Furoic acid 1HC5H3O32 has a Ka value of 6.76 * 10-4 at 25 °C. Calculate the pH at 25 °C of (a) a solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate 1NaC5H3O32 to enough water to form 0.250 L of solution, (b) a solution formed by mixing 30.0 mL of 0.250 M HC5H3O3 and 20.0 mL of 0.22 M NaC5H3O3 and diluting the total volume to 125 mL, (c) a solution prepared by adding 50.0 mL of 1.65 M NaOH solution to 0.500 L of 0.0850 M HC5H3O3.

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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The pH of a 0.10M [tex]C_{6} H_{5} O_{}[/tex] solution is 11.50 which is calculated using the expression of pOH.

[tex]K_{b}[/tex] = [tex]K_{w} / K_{a}[/tex]

    = [tex]1 * 10 ^{-14}[/tex] / [tex]1. 0 * 10 ^{-10}[/tex]

    = [tex]1.0 * 10^{-4}[/tex]

pH is defined as the quantitative measure of the acidity or basicity of aqueous or of other liquid solutions. pH is called as the potential of hydrogen ions where pOH is called as the potential of hydroxide ions.  pH scale is known as a scale which is used to determine the solution's hydrogen ion (H+) concentration. The term pH and pOH that denote the negative log of the concentration of the hydrogen ion or hydroxide ions. High pH indicates that a solution is basic while high pOH means that a solution is acidic.

We know that the expression for pOH is,

pOH = -Log [tex]\sqrt{K_{b} C}[/tex]

Putting all the values in the expression of pOH we get,

pH = 11.50

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The correct question is,

The Ka of C6H5OH is 1.0×10−10 .What is the pH of a 0.10 M C 6H 5O-solution?

Answer: No, Calcium sulfate (CaSO4) is insoluble in water because water dipole strength is too weak to separate the anions and cations of the CaSO4 as both Ca 2+ and SO4 2- ions are big and bigger anion stabilizes bigger cation strongly which makes lattice energy high.

Explanation: Hope this helps!!

The effect of raising the pH value of the solution from pH 4.0 to pH 9.0 on the reaction rate catalyzed by an enzyme with an optimum pH of 7.0 is to increase the reaction.

The pH value is a measure of acidic/basic solution. Characteristic of pH

Enzyme is affected by pH changes. The rate of the enzymatic reaction depends on the pH of the medium.

1) Every enzyme has an optimal pH value at which the rate of enzymatic reaction is maximized.

2) At higher or lower pH, the rate of enzymatic reaction decreases.

3) The optimum pH for most enzymes is in the pH 5 to pH 9 range.

4) With a few exceptions, the pH of pepsin is very acidic and the pH of arginase is very alkaline. Now, the pH of the solution rises from pH 4.0 to pH 9.

0 is the optimal pH of the enzyme 7. Therefore, the reaction rate increases.

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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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There were 0.01 moles of HCl present in the 10.0 mL sample.

The balanced chemical equation for the reaction between HCl and Ba(OH)2 is:

2HCl + Ba(OH)2 -> 2H2O + BaCl2

From the equation, we can see that two moles of HCl are required to react with one mole of Ba(OH)2.

To find the number of moles of HCl in the sample, we need to first calculate the number of moles of Ba(OH)2 that reacted with the HCl.

The number of moles of Ba(OH)2 is given by:

moles of Ba(OH)2 = concentration x volume (in liters)

moles of Ba(OH)2 = 0.4 mol/L x (12.5/1000) L

moles of Ba(OH)2 = 0.005 mol

Since two moles of HCl react with one mole of Ba(OH)2, the number of moles of HCl is:

moles of HCl = 2 x 0.005 mol

moles of HCl = 0.01 mol

Therefore, there were 0.01 moles of HCl present in the 10.0 mL sample.

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the new temperature is 393.75 K.

_______________________________________________________

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

P₁V₁/T₁ = P₂V₂/T₂

where P is the pressure, V is the volume, and T is the temperature.

We are given that the pressure and amount of gas remain constant, so we can simplify the equation to:

V₁/T₁ = V₂/T₂

Substituting the given values, we get:

2.00 L / 315 K = 2.50 L / T₂

Solving for T₂, we get:

T₂ = (2.50 L * 315 K) / 2.00 L = 393.75 K

Therefore, the new temperature is 393.75 K.

The new temperature of the gas is approximately 393.75 K when the volume is changed to 2.50 L.

To find the new temperature of the gas, we can use Charles's Law, which states that the initial volume (V₁) divided by the initial temperature (T₁) is equal to the final volume (V₂) divided by the final temperature (T₂), as long as the pressure and amount of gas remain constant. The formula for Charles's Law is:
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Given the initial volume (V₁) = 2.00 L and the initial temperature (T₁) = 315 K, we want to find the new temperature (T₂) when the volume is changed to 2.50 L (V₂).
First, plug in the known values into the formula:
[tex]\frac{(2.00 )}{(315)} = \frac{(2.50)}{(T_2)}[/tex]
Next, cross-multiply to solve for T₂:
[tex](2.00 ) * (T_2) = (2.50) * (315)[/tex]
Now, divide both sides by 2.00 L to isolate T₂:
[tex]T_2 = \frac{(2.50  * 315 )}{(2.00 )}[/tex]
Finally, perform the calculation to find the new temperature:
[tex]T_2 = \frac{787.5}{2}[/tex]
T₂ ≈ 393.75 K

So, the new temperature of the gas is approximately 393.75 K when the volume is changed to 2.50 L.

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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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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."--

Answer:

A

Explanation:

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)

⁣[tex]5.6 \times 10^1^7[/tex] atoms of phosphorus is equal to [tex]9.32 \times 10^-7[/tex] moles.

Either a positively charged electron or a large number of negatively charged electrons surround the central nucleus of an atom. The positively charged nucleus is made up of two comparatively large particles called protons and neutrons.

As many atoms, molecules, or ions (referred to as entities) are present in one mole of substance as there are in  [tex]12[/tex] grammes of carbon-[tex]12[/tex], it is known as a mole. It is about equal to the Avogadro's number [tex]6.022 \times 10^23[/tex]  , which is entities per mole.

We must divide this amount by Avogadro's number to get the number of moles in  [tex]5.6 \times 10^1^7[/tex] atoms of phosphorus:

Number of moles =[tex](5.6 \times 10^17 atoms) / (6.022 \times 10^23 atoms/mol)[/tex]

Number of moles = [tex]9.32 \times 10^-7[/tex]mol

Therefore, ⁣[tex]5.6 \times 10^1^7[/tex] atoms of phosphorus is equal to [tex]9.32 \times 10^-7[/tex] moles.

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The mass of hydrofluoric acid contained in the sample is approximately 0.1876 g.

To find the mass of hydrofluoric acid (HF) in the sample, follow these steps:

1. Write the balanced chemical equation for the reaction:
 [tex]HF + NaOH → NaF + H2O[/tex]

2. Determine the moles of sodium hydroxide (NaOH) used in the titration:
  [tex]Moles of NaOH = Volume (L) × Molarity[/tex]
 [tex]Moles of NaOH = 25.21 mL × (1 L/1000 mL) × 0.372 M[/tex]
 [tex]Moles of NaOH = 0.00937692 mol[/tex]

3. Determine the moles of hydrofluoric acid (HF) using the stoichiometry from the balanced equation:
  [tex]Moles of HF = Moles of NaOH (1 mol HF / 1 mol NaOH)[/tex]
 [tex]Moles of HF = 0.00937692 mol[/tex]

4. Determine the molar mass of hydrofluoric acid (HF):
  [tex]Molar mass of HF = 1(1.01 g/mol H) + 1(19.00 g/mol F) = 20.01 g/mol[/tex]

5. Calculate the mass of hydrofluoric acid (HF) in the sample:
  Mass of HF = Moles of HF × Molar mass of HF
 [tex]Mass of HF = 0.00937692 mol × 20.01 g/mol[/tex]
  [tex]Mass of HF = 0.1876 g[/tex]

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Approximately 29.5 g of NaCN is required to make 14.7 L of HCN at STP.

To solve this problem, we will use the ideal gas law to calculate the number of moles of HCN produced and then use stoichiometry to determine the mass of NaCN required.

First, we need to determine the number of moles of HCN produced using the ideal gas law:

[tex]PV = nRT[/tex]

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

At STP (standard temperature and pressure), P = 1 atm and T = 273 K. The volume of HCN produced is given as 14.7 L.

Plugging these values into the ideal gas law, we get:

[tex]n = PV/RT = (1 atm) *(14.7 L)/(0.0821 L atm/mol K * 273 K) = 0.603 mol[/tex]

So, 0.603 mol of HCN is produced.

Now we can use stoichiometry to determine the mass of NaCN required. From the balanced chemical equation:

NaCN + HCl → NaCl + HCN

we can see that 1 mole of NaCN produces 1 mole of HCN.

Therefore, the mass of NaCN required can be calculated as:

mass of NaCN = number of moles of NaCN x molar mass of NaCN

The molar mass of NaCN is 49.01 g/mol.

So, the mass of NaCN required is:

mass of [tex]NaCN = 0.603 mol * 49.01 g/mol = 29.5 g[/tex]

Therefore, approximately 29.5 g of NaCN is required to make 14.7 L of HCN at STP.

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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.

[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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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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Answer:

When endothermic reaction takes place, the system gains heat from the surroundings and so the temperature of the surroundings decreases ie. it gets colder

Explanation:

A chemical reaction is exothermic if heat is released by the system into the surroundings.

STUDY WELL!

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 poor specificity of the sulfur reduction test is responsible for not being able to differentiate between H ₂S produced by anaerobic respiration and H ₂S  produced by putrefaction.

What is the sulfur reduction test, The sulfur reduction test is a biochemical test that helps to determine the ability of an organism to reduce sulfur and produce H ₂S (hydrogen sulfide). The test is carried out by inoculating a sulfur-containing medium with the test organism and observing whether the medium changes colour due to the production of H ₂S.

Why is the sulfur reduction test not able to differentiate between H ₂S produced by anaerobic respiration and H ₂S produced by putrefaction, The sulfur reduction test is not able to differentiate between H ₂S produced by anaerobic respiration and H ₂S produced by putrefaction due to the poor specificity of the test.

This means that the test is not able to distinguish between the different sources of H ₂S production and can only detect the presence or absence of H ₂S without providing information on its source.

Therefore, the test has poor specificity but not poor sensitivity since it is able to detect the presence of H ₂S, but cannot distinguish between its different sources.

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4.37 grams of O₂ will be left over after the reaction is complete.

The balanced chemical equation for the reaction is:

4NO₂ + O₂→ 2N₂O₅

From the equation, we can see that 4 moles of NO and 1 mole of O₂ react to form 2 moles of N₂O₅.

To find the amount of each reactant and product in the reaction, we need to first calculate the number of moles of each substance. We can use the molecular weight of each substance to convert the given mass into moles.

The molecular weights of the substances are:

NO₂ = 46.0055 g/mol

O₂ = 31.9988 g/mol

N₂O₅ = 108.0104 g/mol

Number of moles of NO₂ = 5.31 g / 46.0055 g/mol = 0.1156 mol

Number of moles of O₂ = 5.31 g / 31.9988 g/mol = 0.1659 mol

According to the balanced chemical equation, 4 moles of NO₂ react with 1 mole of O₂ to produce 2 moles of N₂O₅.

Therefore, the limiting reactant is NO₂ because there are only 0.1156 mol of it available, while there are 0.1659 mol of O₂ available. This means that all of the NO₂ will be used up, and there will be some excess O₂ left over.

To calculate the amount of N₂O₅ produced, we can use the mole ratio from the balanced chemical equation:

4 mol NO₂ : 1 mol O₂ : 2 mol N₂O₅

Since we know that 0.1156 mol of NO₂ will be used up, we can use the mole ratio to calculate the amount of N₂O₅

produced:

0.1156 mol NO₂ x (2 mol N₂O₅ / 4 mol NO₂) = 0.0578 mol N₂O₅

To find the mass of N₂O₅ produced, we can use the molecular weight:

0.0578 mol N₂O₅ x 108.0104 g/mol = 6.24 g N₂O5

Therefore, 6.24 grams of N₂O₅ will be produced, and there will be some excess O₂ left over. To calculate the amount of O₂ left over, we can use the mole ratio from the balanced chemical equation:

4 mol NO2 : 1 mol O₂ : 2 mol N₂O₅

Since we know that 0.1156 mol of NO₂ will be used up, we can use the mole ratio to calculate the amount of O₂  required:

0.1156 mol NO₂ x (1 mol O₂ / 4 mol NO₂) = 0.0289 mol O₂

Therefore, the amount of O₂ left over is:

0.1659 mol O₂ - 0.0289 mol O₂ = 0.1370 mol O₂

To find the mass of O₂ left over, we can use the molecular weight:

0.1370 mol O₂ x 31.9988 g/mol = 4.37 g O₂

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To find the molar mass of the gas, we can use the ideal gas law:

PV = nRT

where P is the pressure in Pa, V is the volume in m^3, n is the number of moles of gas, R is the ideal gas constant (8.31 J/(mol·K)), and T is the temperature in K.

First, we need to convert the given values to the appropriate units:

12.0 g -> 0.0120 kg

2.8 dm^3 -> 0.0028 m^3

27°C -> 300 K (adding 273 to convert from Celsius to Kelvin)

100 kPa -> 100,000 Pa

Now we can rearrange the ideal gas law to solve for n:

n = PV/RT

n = (100,000 Pa) x (0.0028 m^3) / [(8.31 J/(mol·K)) x (300 K)]

n = 0.001214 mol

Finally, we can calculate the molar mass (M) using the formula:

M = m/n

where m is the mass of the gas (in grams). Since we have the mass in kilograms, we need to multiply by 1000 to convert to grams:

M = (0.0120 kg x 1000 g/kg) / 0.001214 mol

M = 9906.2 g/mol

Therefore, the molar mass of the gas is approximately 9906 g/mol.

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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In order for a titration to be effective, the reaction must produce a precipitate. The correct answer is option B, "reaction must produce a precipitate."

For a titration to be effective, the reaction must be stoichiometric, quantitative, and rapid. A stoichiometric reaction is one in which the amount of reactants is proportional to the amount of products.

A quantitative reaction is one in which all the reactants are consumed, leaving no excess. A rapid reaction is one that occurs quickly and does not take a long time to complete.

However, a reaction producing a precipitate is not necessary for the titration to be effective. Hence option B is correct.

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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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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 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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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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The theoretical dependence between capacitive reactance and frequency. Capacitive reactance (Xc) is the opposition offered by a capacitor to the flow of alternating current (AC) due to its capacitance.

The relationship between capacitive reactance and frequency is inverse and proportional. Mathematically, it is given by the following formula:

Xc = 1 / (2 * π * f * C)

where Xc is the capacitive reactance in ohms, f is the frequency of the AC signal in hertz (Hz), and C is the capacitance of the capacitor in farads (F).

As per this formula, capacitive reactance (Xc) is inversely proportional to the frequency (f) of the AC signal. In other words, as the frequency increases, the capacitive reactance decreases, and as the frequency decreases, the capacitive reactance increases.

This means that a capacitor will act more like a short circuit (i.e. offer less opposition to the flow of current) at higher frequencies and more like an open circuit (i.e. offer more opposition to the flow of current) at lower frequencies.

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