Consider the acid dissociation behavior of carbonic acid, H 2 CO 3 . A p H gradient from 0 to 14 is given. Below a p H equal to p K a 1 which is 6.351, the predominant form is H 2 C O 3. Above a p H equal to p K a 2 which is 10.329, the predominant form is C O 3 2 minus. Between the two p K a values, the predominant form is H C O 3 minus. What is the predominant species present at pH 6.73 ?

Answer:

a) If the white powder didn't dissolve completely, that could have a great effect on the experiment. The concentration of the solution cant get to the point where it needs to get for the rest of the experiment which can skew results.

b) If a solution forms bubbles immediately after an addition of a solid, that could simply mean that gas is formed in that reaction. That has no negative effect on an experiment since that's what is supposed to happen during the reaction.

c) Using a different measuring device can really effect a students calculations later on during the experiment. Different devices are calibrated differently, which can skew the results or calculations later on. Its best to stay consistent with what you are measuring with during an experiment.

Explanation:

If the unknown does not dissolve, it affects the results. Bubbles during dissolution indicate a chemical reaction. Replacing the thermometer does not greatly affect the results.

a) If the unknown white powder failed to dissolve in water, it means that it is insoluble in water. This could affect the experimental results as the solubility of the unknown substance is an important factor in determining its properties.

b) The bubbling observed when the different solid unknown was added to water indicates a chemical reaction. This could affect the experimental results as the reaction may lead to the formation of a new compound, changing the composition of the solution.

c) The breaking of the thermometer and the replacement with a new one should not significantly affect the experimental results. As long as the new thermometer is calibrated and accurate, it can still be used to measure the freezing point of the unknown solution.

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Answer:

n = 0.0022 mol

Explanation:

Moles is denoted by given mass divided by the molar mass ,  

Hence ,  

n = w / m  

n = moles ,  

w = given mass ,  

m = molar mass .  

From the information of the question ,

w = 0.108 g

As we known ,

The molar mass of titanium = 47.867 g / mol

The mole of titanium can be caused by using the above relation , i.e. ,

n = w / m  

n = 0.108 g / 47.867 g / mol

n = 0.0022 mol

Final answer:

To find the number of moles of titanium in 0.108 g, divide the mass by the molar mass of titanium (47.867 g/mol), which yields approximately 0.002256 moles of titanium.

Explanation:

Calculating Moles of Titanium

The student asks: How many moles of titanium is 0.108 g? To answer this, we first need the molar mass of titanium, which is approximately 47.867 g/mol. Using the formula for calculating moles:
 Number of moles = Mass (g) / Molar mass (g/mol)

So for titanium:
 Number of moles of Ti = 0.108 g / 47.867 g/mol

After performing the division:
 Number of moles of Ti = 0.002256 moles

This result signifies that 0.108 g of titanium corresponds to approximately 0.002256 moles of titanium.

The CO2 concentration has been increasing steadily for several decades, with an average annual increase about 1.4 ppm between 1958 and 2014, somewhat lower between 1970 and 1980 and higher between 2004 and 2014. The most recent CO2 concentration measured I found is 414.0 ppm up to July 2021.

It appears that the student's question is missing specific data, however, I can provide information on the average increase in

CO2 concentration

over certain periods. The concentration of CO2 in the atmosphere has been steadily increasing since the beginning of the industrial revolution, with more prominent increases in recent decades due to increased industrial activity and deforestation. For example, between 1958 and 2014, the CO2 concentration in the atmosphere reportedly increased by about 1.4 parts per million (ppm) per year on average. The rate was somewhat lower between 1970 and 1980 (about 1.3 ppm/year), but higher between 2004 and 2014 (about 2 ppm/year). The most recent CO2 concentration recorded I could find is 414.0 ppm up to July 2021. It's important to note that

increasing CO2 concentration

is a significant contributor to global warming and climate change.

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Answer:

Check the explanation

Explanation:

The diagram to the question is in the attached image, and always remember that Constitutional isomers are compounds that have very similar molecular formula and diverse connectivity. To conclude whether two molecules are constitutional isomers, just calculate the figure or amount of every atom in both molecules and see how the atoms are assembled.

Answer:

1. The overall equation is ClO-(aq)+I-(aq) → Cl-(aq)+IO-(aq)

2. The intermediates include: HClO(aq), OH-(aq) and HIO(aq)

3. The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]

4. No,

The rate depends on [OH-], so it's not consistent with the actual rate law

Explanation:

1

Given

(1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]

(2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]

(3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

Add up the three equations

ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]

I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]

OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

Remove all common terms {H2O(l) + I-(aq) +HClO(aq) +OH-(aq) +HIO(aq) => HClO(aq) + OH-(aq) + HIO(aq)

+H2O(l)}

We're left with

ClO-(aq) + I-(aq) => Cl-(aq) + IO-(aq)

2.

There are intermediates generated but they are not visible in the overall equation.

The intermediates include: HClO(aq), OH-(aq) and HIO(aq)

3.

The three steps are bimolecular.

The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]

4. Let K represents equilibrium constant

At step 1,

K1 = [HClO][OH-]/[ClO-]

Simplify;

K1 [ClO-]= [HClO][OH-]

K1[ClO-]/[OH-] = [HClO]

Determine the rate at step 2

= k2[I-][HClO]

= K1k2[I-][ClO-]/[OH-]

= k[ClO-][I-]/[OH-]

The answer is no

Final answer:

The overall equation derived from the given mechanism is ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq), with HClO, OH−, and HIO as intermediates. Molecularity for all steps is bimolecular with specific rate laws for each step, and the mechanism is consistent with the experimental rate law rate = k[ClO−][I−].

Explanation:

The student's question pertains to deriving the overall equation, identifying intermediates, determining molecularity and rate laws for each step, and verifying the consistency of a proposed mechanism with the experimental rate law in a chemical reaction series involving species such as ClO−(aq), H2O(l), I−(aq), and others.

Answers to the Student's Question

The overall equation is ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq).

The intermediates in the reaction mechanism are HClO, OH−, HIO.

For the first step, the molecularity is bimolecular and the rate law is k1[ClO−][H2O]. For the second step, it's bimolecular with a rate law of k2[I−][HClO]. The third step is also bimolecular with a rate law of k3[OH−][HIO].

The mechanism is consistent with the actual rate law, which is rate = k[ClO−][I−], because the rate-determining step involves these reactants.

Explanation:

As we know that HCl is a stronger acid  and NaOH is a stronger base.

And, HF and phenol are weaker acids having [tex]K_{a}[/tex] values of [tex]6.6 \times 10^{-4}[/tex] and [tex]1.3 \times 10^{-10}[/tex] respectively.

In the same way, methyl amine and pyridine are weaker bases with [tex]K_{b}[/tex] values of [tex]4.4 \times 10^{-4}[/tex] and [tex]1.7 \times 10^{-9}[/tex] respectively.

(i)   Volume will be calculated as follows.

           [tex]M_{a}V_{a} = M_{b}V_{b }[/tex]

           [tex]100 \times 0.1 = 0.1 \times V[/tex]

Therefore, volume of NaOH is 100 ml.

(ii)   Similarly, volume of HCl is 100 ml.

(iii)  For Methyl amine,

         [tex][OH]^{-} = \sqrt{4.4 \times 10^{-4} \times 0.1}[/tex]

                     = [tex]6.6 \times 10^{-3}[/tex]

And as, [tex]M_{a}V_{a} = M_{b}V_{b}[/tex]

             [tex]0.1 \times V = 6.6 \times 10^{-3} \times 100[/tex]

Hence, the volume of HCl is 6.6 ml.

(iv) For HF,

           [tex][H]^{+} = \sqrt{6.6 \times 10^{-4} \times 0.1}[/tex]

                      = [tex]8.12 \times 10^{-3}[/tex]

As,    [tex]M_{a}V_{a} = M_{b}V_{b}[/tex]

         [tex]8.12 \times 10^{-3} \times 100 = 0.1 \times V[/tex]

Hence, the volume of NaOH added is 8.12 ml.

Therefore, we can conclude that the increasing order of volums of given titrant is (i) = (ii) > (iv) > (iii).

All titrations (i to iv) involve acid-base reactions with equal molarity and volume of reactants; therefore, the volume of titrant at the equivalence point is the same for each scenario, which is 100.0 mL.

To rank the titrations in order of increasing volume of titrant added to reach the equivalence point, we must consider the nature of the reactants in each titration. Strong acids and bases will react in a 1:1 ratio, meaning an equal volume of titrant is needed to reach the equivalence point if their concentrations are the same. Thus, titrations i and ii, which involve the strong acid HCl and the strong base NaOH, will have the same volume of titrant at the equivalence point.

Titeration iii, which involves the weak base CH₃NH₂ and the strong acid HCl, will have a different volume compared to a strong acid/strong base titration due to the possibility of incomplete dissociation of the weak base. However, since the concentrations are equal, 100.0 mL of titrant is still expected to be needed to reach the equivalence point.

Titeration iv, which includes the weak acid HF and strong base NaOH, also involves a 1:1 stoichiometry at the equivalence point, but weak acids can exhibit buffering effects which may slightly alter the volume needed to reach the equivalence point. Nevertheless, with equal concentrations, the same volume of titrant is expected to be used as in the previous cases.

Therefore, because all titrations have equal molarities and volumes of both the titrants and analytes, we'd expect the volume of titrant needed to reach the equivalence point to be the same for each scenario, which is 100.0 mL.

To calculate the standard enthalpy change for the given chemical reaction, balance the equation, then apply standard enthalpies of formation for the reactants and products, and use Hess's Law. The calculated standard enthalpy change for this reaction is -433.37 kJ/mol.

To calculate the standard enthalpy change ([tex]H_orxn[/tex]) for the reaction [tex]CH_4(g) + Cl_2(g) \rightarrow CCl_4(l) + HCl(g)[/tex], we need to first balance the reaction:

[tex]CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)[/tex]

Then we use the standard enthalpies of formation ( Hf) for each substance to calculate [tex]H_orxn[/tex] using Hess's Law:

[tex]H_o = [ H_f[CCl_4(l)] + 4 H_f[HCl(g)]] - [ H_f[CH_4(g)] + 4 H_f[Cl_2(g)]][/tex]

[tex]H_o = [-139 kJ/mol + 4(-92.31 kJ/mol)] - [-74.87 kJ/mol + 4(0 kJ/mol)][/tex]

[tex]H_o = -139 kJ/mol - 369.24 kJ/mol - (-74.87 kJ/mol)[/tex]

[tex]H_o = -433.37 kJ/mol[/tex]

ΔH°rxn for CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g), is -433.37 kJ

To calculate the enthalpy change (ΔH°rxn) for the reaction CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g), we first need to balance the equation and use the standard enthalpies of formation (ΔH°f) provided for each compound.

Balance the chemical equation:

The balanced equation is: CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g)

Write the enthalpy of formation values:

[tex]\Delta H^\circ_f[\text{CH}_4(g)] &= -74.87 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{CCl}_4(\ell)] &= -139 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{HCl(g)}] &= -92.31 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{Cl}_2(g)] &= 0 \, \text{kJ/mol} \quad (\text{by convention})[/tex]

Calculate the total ΔH°f for the products:

[tex]\text{For CCl}_4(\ell) \text{ and HCl(g):}\\\\ & {1 \, \text{mol of CCl}_4(\ell) \times (-139 \, \text{kJ/mol})} + {4 \, \text{mol of HCl(g)} \times (-92.31 \, \text{kJ/mol})} \\& = -139 \, \text{kJ} + (-369.24 \, \text{kJ}) \\& = -508.24 \, \text{kJ} \\[/tex]

Calculate the total ΔH°f for the reactants:

[tex]\text{For CH}_4(g) \text{ and Cl}_2(g): \\\\& {1 \, \text{mol of CH}_4(g) \times (-74.87 \, \text{kJ/mol})} + {4 \, \text{mol of Cl}_2(g) \times (0 \, \text{kJ/mol})} \\& = -74.87 \, \text{kJ} \\[/tex]

Determine ΔH°rxn:

[tex]\Delta H^\circ_\text{rxn}: & = \Sigma \Delta H^\circ_f(\text{products}) - \Sigma \Delta H^\circ_f(\text{reactants}) \\\\& = -508.24 \, \text{kJ} - (-74.87 \, \text{kJ}) \\& = -433.37 \, \text{kJ}[/tex]

Therefore, the enthalpy change for the reaction ΔH°rxn is -433.37 kJ.

Answer:

Asking for and receiving a paper that someone who is in another lab section wrote is an honor violation. True

An oral discussion with a classmate regarding a paper's topics and format is cheating. False

Purchasing papers from tutoring companies is allowed. False

Sending a paper that you wrote to a student who is currently in another section is an honor violation. True

Misconduct in research or scholarship includes fabrication, falsification, or plagiarism in proposing, performing, reviewing, or reporting research. It does not include honest error or honest differences in interpretations or judgements of data. True

Discussing an exam or laboratory practical with anyone prior to the conclusion of the week that final exams and laboratory practicals are given would be considered an honor violation. True

Explanation:

The Aggie code of honour is a code of academic integrity of the Texas A&M University. It spells out the codes of academic integrity and responsible research. Instructors are to include this code in all syllabi. It targets the application of the highest degree of integrity in academic research and forbids misconducts such as cheating, plagiarism, falsification et cetera.

The statements regarding the Aggie Honor System and Academic Integrity rules for CHEM 111/112/117 are generally correct, with exceptions for discussions about paper topics, which is not innately cheating, and purchasing of papers, which is not allowed.

Based on the Aggie Honor System Rules and the Academic Integrity statement on the CHEM 111/112/117 syllabus, the following assessments can be made about the statements:

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Answer: Total pressure inside of a vessel is 0.908 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual partial pressures. exerted by each gas alone.

[tex]p_{total}=p_1+p_2+p_3[/tex]

[tex]p_{N_2}[/tex] = partial pressure of nitrogen = 0.256 atm

[tex]p_{He}[/tex] = partial pressure of helium = 203 mm Hg = 0.267 atm  (760mmHg=1atm)

[tex]p_{H_2}[/tex] = partial pressure of hydrogen =39.0 kPa = 0.385 atm  (1kPa=0.00987 atm)

Thus [tex]p_{total}=p_{H_2}+p_{He}+p_{H_2}[/tex]

[tex]p_{total}[/tex] =0.256atm+0.267atm+0.385atm =0.908atm

Thus total pressure (in atm) inside of a vessel is 0.908

Answer:

The total pressure inside the vessel is 0.908 atm

Explanation:

Step 1: Data given

Partial pressure N2 = 0.256 atm

Partial pressure He = 203 mmHg = 0.267105 atm

Partial pressure H2 = 39.0 kPa = 0.3849 atm

Step 2: Calculate the total pressure

Total pressure  =  p(N2)  + p(He)  +  p(H2)

Total pressure = 0.256 atm + 0.267105 atm + 0.3849 atm

Total pressure = 0.908 atm

The total pressure inside the vessel is 0.908 atm

The question is incomplete, here is the complete question:

Ethylene is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10¹⁰ of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas.

Suppose an engineer studying ethane cracking fills a 30.0 L reaction tank with 38.0 atm of ethane gas and raises the temperature to 400°C. He believes Kp = 0.4 at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture.

Answer: The mass percent of ethylene gas is 9.20 %

Explanation:

We are given:

Initial partial pressure of ethane gas = 38.0 atm

The chemical equation for the dehydrogenation of ethane follows:

                  [tex]C_2H_6\rightleftharpoons C_2H_4+H_2[/tex]

Initial:             38

At eqllm:      38-x         x       x

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}[/tex]

We are given:

[tex]K_p=0.40[/tex]

Putting values in above equation, we get:

[tex]0.40=\frac{x\times x}{38-x}\\\\x=-4.10,3.70[/tex]

Neglecting the negative value of 'x' because partial pressure cannot be negative

So, equilibrium partial pressure of ethane = 38 - x = 38 - 3.70 = 34.30 atm

Equilibrium partial pressure of ethene = x = 3.70 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]         ..........(1)

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(2)

We are given:

[tex]P=34.3atm\\V=30.0L\\R=0.0821\text{ L atm }mol^{-1}K^{-1}\\T=400^oC=[400+273]=673K[/tex]

Putting values in equation 1, we get:

[tex]34.3atm\times 30.0L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 673K\\\\n=\frac{34.3\times 30.0}{0.0821\times 673}=18.62mol[/tex]

Molar mass of ethane gas = 30 g/mol

Moles of ethane gas = 18.62 mol

Putting values in equation 2, we get:

[tex]18.62mol=\frac{\text{Mass of ethane}}{30g/mol}\\\\\text{Mass of ethane gas}=(18.62mol\times 30g/mol)=558.6g[/tex]

We are given:

[tex]P=3.70atm\\V=30.0L\\R=0.0821\text{ L atm }mol^{-1}K^{-1}\\T=400^oC=[400+273]=673K[/tex]

Putting values in equation 1, we get:

[tex]3.70atm\times 30.0L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 673K\\\\n=\frac{3.70\times 30.0}{0.0821\times 673}=2.01mol[/tex]

Molar mass of ethylene gas = 28 g/mol

Moles of ethylene gas = 2.01 mol

Putting values in equation 2, we get:

[tex]2.01mol=\frac{\text{Mass of ethylene}}{28g/mol}\\\\\text{Mass of ethylene gas}=(2.01mol\times 28g/mol)=56.28g[/tex]

[tex]\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100[/tex]

Mass of ethylene = 56.28 g

Mass of mixture = [558.6 + 56.28] g = 641.88 g

Putting values in above equation, we get:

[tex]\text{Mass percent of ethylene}=\frac{56.28g}{641.88g}\times 100=9.20\%[/tex]

Hence, the mass percent of ethylene gas is 9.20 %

Answer:

The chemical formula of this residue: [tex]Na_2CrO_4.10H_2O[/tex]

Explanation:

Mass of residue = 19.12

Let the formula of the residue be [tex]Na_2CrO_4.xH_2O[/tex]

Moles of residue: n

[tex]n=\frac{19.12}{162 g/mol+x\times 18 g/mol}[/tex]

Moles of sodium chromate = n'

Molarity of sodium chromate = 0.1035 M

Volume of sodium chromate solution = 540.0 mL = 0.540 L

1 mL = 0.001 L

[tex]Concentration=\frac{moles}{Volume(L)}[/tex]

[tex]0.1035 M=\frac{n}{0.540 L}[/tex]

[tex]n'=0.1035 M\times 0.540 L=0.05589 mol[/tex]

[tex]Na_2CrO_4+xH_2O\rightarrow Na_2CrO_4.xH_2O[/tex]

According to reaction,1 mole of  [tex]Na_2CrO_4[/tex] gives 1 mole of[tex]Na_2CrO_4.xH_2O[/tex]

So, n = n'

[tex]\frac{19.12}{162 g/mol+x\times 18 g/mol}=0.05589 mol[/tex]

x = 10

The chemical formula of this residue: [tex]Na_2CrO_4.10H_2O[/tex]

The methyl protons of (CH3)2Mg are in a more electron rich environment than the methyl protons of TMS. Thus the protons of (CH3)2Mg show a signal upfield from TMS.

Explanation:

Answer : The empirical of the compound is, [tex]C_{11}H_{12}N_2[/tex]

Explanation :

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

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

Mass of C = 76.71 g

Mass of H = 7.02 g

Mass of N = 16.27 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{76.71g}{12g/mole}=6.39moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.02g}{1g/mole}=7.02moles[/tex]

Moles of N = [tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.27g}{14g/mole}=1.16moles[/tex]

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

For C = [tex]\frac{6.39}{1.16}=5.5[/tex]

For H = [tex]\frac{7.02}{1.16}=6.0\approx 6[/tex]

For N = [tex]\frac{1.16}{1.16}=1[/tex]

The ratio of C : H : N = 5.5 : 6 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : N = 11 : 12 : 2

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

The Empirical formula = [tex]C_{11}H_{12}N_2[/tex]

Therefore, the empirical of the compound is, [tex]C_{11}H_{12}N_2[/tex]

To find the volume of oxygen at STP necessary to react with 1.0 gal of gasoline we perform a series of conversions: from volume of gasoline to mass using density from mass of gasoline to moles using molar mass of octane from moles of octane to moles of oxygen using the stoichiometric ratio from the balanced chemical equation and from moles of oxygen to volume using the molar volume of gas at STP. This series of conversions yields the answer.

The question you asked pertains to how gasoline, or octane (C8H18), reacts with oxygen to produce carbon dioxide and water, and how much oxygen volume at STP is necessary for this reaction with 1.0 gal of gasoline. To solve this task, we first convert gasoline volume to grams using its given density. Knowing that 1 gal of gasoline equals 3.78 L and the density of gasoline is 0.81 g/mL, we multiply these values to get the total mass of gasoline. Then, we consider the balanced chemical equation of the combustion of octane: 2C8H18 + 25O2 -> 16CO2 + 18H2O. This equation tells us that 25 moles of oxygen react with 2 moles of octane. Using octane's molar mass, we convert the mass of gasoline to moles. Once we have the moles of octane, we use the stoichiometric ratio from the balanced equation to find the moles of oxygen necessary for the reaction. Finally using the molar volume of a gas at STP which is approximately 22.4 L, we convert the moles of oxygen to volume. This lengthy chemical calculation processes yields the volume of oxygen necessary to react with 1 gal of gasoline.

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Answer:

C.sterols

Explanation:

Sterols or steroid alcohols are type of lipids. In plants they are present as phytosterols and in animals as zoosterols. They maintain the fluidity of cell membrane, act as signalling molecules and also form the skin oils in animals.

Cholesterol is an important zoosterol. It is a fatty waxy substance. It is present in cell membrane. It is also a precursor for vitamin D. It is precursor for steroid hormones like cortisol and aldosterone. When it is non esterified, it gets converted to bile.  

Answer:

Option-C

Explanation:

Steroids are the hydrophobic molecules that can be structurally characterised by the fused rings and -OH group. The steroids since are hydrophobic therefore are kept with the phospholipids.

A sterol which is present as a component of the lipid layer is known as the cholesterol. The cholesterol acts as a precursor to a variety of steroid hormone-like estrogens,  glucocorticoids, progestagens and many others.  The cholesterol also acts as precursor to vitamin D and bile acids in the body.

Thus, Option-C is correct.

Answer: The amount of heat required for melting of benzene is 20.38 kJ

Explanation:

The processes involved in the given problem are:

[tex]1.)C_6H_6(s)(5.5^oC)\rightleftharpoons C_6H_6(l)(5.5^oC)\\2.)C_6H_6(l)(5.5^oC)\rightleftharpoons C_6H_6(l)(60.6^oC)[/tex]

To calculate the amount of heat required to melt the benzene at its melting point, we use the equation:

[tex]q_1=m\times \Delta H_{fusion}[/tex]

where,

[tex]q_1[/tex] = amount of heat absorbed = ?

m = mass of benzene = 91.5 g

[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 127.40 J/g

Putting all the values in above equation, we get:

[tex]q_1=91.5g\times 127.40J/g=11657.1J[/tex]

To calculate the amount of heat absorbed at different temperature, we use the equation:

[tex]q_2=m\times C_{p}\times (T_{2}-T_{1})[/tex]

where,

[tex]C_{p}[/tex] = specific heat capacity of benzene = 1.73 J/g°C

m = mass of benzene = 91.5 g

[tex]T_2[/tex] = final temperature  = 60.6°C

[tex]T_1[/tex] = initial temperature = 5.5°C

Putting values in above equation, we get:

[tex]q_2=91.5\times 1.73J/g^oC\times (60.6-(5.5))^oC\\\\q_2=8722.05J[/tex]

Total heat absorbed = [tex]q_1+q_2[/tex]

Total heat absorbed = [tex][11657.1+8722.05]J=20379.2=20.38kJ[/tex]

Hence, the amount of heat required for melting of benzene is 20.38 kJ

Final answer:

The amount of heat produced by the combustion of the benzene sample is 6.562 kJ.

Explanation:

To calculate the amount of heat produced by the combustion of the benzene sample, we can use the heat capacity of the bomb calorimeter and the change in temperature. The heat produced can be calculated using the formula: q = C × ΔT, where q is the heat produced, C is the heat capacity of the bomb calorimeter, and ΔT is the change in temperature. In this case, the heat capacity of the bomb calorimeter is 784 J/°C and the change in temperature is 8.39 °C.

First, we calculate the heat produced by the bomb calorimeter using the formula: q = C × ΔT. q = 784 J/°C × 8.39 °C = 6561.76 J. Next, we convert this to kilojoules by dividing by 1000: 6561.76 J ÷ 1000 = 6.562 kJ.

Therefore, the amount of heat produced by the combustion of the benzene sample is 6.562 kJ.

Answer:

Yes.

Explanation:

It should be noted that the meaning of molarity is the ratio of moles of solute per liter of solution.

It should be understood that when determining or finding the molarity of an unknown compound ,the process should be performed or carried out at least 3 times. This is done to remove any form of doubt.

The first calculated value for the concentration of the compound will be regarded as rough value, while the second and the third will be regarded as the first and second values respectively.

In this case, the third value for the concentration of HCl will be calculated to for confirmation of other value, that is to be finally sure of its concentration.  

Answer:

Yes.

Explanation:

For a typical experiment, you should plan to repeat it at least three times (more is better).

The value gotten in the first trial is not close to the value gotten from the second trial.

The solution to these problems is to do repeated trials. Repeated trials are when you do a measurement multiple times - at least three, commonly five, but the more the better. When you measure something once, the chance that the number you get is accurate is much lower. But if you measure it several times, you can take an average of those numbers and get a result that is much closer to the truth.

Repeating a trial gives a RELIABLE result.

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

Answer: The mass of oxygen that will react with same amount of carbon as in carbon monoxide is 23.31 grams.

Explanation:

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: [tex]Cu_2O\text{ and }CuO[/tex]

We are given:

Mass of oxygen in CO = 13.6 grams

Applying unitary method:

16 grams of oxygen are reacting with 12 grams of carbon

So, 13.6 grams of oxygen will be reacting with = [tex]\frac{12}{16}\times 13.6=10.2g[/tex] of carbon

Applying unitary method:

12 grams of carbon are reacting with 32 grams of oxygen

So, 10.2 grams of carbon will be reacting with = [tex]\frac{32}{12}\times 10.2=23.31g[/tex] of oxygen

Hence, the mass of oxygen that will react with same amount of carbon as in carbon monoxide is 23.31 grams.

In the given question, 27.2 grams of oxygen will react with 10.2 grams of carbon to give 39.1 grams of carbon dioxide.

The mass of oxygen given in the question is 13.6 grams. The molecular mass of oxygen or O2 is 32 g/mol.

The reactions taking place in the given case are:

Now the number of moles of O₂ will be,

[tex]Moles of O2= \frac{Weight}{Molecular mass}[/tex]

[tex]Moles = \frac{13.6}{32} \\Moles = 0.425 moles[/tex]

Based on the reaction, it can be seen that with 1 mole of O₂, 2 moles of C react to produce 2 moles of CO.

Now for 0.425 moles of oxygen,

The moles of C, that is, 0.425 × 2 = 0.850 moles will react 0.425 moles of O₂ to produce 0.850 moles of CO

Now the number of moles of C is 0.850 moles, and the molecular mass of C is 12 g/mol.

The mass of C will be,

Mass = Number of moles × Molecular mass

Mass = 0.850 × 12 = 10.2 grams

Now in the second reaction, that is, in the formation of CO₂, 1 mole of C will react with 1 mole of O₂ to produce 1 mole of CO₂.

Therefore, 0.850 moles of C will react with 0.850 moles of O₂ to give 0.850 moles of CO₂.

Now the molar mass of CO₂ is 46 g/mol, the mass of CO₂ will be,

Mass = 0.850 moles × 46 = 39.1 grams

The moles of O₂ is 0.850 moles, the molar mass of O₂ is 32 g/mol. Now the mass of O₂ is,

Mass = Number of moles × Molar mass

Mass = 0.850 × 32 g/mol

Mass = 27.2 grams

Thus, 27.2 grams of oxygen will react with 10.2 grams of carbon to produce 39.1 grams of CO₂.

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Answer:

[tex]\boxed{\text{36 $\mu$mol}}[/tex]

Explanation:

Data:

c = 1.5 x 10⁻⁴ mol/L

V = 240.0 mL

Calculations:

[tex]\text{Moles} = \text{0.2400 L} \times \dfrac{1.5 \times 10^{-4}\text{ mol}}{\text{1 L}} \times \dfrac{10^{6}\text{ $\mu$mol}}{\text{1 mol}} = \textbf{36 $\mu$mol}\\\\\text{The chemist has added $\boxed{\textbf{36 $\mu$mol}}$ of magnesium fluoride to the flask.}[/tex]

The pH changes significantly through the different stages of titration of triethylamine with HCl. After adding 11.00 mL of HCl, the pH is approximately 11.30, demonstrating a basic solution. At 20.60 mL and 25.00 mL of HCl added, the pH drops to approximately 2.83 and 1.95, respectively, indicating an acidic solution due to the excess HCl.

To find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, [((CH₃CH₂)₃N)] , with 0.1000 M HCl solution, we need to follow the steps below at different volumes of HCl addition:

(a) After adding 11.00 mL of HCl

Calculate moles of triethylamine:
moles of [((CH₃CH₂)₃N)] = 0.1000 M x 0.0200 L = 0.00200 mol

Calculate moles of HCl added:
moles of HCl = 0.1000 M * 0.0110 L = 0.00110 mol

Moles of triethylamine remaining:
0.00200 mol - 0.00110 mol = 0.00090 mol

Volume of solution after adding HCl:
20.00 mL + 11.00 mL = 31.00 mL or 0.0310 L

Concentration of triethylamine:
[((CH₃CH₂)₃N)] = 0.00090 mol / 0.0310 L ≈ 0.0290 M

Using  Kᵇ, find [OH⁻]:
Kᵇ = 5.2 x 10⁻⁴, set up ICE table for equilibrium calculation, estimate [OH⁻].

Find pOH and then pH:
pOH = -log[OH⁻]; pH = 14 - pOH

pH ≈ 11.30

(b) After adding 20.60 mL of HCl

Moles of HCl added:
0.1000 M x 0.0206 L = 0.00206 mol

Moles of triethylamine remaining:
0.00200 mol - 0.00206 mol = -0.00006 mol. This indicates an excess of HCl.

Moles of excess HCl:
0.00006 mol

Volume of solution:
20.00 mL + 20.60 mL = 40.60 mL or 0.0406 L

Concentration of H⁺:
[H⁺] = 0.00006 mol / 0.0406 L ≈ 0.00148 M

Find pH:
pH = -log[H⁺]

pH ≈ 2.83

(c) After adding 25.00 mL of HCl

The pH values during the titration are (a) 10.65 after adding 11.00 mL of HCl, (b) 3.97 at equivalence point (20.60 mL), and (c) 1.95 after adding 25.00 mL of HCl.

To find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH₃CH₂)₃N (Kb = 5.2 × 10⁻⁴), with 0.1000 M HCl solution after different volumes of titrant have been added, follow these steps:

Initial pH Calculation (before titration)

1. Calculate the pOH from Kb and the initial concentration of triethylamine:

Kb = 5.2 × 10⁻⁴

[B] = 0.1000 M

Using the formula:

[tex]& K_b = \frac{[\text{OH}^-][\text{BH}^+]}{[B]} \\[/tex]

We assume that [OH⁻] = [BH⁺]. Thus, [tex]K_b = \frac{[\text{OH}^-]^2}{[B]} \\[/tex]

[tex]& [\text{OH}^-] = \sqrt{K_b \cdot [B]} = \sqrt{5.2 \times 10^{-4} \cdot 0.1000} = 0.0072 \, \text{M} \\\\[/tex]

[tex]& \text{pOH} = -\log[\text{OH}^-] = -\log(0.0072) \approx 2.14 \\\\ & \text{pH} = 14 - \text{pOH} = 14 - 2.14 = 11.86 \\[/tex]

After Adding 11.00 mL of HCl

2. Calculate the moles of HCl added:

n(HCl) = M * V = 0.1000 M * 0.01100 L = 0.001100 mol

Initial moles of (CH₃CH₂)₃N = 0.1000 M * 0.02000 L = 0.002000 mol

Remaining (CH₃CH₂)₃N = 0.002000 mol - 0.001100 mol = 0.000900 mol

Moles of (CH₃CH₂)₃NH⁺ formed = 0.001100 mol

Total volume = 20 mL + 11 mL = 31 mL = 0.031 L

Compute the concentrations:

[B] = 0.000900 mol / 0.031 L ≈ 0.029 M

[HB+] = 0.001100 mol / 0.031 L ≈ 0.035 M

Using the Henderson-Hasselbalch equation:

[tex]pH = pKa + \log\left(\frac{[B]}{[\text{HB}^+]}\right)\\\\pKa = 14 - \text{pKb} = 14 - 3.28 = 10.72\\\\pH = 10.72 + \log\left(\frac{0.029}{0.035}\right) = 10.72 + \log(0.829) \approx 10.65[/tex]

At Equivalence Point (20.60 mL)

3. Calculate moles at equivalence point:

Moles of  (CH₃CH₂)₃N = 0.002000 mol

Equivalence moles of HCl = 0.002060 mol

All the base has been neutralized:

(CH₃CH₂)₃NH⁺ in solution = 0.002060 / 0.04060 = 0.0507 M

Calculate pH:

[H+] from the equilibrium of (CH₃CH₂)₃NH⁺ :

[tex]K_a \text{ for } (CH_3CH_2)_3NH^+ = \frac{1}{K_b} = \frac{1}{5.2 \times 10^{-4}} = 1.923 \times 10^3[/tex]

[tex]& [\text{H}^+] = \sqrt{1.923 \times 10^3 \times 0.0507 \, \text{M}} = 0.09825 \, \text{M} \\\\& \text{pH} = -\log(0.09825) \approx 3.97 \\[/tex]

After Adding 25.00 mL of HCl

4. Calculate moles of excess HCl:

HCl added (total) = 0.002500 moles

Excess HCl = 0.002500 - 0.002000 = 0.000500 mol

Total volume = 20 mL + 25 mL = 45 mL = 0.045 L

[tex]& [\text{H}^+] = \frac{0.000500 \, \text{mol}}{0.045 \, \text{L}} \approx 0.0111 \, \text{M} \\[/tex]

[tex]& \text{pH} = -\log(0.0111) \approx 1.95 \\[/tex]

In summary, the pH values during the titration are 10.65 after adding 11.00 mL of HCl, 3.97 at equivalence point (20.60 mL), and 1.95 after adding 25.00 mL of HCl.

Answer:

0.089L

Explanation:

Mass of copper= 800g

Density of copper= 8.96g/ml or 8960g/L

Density = mass/volume

Volume = mass/density = 800/8960= 0.089L

The conversion of density from g/ml to g/L units was necessary because the volume was required in liters according to the statement in the question

Answer:

0.1724 M is the concentration of magnesium ions in this solution.

Explanation:

Mass of magnesium hydroxide gas = 10.00 g

Molar mass of magnesium hydroxide = 58 g/mol

Moles of magnesium hydroxide = [tex]\frac{10.00 g}{58 g/mol}=0.1724 mol[/tex]

Volume of the solution = V = 1.00 L

Molarity or concentration of the magnesium hydroxide:M

[tex]M=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex]

[tex]M=\frac{0.1724 mol}{1.00 L}=0.1724 M[/tex]

1 mole of magnesium hydroxide contains 1 mole of magnesium ion.Then 0.1724 M of magnesium hydroxide will have :

[tex][Mg^{2+}]=1\times 0.1724 M=0.1724 M[/tex]

0.1724 M is the concentration of magnesium ions in this solution.

The concentration of magnesium hydroxide in the solution is 0.1714 Molar. This is calculated by dividing the number of moles of magnesium hydroxide by the volume of the solution in the volumetric flask.

The subject of this question falls under Chemistry, and it's asking about calculating concentration, specifically for a solution of magnesium hydroxide. In chemistry, concentration is commonly expressed in moles per liter (M). It's necessary to convert the mass of magnesium hydroxide to moles by using its molar mass.

First, to solve this question, we need to know the molar mass of magnesium hydroxide (Mg(OH)2), which is about 58.3197 grams/mole. Dividing the given mass of magnesium hydroxide (10.00g) by the molar mass gives us the number of moles. Thus, 10.00 g / 58.3197 g/mol ≈ 0.1714 mol of Mg(OH)2.

The whole solution was made up in a 1.00 L volumetric flask. Concentration is defined as the amount of solute per unit volume of solvent. Hence, the concentration of Mg(OH)2 is 0.1714 mol / 1.00 L = 0.1714 M (molar).

On note, the temperature given in the question (25◦C) does not affect the calculation of the concentration in this case, nor does the pH of the water used.

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Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

[tex]2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)[/tex]

[tex]\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles[/tex]

[tex]\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol[/tex]    

According to stoichiometry :

2 moles of [tex]NaCl[/tex] are formed from = 2 moles of [tex]Na[/tex]

Thus 16 moles of [tex]NaCl[/tex] are formed from=[tex]\frac{2}{2}\times 16=16moles[/tex]  of [tex]Na[/tex]

Mass of [tex]Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g[/tex]

Thus 368 grams of sodium reacted.

The question is incomplete, complete question is ;

Allura Red (AR) has a concentration of 21.22 ppm. What is this is micro moles per liter? Report the precise concentration of the undiluted stock solution #1 of AR in micromoles per liter. This is your most concentrated (undiluted) standard solution for which you measured the absorbance. Use 3 significant figures. Molarity (micro mol/L) =

Answer:

The molarity of the solution of allura red is 42.75 micro moles per Liter.

Explanation:

The ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

[tex]\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6[/tex]

Both the masses are in grams.

We are given:

The ppm concentration of allura red = 21.22 ppm

This means that 21.22 mg of allura red was present 1 kg of solution.

Mass of Allura red = 21.22 mg = [tex]21.22\times 0.001 g[/tex]

1 mg = 0.001 g

Mass of solution = 1 kg = 1000 g

Density of the solution = Density of water = d = 1.00 g/mL

( since solution has very small amount of solute)

Volume of the solution :

[tex]=\frac{1000 g}{1.00 g/mL}=1000 mL[/tex]

1000 mL = 1 L

Volume of the solution, V = 1 L

Moles of Allura red = [tex]\frac{21.22\times 0.001 g}{496.42 g/mol}=4.275\times 10^{-5} mol=4.275\times 10^{-5}\times 10^{6} \mu mole[/tex]

Molarity of the solution ;

[tex]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}[/tex]

[tex]M=\frac{4.275\times 10^{-5}\times 10^6 \mu mol}{1 L}=42.75 \mu mol/L[/tex]

The molarity of the solution of allura red is 42.75 micro moles per Liter.

Mitochondria is the site of ATP synthesis in eukaryotic cell during aerobic respiration. It regulates the biochemical processes of the cell and helps in generation of co enzymes as sulphur and iron.

Explanation:

Mitochondria is a cell membrane bound organelles present in cytoplasm of eukaryotic cells.

The major roles of mitochondria in eukaryotic cell is to synthesize ATP by cellular respiration. It also regulates the metabolism or biochemical processes of the cell.

Mitochondria also aids in generating the iron and sulphur which acts as a coenzyme in several biochemical reactions.

The electron transport chain takes place in inner mitochondrial membrane.

There are 2 aerobic phases of cellular respiration those are Kreb's cycle and electron transport chain. The machinery for Kreb's Cycle is in matrix. The ETC is present as embedded in the inner membrane of matrix.

The cellular compartment of mitochondria performs following functions:

The space in intermembrane has electron transport chain and holds ATP synthase (enzyme required for

The cristae is a finger like projection which increases surface area for increase ATP synthesis.

In matrix of the mitochondria the Electron transport takes place.

In Kreb's cycle mitochondria aids in production of NADH and GTP. Also, synthesis of phospholipid takes place in mitochondria.

Answer:

328.4KJ

Explanation:

Before we move on to calculate enthalpy change, we calculate the amount of heat Q

Q= mcΔT

m = density * volume = 250 * 1.25 = 312.5g

c = 3.74J/g.k

ΔT = 7.80 + 273.15K = 280.95K

Q= 312.5 * 3.74 * 280.95 = 328,360.312 J= 328.4KJ(1000J = 1KJ, so divide by 1000)

The enthalpy change in the reaction is same as amount of heat transferred = 328.4KJ

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = [tex]2 \times 0.475[/tex]

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = [tex]2 \times 0.025[/tex]

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         [tex]C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O[/tex]

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = [tex]pK_{a} + log \frac{Base}{Acid}[/tex]  

              = [tex]4.2 + log \frac{0.05}{0.045}[/tex]

              = 4.245

For,  

         [tex]HCOOH + NaOH \rightarrow HCOONa + H_{2}O[/tex]

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = [tex]pK_{a} + log \frac{Base}{Acid}[/tex]  

          4.245 = 3.75 + [tex]log \frac{Base}{Acid}[/tex]

      [tex]log \frac{Base}{Acid}[/tex] = 0.5

    [tex]\frac{Base}{Acid}[/tex] = 3.162

Now,

        [tex]\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}[/tex] = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

We can arrive at this answer through the following calculation:

Amount of moles of NaOH: [tex]2 * 0.025 = 0.05 mol[/tex]

Amount of moles of benzoic acid: [tex]2*0.475=0.095mol[/tex]

[tex]pH=pK{a}+log\frac{base}{acid}[/tex]

[tex]4.2+log\frac{0.05}{0.045}=4.245[/tex]

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows:

[tex]pH=pK{a} +log\frac{base}{acid} \\4.245=3.75+log\frac{base}{acid} \\log\frac{base}{acid}=0.5\\\frac{base}{acid} = 3.162[/tex]

[tex]\frac{2(0.5-x)}{0.2x-2(0.5-x)} =0.464L[/tex]

NaOH volume:

[tex](0.5-0.464) L\\0.036L ----- 36mL[/tex]

HCOOH volume:

[tex]500 mL-36mL = 464 mL[/tex]

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Answer:

81.7 %

Explanation:

Equation of the decomposition of NaHCO₃

2 NaHCO₃ → Na₂CO₃ + CO₂ + H₂O

2 mole of NaHCO₃ yielded 1 mole of CO₂ and 1 mole of  H₂O

molar mass of CO₂ = 44 g

molar mass of H₂O = 18 g

1 mole of CO₂ : 1 mole H₂O  = ( mass of CO₂ / molar mass of CO₂) : ( mass of H₂O / molar mass of  H₂O

also  mass of CO₂ + mass of H₂O = ( 0.654 - 0.456 ) g = 0.198 g

1 = ( mass of CO₂/ 44g) : (  mass of H₂O / 18g)

44 mass of H₂O = 18 mass of CO₂

1 mass of CO₂ = 44 / 18  mass of H₂O

substitute into equation 1

mass of CO₂ + mass of H₂O = 0.198 g

2.44 mass of H₂O + mass of H₂O = 0.198 g

3.44 mass of H₂O = 0.198 g

mass of H₂O = 0.198 g / 3.44 = 0.0576 g

mass of CO₂ = 0.198 g  - 0.0576 g  = 0.140 g

2 mole of  NaHCO₃ yielded 1 mole of CO₂

168 g of NaHCO₃ yielded 44 g of CO₂

unknown mass of NaHCO₃ yielded 0.140 g CO₂

unknown mass of NaHCO₃ = 168 g × 0.140 g / 44 g = 0.535 g

mass percent of NaHCO₃ = 0.535 g / 0.654 g = 81.7 %

Answer:

Explanation:

Mass of impure NaHCO3 = 0.654g

Mass of residue= 0.456g

Mass loss of NaHCO3= 0.654-0.456= 0.198g

Balanced reaction equation:

2NaHCO3(s)-------> Na2CO3(s) + H2O(g)+CO2(g)

Note CO2 and water vapour produced in the decomposition combines to give H2CO3

84g of NaHCO3 yields 62g of H2CO3

Xg of NaHCO3 yields 0.198g of H2CO3

Therefore X= 84 × 0.198/ 62

=0.268g

Mass%= 0.268/0.654 × 100

=41% NaHCO3

Answer: the correct stoichiometric coefficients will be

3:2= 1:6

Explanation:

3Ba(NO3)2 (aq) + 2Na3PO4 (aq) → Ba3(PO4)2 (s) + 6NaNO3 (aq)

From the equation of reaction,

3 moles of Ba(NO3)2 (aq) will require 2 moles of Na3PO4 , to give 1 mole of Ba3(PO4)2 (s) and 6moles of NaNO3 (aq)