 0.450 mol

The new volume of the solution is 400 mL + 25.0 mL = 0.425 L. Therefore, the new concentration of HX is:

[HX] = n(HX) / V = 0.450 mol / 0.425 L = 1.06 M

The new concentration of X- can be calculated from the buffer equation:

Ka = [H+][X-] / [HX]

Ka = 10⁻pKa = 10⁻⁴.925 = 7.09 x 10⁻⁵

[H+] = Ka [HX] / [X-] = 7.09 x 10⁻⁵ x 1.06 M / 0.50 M = 0.151 mM

[pH = -log[H+] = 2.82]

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We can use the formula for heat released:

Q = mcΔT

where Q is the heat released, m is the mass of the sample, c is the specific heat of the material, and ΔT is the change in temperature.

Plugging in the given values, we have:

113.8 J = m(0.240 J/g°C)(-45.0°C)

Simplifying:

m = 113.8 J / (0.240 J/g°C * -45.0°C)

m = 106.5 g

Therefore, the mass of the silver sample is 106.5 g.

The substance that will give the solution with the lowest electrical conductivity when dissolved in water at equal molar concentrations is C6H12O6 (glucose). The correct option is D.

a. CaCl2: Calcium chloride is an ionic compound that dissociates into ions (Ca2+ and 2Cl-) when dissolved in water, which increases electrical conductivity.

b. HNO3: Nitric acid is a strong acid that dissociates completely into ions (H+ and NO3-) when dissolved in water, which also increases electrical conductivity.

c. NH3: Ammonia is a weak base that partially forms ions (NH4+ and OH-) when dissolved in water, contributing to some electrical conductivity.

d. C6H12O6: Glucose is a covalent compound that does not dissociate into ions when dissolved in water, so it will not increase electrical conductivity.

e. CO2: Carbon dioxide is a covalent compound that dissolves in water to form a weak acid (H2CO3), which partially dissociates into ions (H+ and HCO3-), contributing to some electrical conductivity.

Since glucose (C6H12O6) does not dissociate into ions, it results in the lowest electrical conductivity among the listed substances.

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The dependent variable among the following components in the experiment is temperature as it is dependent on the others.

The temperature will be dependent on the other variables, such as the oxygen, vinegar, and steel wood. This experiment is designed to measure how the temperature changes when the other variables are altered. Temperature is the outcome that will be affected by changes in the independent variables, and changes in the dependent variable will provide insight into the effect of the independent variables. For example, when the oxygen is increased, the temperature may increase as well. Similarly when the vinegar is decreased, the temperature may decrease and the steel wood may also affect the temperature, though this will depend on the specific experiment.

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complete question: Julianne and Thabo are investigating temperature changes in chemical reactions. They have heard that vinegar removes the protective coating from steel wool, allowing it to rust. When iron combines with oxygen, heat is released. They use a Styrofoam cup with a lid to investigate this report.

Julianne records the initial temperature by wrapping the steel wool around the thermometer and placing it inside the empty Styrofoam cup. She records the temperature after 2 min. Thabo soaks the piece of steel wool in vinegar for 1 min and then squeezes out the excess vinegar. He wraps the wool around the thermometer, placing it back in the Styrofoam cup and seals the lid. He records the temperature after 15 min. The temperature has increased from 22 °C to 26 °C.

what is the dependent variable in this experiment?

A.) oxygen

B.) vinegar

C.) steel wood

D.) temperature

To find the pH of 0.33 M HCOOK, we need to first determine the concentration of HCOO- ions in the solution. When HCOOK dissolves in water, it dissociates to form HCOO- and K+ ions.

Since HCOOK is a salt of a weak acid, it will undergo hydrolysis in water, and the HCOO- ions will react with water to form HCOOH and OH- ions.

The balanced equation for this reaction is: HCOO- (aq) + H2O (l) ⇌ HCOOH (aq) + OH- (aq)

Using the given K₂ value for HCOOH, we can calculate the equilibrium concentrations of HCOO- and HCOOH:

K₂ = [HCOOH][OH-] / [HCOO-]

1.78 x 10⁻⁴ = [HCOOH][OH-] / [0.33]

[HCOOH][OH-] = 5.874 x 10⁻⁵

Assuming x is the concentration of HCOOH and OH- formed, we can set up an ICE table:

HCOO⁻ (aq) + H2O (aq) = HCOOH (aq) + OH^- (aq)

I 0.33 M 0 0

C - x x x

E 0.33 - x x x

Substituting the equilibrium concentrations into the K₂ expression, we get:

1.78 x 10⁻⁴ = x⁻² / (0.33 - x)

Since x is small compared to 0.33, we can approximate (0.33 - x) as 0.33:

1.78 x 10⁻⁴ = x⁻² / 0.33

Solving for x, we get x = 2.49 x 10⁻³ M

So, [HCOOH] = [OH-] = 2.49 x 10⁻³ M

To calculate the pH, we can use the equation: pH = 14 - pOH

pOH = -log[OH-] = -log(2.49 x 10⁻³) = 2.60

Therefore, pH = 14 - 2.60 = 11.40 (rounded to two significant figures)

Hence, the pH of 0.33 M HCOOK is 11.40.

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Humic substances are recalcitrant organic molecules. Option E is correct.

Humic substances are complex organic compounds that are formed by the decomposition of dead plant and animal matter. They are found in soil, water, and sediments, and are an important component of organic matter in the environment. Humic substances are recalcitrant organic molecules that are resistant to further decomposition, and play a critical role in nutrient cycling, water retention, and soil structure.

They are composed of three main fractions: humic acid, fulvic acid, and humin. Humic acid is the largest and most complex fraction, and is insoluble in water at acidic pH. Fulvic acid is the smallest and most soluble fraction, and is soluble in water at all pH values.

Humin is the fraction that remains after the extraction of humic and fulvic acids, and is relatively insoluble in both water and organic solvents. Humic substances are known to have a wide range of beneficial properties, including soil improvement, plant growth enhancement, and water quality improvement.

Hence, E. is the correct option.

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--The given question is incomplete, the complete question is

"What are humic substances? group of answer choices A) carbonates and phosphates B) glucose molecules C) clay particles D) silt and sand E) recalcitrant organic molecules."--

The pH of the buffer solution after 0.024 mol of Ba(OH)₂ are added is approximately 4.90.

To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the weak acid and its conjugate base;

pH = pKa + log([base]/[acid])

where pKa will be the dissociation constant of the weak acid, [base] is the concentration of the conjugate base, as well as [acid] is the concentration of the weak acid.

From the given information, we can determine the pKa of the weak acid;

pH = 8.87 = pKa + log([base]/[acid])

pKa = 8.87 - log([base]/[acid])

pKa = 4.83

We can also determine the initial concentrations of the weak acid and its conjugate base.

[base] = 0.49 M

[acid] = 0.45 M

Now, we can use the balanced chemical equation for the reaction of Ba(OH)₂ with BH;

Ba(OH)₂ + 2BH → BaB₂ + 2H₂O

The moles of BH that react with the added Ba(OH)₂ is equal to half the moles of Ba(OH)₂ added, because the stoichiometry of the reaction is 2:1. Therefore, moles of BH consumed = 0.024 mol / 2 = 0.012 mol

The volume of the solution is given as 0.62 L, so the new concentration of BH is;

[acid] = (0.45 M x 0.62 L - 0.012 mol) / 0.62 L

[acid] = 0.441 M

The new concentration of B can be determined from the conservation of mass equation:

[base] + [BaB2] = constant

[base] + 0.012 mol / 0.62 L = constant

[base] = 0.49 M - 0.012 mol / 0.62 L

[base] = 0.470 M

Now, we can use the Henderson-Hasselbalch equation to determine the new pH of the buffer solution

pH = pKa + log([base]/[acid])

pH = 4.83 + log(0.470/0.441)

pH = 4.90

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The energy that would be released for the formation of 25 moles of liquid water is 6130kJ.

Given the number of moles of liquid water = 25

Let the energy released = E

The formation of 25 moles of liquid water requires the input of energy and results in the release of energy.

This can be calculated as follows:

Energy required for formation of 25 moles of liquid water:

[tex]H_2 + 1/2O_2 -- > H_2O(l)[/tex]

[tex]H_2O (l) -- > H_2O(g)[/tex]

The enthalpy of formation of H2 = 0kJmol

The enthalpy of formation of O2 = 0kJmol

The enthalpy of formation of liquid H2O = -286kJ/mol

The enthalpy of sublimation of liquid H2O to gaseous H2O = 40.8kJ

The enthalpy of formation of gaseous H2O = -286kJ/mol + 40.8kJ = -245.2kJ/mol

For 25 moles the energy released = 25 * 245.2kJ = 6130kJ

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The balanced reaction that shows incomplete combustion is  2C4H10 + 5O2 → 4CO + 6H2O.

option D.

The balanced equation for incomplete combustion involves a reactant, usually a hydrocarbon fuel, reacting with a limited supply of oxygen to produce carbon monoxide (CO) instead of carbon dioxide (CO2) and water (H2O).

Out of the options given, the balanced equation that shows incomplete combustion is:

D. 2C4H10 + 5O2 → 4CO + 6H2O

This equation shows incomplete combustion because it produces carbon monoxide (CO) instead of carbon dioxide (CO2). The equation is balanced because there are equal numbers of atoms of each element on both the reactant and product sides of the equation.

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To find the volume of ethane , we can use the Ideal gas law. Which states -

[tex] \:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{PV=nRT} \\[/tex]

Where:-

As per question, we are given that-

Now that we have all the required values, so we can put them all in the Ideal gas law formula and solve for Volume -

[tex] \:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{PV=nRT} \\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf 0.9035 \times V= 5.2643\times 0.0821 \times 779.25\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf 0.9035 \times V= 336.791\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf V= \dfrac{336.791}{0.9035}\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf V= 372.762......\\[/tex]

[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{V= 372.762 \: L}\\[/tex]

The chemical reaction between oxygen and nitrogen dioxide has the following balanced chemical equation: 2Nitrogen dioxide + Oxygen -> 2Nitrogen dioxide

Hence volume of oxygen gas measured at 0oC and 1atm, needed to burn completely 1L of propane gas under the same conditions is 5L.

According to the equation, 2 moles of Nitrogen dioxide and 1 mole of Oxygen combine to form 2 moles of Nitrogen dioxide.

We must first determine how many moles of Nitrogen dioxide 55.2 grams will produce:

mass / molar mass equals moles of Nitrogen dioxide.

Nitrogen dioxide moles are equal to 55.2 g/46.0055 g/mol.

1.200 mol Equals 1 mole of Nitrogen dioxide.

We just require half as many moles of Oxygen since 1 mole of Oxygen produces 2 moles of Nitrogen dioxide:

1.200 moles of Oxygen are equal to 0.600 moles when divided by two.

Now, we can calculate the volume of Oxygen needed using the ideal gas law:

PV = nRT

P equals pressure where= 0.890 atm

V = volume (in liters)

n = number of moles = 0.600 mol

R = gas constant = 0.08206 L·atm/mol·K

T = temperature = 10.0 + 273.15 = 283.15 K

Solving for V:

V = nRT / P

V = (0.600 mol)(0.08206 L·atm/mol·K)(283.15 K) / 0.890 atm

V = 14.2 L

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5.00 moles of Z would be produced when 2.50 moles of X are reacted with Y.

From the balanced chemical equation:

2 X + 3Y → 4 Z

We can see that the molar ratio of X to Z is 2:4 or 1:2.

Therefore, if 2.50 moles of X are reacted, we can calculate the number of moles of Z produced using this ratio:

2.50 moles X × (2 moles Z / 1 mole X) = 5.00 moles Z.

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The trans isomer (E) would be the major product if you only take into account product stability.

As per the statement "Both the e and z forms of the alkene will form in this reaction. If you only take into account product stability, which one would you expect to be the major product?", the stability of the alkene is an important factor to determine the major product of a reaction. In the E/Z system, the two highest priority groups on each carbon atom in a double bond are placed in relation to each other.

The E/Z notation is based on the stereochemistry of alkenes or cycloalkenes. If the two highest priority groups are on the same side of the double bond, it is termed as Z (zusammen, German for "together"), while if they are on the opposite side, it is termed as E (entgegen, German for "opposite").For example:If you take into account product stability, then the major product would be the one with more stability.

In general, trans isomer (E) is more stable than the cis isomer (Z) because of the steric-hindrance caused by the substituent groups attached to the double bond. The greater the degree of steric hindrance, the lower the stability of the molecule.The trans isomer (E) has a linear arrangement of the carbon atoms around the double bond, whereas the cis isomer (Z) has a bent arrangement. The linear structure of the trans isomer is energetically more favorable than the bent structure of the cis isomer because it causes less steric hindrance. Hence, trans isomer is more stable than cis isomer.

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The volume of the result is  9.03 L when 5.00 l of a gas is known to contain 0.965 mol. if the amount of gas is increased to 1.80 mol.

The best gas regulation condition, PV = nRT, relates the strain (P), volume (V), measure of substance (n), and temperature (T) of a gas. Since the temperature is held consistent in this issue, we can utilize the accompanying type of the best gas regulation:

P1V1 = n1RT and P2V2 = n2RT

where P1, V1, n1, and P2, n2, and V2 are the underlying strain, volume, and measure of substance and last tension, volume, and measure of substance, individually.

We are given that the underlying volume V1 is 5.00 L and the underlying measure of substance n1 is 0.965 mol. We are likewise given that the last measure of substance n2 is 1.80 mol. To find the last volume V2, we can revamp the best gas regulation condition and address for V2:

V2 = (n2RT)/P2

We can utilize the underlying circumstances to find the underlying strain, which is:

P1 = (n1RT)/V1

We can then utilize the last measure of substance and the underlying strain to find the last tension, which is:

P2 = (n2RT)/V1

Subbing these qualities into the situation for V2 gives:

V2 = (n2RT * V1)/(n1RT + P2V1)

We can work on this articulation by offsetting the R and T terms, and connecting the given qualities:

V2 = (1.80 mol * 5.00 L)/(0.965 mol + (1.80 mol * 5.00 L * (0.965 mol/5.00 L)))

Subsequent to rearranging, we get:

V2 = 9.03 L

Thusly, the new volume will be 9.03 L when how much gas is expanded to 1.80 mol.

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To calculate (a) the (molar) Gibbs energy of mixing and (b) the (molar) entropy of mixing when the two major components of air (nitrogen and oxygen) are mixed to form air, the mole fractions of N2 and O2 are needed.

Given that mole fractions of N2 and O2 are 0.78 and 0.22, respectively. The formula for calculating Gibbs energy of mixing and entropy of mixing is as follows.∆Gmix=∆Hmix−T∆SmixΔ G mix = Δ H mix - T Δ S mixΔSmix=−RΣxi ln xiΔ S mix = - RΣ x i ln x iWhere,ΔHmix = Enthalpy of mixing of the two gasesΔGmix = Gibbs energy of mixing of the two gasesΔSmix = Entropy of mixing of the two gasesT = TemperatureR = Gas constantxi = Mole fraction of gas i.

(a) The Gibbs energy of mixing is given as,∆Gmix=∆Hmix−T∆Smix=0.2095 kJ/mol(b) The entropy of mixing is given as,ΔSmix=−RΣxi ln xi=-0.193 J/K mol The value of Gibbs energy of mixing is positive indicating that the mixing process is not spontaneous. However, the value of entropy of mixing is negative indicating that the mixing process is spontaneous.

Therefore the giving process is spontaneous.

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Calcium carbonate (CaCO3) reacts with hydrochloric acid (HCl)  according to the following balanced chemical equation. Therefore, 400 mg of Tums (containing 40.0% CaCO3 by mass) can neutralize 31.98 mL of 0.100 M HCl.

CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

From this equation, we can see that one mole of CaCO3 reacts with two moles of HCl. Therefore, we need to calculate the number of moles of CaCO3 in 400 mg of Tums:

mass of CaCO3 = 0.4 g × 0.4 = 0.16 g

number of moles of CaCO3 = mass / molar mass = 0.16 g / 100.09 g/mol = 0.001599 mol

To neutralize this amount of CaCO3, we will need twice as many moles of HCl,  or

number of moles of HCl = 2 × 0.001599 mol = 0.003198 mol

Now, we can use the concentration of the hydrochloric acid solution (0.100 M) and the number of moles of HCl to calculate the volume of HCl required to neutralize the CaCO3:

number of moles of HCl = concentration × volume

volume = number of moles of HCl / concentration = 0.003198 mol / 0.100 mol/L = 0.03198 L

Finally, we can convert the volume to milliliters:

0.03198 L × 1000 mL/L = 31.98 mL

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The initial volume of HCl is 5ml whose initial concentration is 12M and is required to make 20.0 mL of 3.0 M HCI solution.

Given the concentration of HCl (M1) = 12.0M

Let the initial volume of HCl = V1

The final volume of HCl (V2) = 20mL = 0.02L

The final concentration of HCl (M2) = 3.0M

Molarity is a measure of concentration of a solution, expressed as moles of solute per liter of solution. It is represented as M or mol/L.

We know that M1V1 = M2V2 where molarity is constant before and after such that:

12 * V1 = 0.02 * 3

V1 = 0.005L

Hence the initial volume of HCl is 5.0mL

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[tex]M_{2} V_{2}[/tex] The volume of water needed to create a 0.330 M [tex]H_{2} SO_{4}[/tex] solution from a 1.00 M solution is 91.36 mL.

To calculate the volume of water needed to create a 0.330 M [tex]H_{2} SO_{4}[/tex] solution from a 1.00 M solution, we can use the dilution formula:

[tex]M_{1} V_{1}[/tex] = [tex]M_{2} V_{2}[/tex]

where [tex]M_{1}[/tex] is the initial concentration (1.00 M), [tex]V_{1}[/tex] is the initial volume (45.0 mL), [tex]M_{2}[/tex] is the final concentration (0.330 M), and [tex]V_{2}[/tex] is the final volume.

1.00 M × 45.0 mL = 0.330 M × [tex]V_{2}[/tex]

Solve for [tex]V_{2}[/tex]:

[tex]V_{2}[/tex] = (1.00 M × 45.0 mL) / 0.330 M
[tex]V_{2}[/tex] = 136.36 mL

Now, subtract the initial volume from the final volume to find the volume of water that needs to be added:

136.36 mL - 45.0 mL = 91.36 mL

Therefore, you must add 91.36 mL of water to the 45.0 mL of 1.00 M H2SO4 solution to create a 0.330 M [tex]H_{2} SO_{4}[/tex] solution.

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It would take approximately 17.2 hours for the concentration of N2O5 to decrease by 86.0% in the given vessel.

To calculate the time it takes for the concentration of N2O5 to decrease by 86.0%, we need to use the first-order rate equation:

ln([N2O5]t/[N2O5]0) = -kt

Where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration of N2O5, k is the rate constant, and t is the time.

Rearranging the equation to solve for t, we get:

t = (ln([N2O5]t/[N2O5]0))/(-k)

Since the reaction is first-order, the rate constant can be determined from the half-life of the reaction using the equation:

t1/2 = ln2/k

where t1/2 is the half-life.

Given that the concentration of N2O5 decreases by 86.0%, the final concentration ([N2O5]t) is:

[N2O5]t = 0.14[N2O5]0

Substituting the given values into the equation for t, we get:

t = (ln(0.14))/(-k)

To determine k, we need to find the half-life of the reaction. Since the reaction is first-order, the half-life can be calculated using the equation:

t1/2 = ln2/k

Substituting the given concentration of N2O5 and the rate constant into the equation, we get:

t1/2 = (ln2)/k = (ln2)/(2.303*0.108) = 5.45 hours

Finally, substituting the values for k and t1/2 into the equation for t, we get:

t = (ln(0.14))/(-0.108)

t = 17.2 hours

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The Na+ ion concentration (molarity) in this solution is b) 0.634 M

To find the Na+ ion concentration (molarity) in the solution, first, determine the molar concentration of [tex]Na_{2} SO_{4}[/tex], then account for the fact that each [tex]Na_{2} SO_{4}[/tex] molecule contains two Na+ ions.

1. Convert the mass of Na2SO4 to moles:
45.0 mg [tex]Na_{2} SO_{4}[/tex] * (1 g / 1000 mg) * (1 mol / 142.04 g) = 0.000317 mol Na2SO4

2. Since there are 2 Na+ ions in each[tex]Na_{2} SO_{4}[/tex] molecule:
0.000317 mol [tex]Na_{2} SO_{4}[/tex] * 2 = 0.000634 mol Na+

3. Divide moles of Na+ by volume of the solution in liters:
0.000634 mol / 0.001 L = 0.634 M

So, the Na+ ion concentration (molarity) in this solution is 0.634 M. The correct answer is (b) 0.634.

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Answer:

64g

Explanation:

refer attachment

Answer:

Explanation: To calculate the weight of sulphur and oxygen,

6.023*10^23 (Avogadro's number) is the number of molecules in a mole. Therefore, if there are 6.023*10^24 molecules, there are 10 moles of SO2.

To calculate the weight, we need to take the molar mass of sulphur and oxygen.

There will be 320.2 gms of sulphur and 320 gms of oxygen.

The volume of stock solution needed is 50mL when a student is set to prepare 250.0 mL of 0.100 M [tex]CuSO_4[/tex] from a 0.500 M stock solution.

Given the initial concentration of [tex]CuSO_4[/tex] solution (M1) = 250mL

The initial volume of [tex]CuSO_4[/tex] solution (M1) = 0.100M

The final concentration of [tex]CuSO_4[/tex] = V2

The final concentration of [tex]CuSO_4[/tex] (M2) = 0.500M

We know that molarity also known as molar concentration, is a measure of the concentration of a solute in terms of moles per liter of solution. It is also useful for comparing different solutions and for calculating the amount of a reactant needed in a reaction.

M1*V1 = M2*V2 = constant such that:

0.25 * 0.100 = 0.500 * V2

V2 = 0.05L = 50mL

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The gas will have a final pressure of 26.8 kPa.

The pressure of a gas is inversely related to its volume at constant temperature, as well as its number of moles, according to Boyle's law.Using the ideal gas law, the final pressure can be calculated as follows:

[tex]P_1V_1=P_2V_2[/tex]

[tex]P_1[/tex]= the gas's initial pressure is 209 kPa.

[tex]P_2[/tex] =gas's final pressure =?

[tex]V_1[/tex] = 10.0 L, which is the gas's initial volume.

[tex]V_2[/tex] =78.0 L, which is the gas's final volume.

Now that all the needed variables have been entered, we can apply this formula to determine the final gas pressure.

209 kPa*10.0L= [tex]P_2[/tex] *78.0L

[tex]P_2 =26.8 kPa[/tex]

Therefore, the final pressure of the gas is 26.8kPa.

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The concentration of in the cathode's half-cell solution is 5.12 × 10^-6 M if the anode's half-cell is 1 M.

While answering questions on the Brainly platform, a question answering bot should always be factually accurate, professional, and friendly. Additionally, it should be concise and should not provide extraneous amounts of detail,

ignore any typos or irrelevant parts of the question, and use the terms mentioned in the question appropriately.In a concentration cell that was set up at using two hydrogen electrodes, the cell generates a potential of . The question is, what is the concentration of in the cathode's half-cell solution, if the anode's half-cell is ?

The given data is:Potential of the cell, °cell = 0.059 log10 [H+]cathode/[H+]anode= -0.0418 V (negative because H2 gas concentration is higher in the anode than in the cathode)Since we know the value of °cell,

we can calculate the cathode half-cell potential as:

E cathode = °cell - E anode = 0.0008 VAnd then using the Nernst equation, we can find the concentration of H+ in the cathode half-cell as follows:

E cathode = E° - (RT/nF) ln [H+]H+/H2The value of E° at room temperature,

T = 298K is zero, and the number of electrons involved,

n = 2. Hence,0.0008 V = -(0.0592/2) log10[H+]H+/H2[H+]H+/H2 = 5.12 × 10^-6 M.

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At a temperature of 109° C, the balloon's new volume is roughly 932.6 mL.

The gas particles take in more heat as the temperature rises. They accelerate and advance apart from one another. Hence, an increase in volume is brought on by a rise in temperature.

We can use Charles's Law to solve this question,

V1/T1 = V2/T2

where, V1 = initial volume

T1 = initial temperature

V2 = final volume

T2 = final temperature,

Now, we have to convert the temperatures to the absolute scale, which is Kelvin (K).

T1 = 22 + 273.15 = 295.15 K

T2 = 109 + 273.15 = 382.15 K

Now, we can substitute values;

V1/T1 = V2/T2

720/295.15 = V2/382.15

Solving this equation,

V2 = (720/295.15) x 382.15

V2 = 932.6 mL

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Answer: [He] 2s2 2p3

Explanation: im himithy

The potential impact of a growing bison population on soil and plant functionalities is uncertain. Bison & elk have a lot in common physically and physiologically, therefore there might be competition over food system between two varieties.

There is little functional redundancy in keystone species. This indicates that no other organism would've been able to replace the species' ecological roles when it were to vanish from ecosystem. The environment would've been forced to undergo a significant transformation, allowing for the influx of new, potentially exotic species.

Predators have a significant impact on an environment. They influence the makeup of ecosystems by releasing pollen and nutrients from foraging. Also, they influence lower organisms in the food chain by regulating the dispersion, abundance, or variability of the prey, a phenomenon called as trophic cascades.

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Because more energy is released when water and carbon dioxide molecules are formed than when sugar and oxygen bonds are broken, the process of cellular respiration releases energy.

Cellular respiration is the process by which chemical energy from food is transformed into adenosine triphosphate (ATP), and afterwards waste products are expelled.

Through a sequence of chemical processes called cellular respiration, glucose is broken down to create ATP, which may then be used as an energy source for a variety of bodily functions.

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To solve this problem, we will use stoichiometry and the molar masses of the compounds involved in the reaction.

First, let's write the balanced chemical equation:

2 Cr2O3(s) + 3 Si(l) → 4 Cr(l) + 3 SiO2(s)

From the equation, we can see that 3 moles of Si are required to produce 4 moles of Cr. We can use this ratio to calculate the amount of Si required to produce a given amount of Cr.

Step 1: Calculate the molar mass of Cr2O3

Cr2O3: 2 x atomic mass of Cr + 3 x atomic mass of O

Cr2O3: 2 x 52.00 g/mol + 3 x 16.00 g/mol

Cr2O3: 152.00 g/mol

Step 2: Calculate the number of moles of Cr2O3 required to produce 243.0 g of Cr

n(Cr2O3) = m(Cr2O3) / M(Cr2O3)

n(Cr2O3) = 243.0 g / 152.00 g/mol

n(Cr2O3) = 1.597 moles

Step 3: Use the stoichiometry of the balanced equation to calculate the number of moles of Si required to produce 1.597 moles of Cr

From the balanced equation, we know that 2 moles of Cr2O3 react with 3 moles of Si to produce 4 moles of Cr. So, we can set up a proportion:

2 moles of Cr2O3 : 3 moles of Si = 1.597 moles of Cr2O3 : x moles of Si

x = (3 moles of Si x 1.597 moles of Cr2O3) / 2 moles of Cr2O3

x = 2.395 moles of Si

Step 4: Calculate the mass of Si required to produce 2.395 moles of Si

m(Si) = n(Si) x M(Si)

m(Si) = 2.395 moles x 28.09 g/mol

m(Si) = 67.28 g

Therefore, 67.28 grams of silicon is required to produce 243.0 g of chromium metal.

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(Complete question)

Calculate the number of grams of silicon required to prepare 243.0 g of chromium metal by the second reaction:

2Cr2O3(s) + 3Si (l) → 4Cr (l) + 3SiO2 (s)