Mercury and oxygen react to form mercury(II) oxide, like this: 2 Hg(l)+02(g)--HgO(s) At a certain temperature, a chemist finds that a 6.9 L reaction vessel containing a mixture of mercury, oxygen, and mercury(II) oxide at equilibrium has the following composition: compound amount Hg 16.9 g O 10.9 g HgO 23.8 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer:

1.315x10⁻³M = [Ca²⁺]

Explanation:

Based in the reaction:

Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)

Solubility product, ksp, is defined as:

ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²

From 1 mole of hydroxyapatite are produced  10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:

6/10 Ca²⁺ = PO₄³⁻

Replacing in ksp formula:

ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²

As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:

2.34x10⁻⁵⁹ =  [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²

3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶

1.315x10⁻³M = [Ca²⁺]

I hope it helps!

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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Final answer:

To calculate the pH of the given aqueous solutions at 25°C, we can use the formula pH = -log[H3O+]. For NaOH, LiOH, and Ba(OH)2, the concentration of hydroxide ions is equal to the concentration of the base, which allows us to find the pH.

Explanation:

To calculate the pH of an aqueous solution, we need to determine the concentration of the hydronium ions (H3O+). Since NaOH, LiOH, and Ba(OH)2 are strong bases that ionize completely, we can assume that the concentration of the hydroxide ions (OH-) is equal to the concentration of the base. To find the pH, we can use the formula:

pH = -log[H3O+]

For example:

For 9.5 × 10^(-8) M NaOH, the concentration of hydroxide ions is 9.5 × 10^(-8) M. Taking the negative logarithm, the pH is approximately 7.02.

For 6.3 × 10^(-2) M LiOH, the concentration of hydroxide ions is 6.3 × 10^(-2) M. Taking the negative logarithm, the pH is approximately 11.20.

For 6.3 × 10^(-2) M Ba(OH)2, the concentration of hydroxide ions is twice the concentration of the base, so it is 2 * 6.3 × 10^(-2) M = 1.26 × 10^(-1) M. Taking the negative logarithm, the pH is approximately 1.90.

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: 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.

Find out more information about calculations based on the law of multiple proportions here:

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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.

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.

Explanation:

We will calculate the specific gravity of the block as follows.

    Specific gravity of block = [tex]\frac{\text{block vol below}}{\text{total block vol } \times \text{specific gravity of water}}[/tex]

                   = [tex]\frac{1.5}{(1.5 + 0.5) \times 1}[/tex]

                   = 0.75

And, the specific gravity of the solution is as follows.

 Specific gravity of solution = [tex]\frac{\text{total block vol}}{\text{block vol below} \times \text{specific gravity of block}}[/tex]

            = [tex]\frac{2}{1 \times 0.75}[/tex]

            = 1.5

As the given block is rectangular, and its volume is directly proportional to its height and A is not needed in these calculations.

Also, we can note here that the term specific gravity is no longer approved and has been replaced by the term relative density.

The specific gravity of the block is 0.75 and the specific gravity of the solution is also 0.75. This is calculated using the principles of fluid dynamics and buoyancy, specifically Archimedes' principle of equal weight displacement.

When analysing fluid dynamics and buoyancy, the density of an object and the fluid it's submerged in is key. According to Archimedes' principle, the weight of the block equals the weight of the water displaced when it is fully submerged. Therefore, mass of the block is the total volume of the block (2 in) times density of the block, and mass of the water displaced is (1.5 in) times the density of the water. Thus, density of the block would be (0.75/1) times density of the water, making the specific gravity of the block 0.75. In the aqueous solution, the block floats with 1 in below the surface, meaning specific gravity in relation to the solution equals 1 (since 1 in/1 in = 1), therefore, the specific gravity of the aqueous solution equals 0.75, the specific gravity of the block.

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Explanation:

For the given reaction equation, we will write the state of each specie as follows.

       [tex]Ag_{2}S(s) \rightarrow 2Ag^{+}(aq) + S^{2-}(aq)[/tex]

Since, silver sulfide ([tex]Ag_{2}S[/tex]) will remain in solid state. Therefore, it acts as a precipitate, that is, insoluble solid. Hence, it is insoluble.

And, expression for the equilibrium constant of this reaction is as follows.

         [tex]K_{eq} = \frac{[Ag^{+}]^{2}[S^{2-}]}{[AgS]^{2}}[/tex]

For solids, it is considered to be equal to 1. Hence, the equilibrium constant expression will be as follows.

            [tex]K_{eq} = [Ag^{+}]^{2}[S^{2-}][/tex]

Therefore, we can conclude that its equilibrium constant for this reaction will be greater than 1.

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.

Explanation:

As Iron (III) nitrate is an ionic compound so when it is dissolved in water then it will dissociate to give nitrate ions and ferric ions. Also, we know that like dissolves like and here iron (III) nitrate being an ionic compound will readily dissolve in water (polar solvent).

The chemical equation for this reaction is as follows.

   [tex]Fe(NO_{3})_{3}(s) + H_{2}O(l) \rightarrow Fe^{3+}(aq) + 3NO^{-}_{3}(aq)[/tex]  

Hence, iron(III) nitrate considered soluble in water.

Now, equilibrium constant expression for this reaction is as follows.

           [tex]K_{eq} = [Fe^{3+}][(NO_{3})_{3}]^{3}[/tex]

Therefore, the equilibrium constant for this reaction will be greater than 1.

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)

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]

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:

[tex]NH^{+} _{4}[/tex]  ammonium ion.

Explanation:

Check the box next to each molecule on the right that has the shape of the model molecule on the left: model molecules (check all that apply X 5 ? | O CH20 CNH, You can drag the slider to rotate the model molecule. + 1 O Brf 4 CH2Cl2 Note for advanced students: the length of bonds and size of atoms in the model is not necessarily realistic. The model is only meant to show you the general geometry and 3D shape of the molecule.

when you look at the diagram from the source page

one can conclude that the diagram is tetrahedral and the angle between the molecules is 109.5 deg

There is one central atom bonded to four atoms in a tetrahedral molecule .it has no lone electron pairs.m

NH4+ is the answer

Other molecules that are tetrahedral in shape are methane ion and phosphate ion.

Answer:

CH3O- , BrF4- and NH4+ have tetrahedral geometry on the basis of their electron domain geometry..

Explanation:

The object on the picture as shown on the fig below describes a compound with a total of 4 pair electron domain with it's electron domain typically described as tetrahedral.

The task is to sort out which of those in the options fort into the category.

Although NH3 and NH4+ ion both have the SP3 hybridization their electron pair geometry differs. In the NH3 molecule one lone pair and three bond pairs are present. While Distortion is caused by repulsion between lone pair and bond pair the geometry of NH3 causes it to become pyramidal in NH4+ irrespective of possessing the sp3 hybridization.

Its resulting trigonal pyramidal geometry is thus described as tetrahedral. It consequently has 3 bonding domains and 1 nonbinding domain.

CH3O- is also tetrahedral with an idealized bond angle of 109.5°.

BrF4- It has 4 bond pair present hence the tetrahedral geometry. The presence of two lone pair makes it square planar described sometimes as AE2X4. It has 6 electron regions.

C2Cl4 has a linear geometry it has one triple bond and two single bonds this giving hints that its coordinate and steric number is 2 and its bond angle is 180°.

So, CH3O- , BrF4- and NH4+ have tetrahedral geometry

Answer:

Unemployed

Explanation:

(of a person) without a paid job but available to work.

"I was unemployed for three years"   - ----- Example

Similar:

jobless

out of work

out of a job

not working

between jobs .

Answer:

sorry but I am just answering the questions because I need points

Explanation:

thank you

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.

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after [tex]4.53\times 10^{10} s[/tex] is 0.00345 mol/L.

Explanation:

[tex]2HI(g)\rightarrow H_2(g)+I_2(g) [/tex]

Rate Law: [tex]k[HI]^2 [/tex]

Rate constant of the reaction = k = [tex]6.4\times 10^{-9} L/mol s[/tex]

Order of the reaction = 2

Initial rate of reaction = [tex]R=1.6\times 10^{-7} Mol/L s[/tex]

Initial concentration of HI =[tex][A_o][/tex]

[tex]1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2[/tex]

[tex][A_o]=5 mol/L[/tex]

Final concentration of HI after t = [A]

t = [tex]4.53\times 10^{10} s[/tex]

Integrated rate law for second order kinetics is given by:

[tex]\frac{1}{[A]}=kt+\frac{1}{[A_o]}[/tex]

[tex]\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}[/tex]

[tex][A]=0.00345 mol/L[/tex]

The concentration of HI after [tex]4.53\times 10^{10} s[/tex] is 0.00345 mol/L.

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]

Question:

Imagine a six-pound bowling ball. That's generally the lightest you'll find in a bowling alley.  Also imagine six pounds of feathers.  Your average feathered bedroom pillow weighs about 2 pounds, so imagine three pillows in a stack. You now have six pounds of bowling ball and six pounds of feathers. If you were to take each and throw them into a swimming pool, you would expect the bowling ball to sink and the feathers to float, but why?First, let's figure out their volume.A bowling ball has a radius of 4.25 in, which converts to 10.8 cm.  Using the formula to find the volume of a sphere:

Note; The result is 322 cm³(= error by calculating in 4.25 cm instead).  

Next, the standard size for a bedroom pillow is 26 in x 20 in x 4 in.  When we convert to centimeters, our stacked pillows are 66 cm x 51 cm x 10 cm.  The volume is then 33,660 cm³. We already know the mass of each object (six pounds), but remember that mass in this formula is measured in grams.  You'll first need to convert six pounds to grams.  

One pound converts to 453.6 g.  Round your calculation to the nearest full gram and write it down.Now use the density formula to calculate the density of each object.  Round your answers to the nearest one-tenth of a gram (one decimal place).

Bowling Ball: 0.1 g/cm³Feathers: 8.5 g/cm³ Correct!

Bowling Ball: 8.5 g/cm³ Feathers: 0.1 g/cm³

Bowling Ball: 0.1 lb/in³ Feathers: 8.5 lb/in³

Bowling Ball: 8.5 g/in³ Feathers: 0.1 g/in³

Answer:

The correct answer to the question is

b. Bowling Ball: 8.5 g/cm³ Feathers: 0.1 g/cm³.

Explanation:

To solve the question, we note  that both the sphere and the Bowling Ball have a mass of 6 lb which is equal to 2721.554 g

The radius of a bowling ball is 4.25 cm = ‬ 1.67 in which gives its volume as

322 cm³

The mass of the bowling ball 6 lb while

The volume of a feathered pillow = 26 in × 20 in × 4 in  = 2080 in³  = 58899040.911 cm³

The density of the Bowling Ball = mass/volume =  2721.554 g/322 cm³

= 8.452 g/cm³

The density of the feathers = mass/volume

=  2721.554 g/34085 cm³ = 0.0798 g/cm³ which to one decimal place

= 0.1 g/cm³

Therefore the density of the Bowling Ball to the nearest one-tenth

= 8.5 g/cm³     and                                        

The density of the feathers to the nearest one-tenth = = 0.1 g/cm³

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:

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:

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:

ClO4-, and Ba2+.

Explanation:

A spectator ion is an ion that exists as a reactant and a product in a chemical equation.

Equation of the reaction.

2HClO4 + Ba(OH)2 --> Ba(ClO4)2 + 2H2O

ClO4- + OH- --> 2ClO4- + H2O

The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ClO₄⁻ and Ba⁺². These ions do not participate in the reaction and remain unchanged in the solution after the reaction.

The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ions that do not participate in the chemical reaction and do not change during the course of the reaction. Aqueous solutions of acids and bases typically dissociate into their constituent ions in water. For perchloric acid (HClO₄), it dissociates into H⁺ and ClO⁴⁻ ions, while barium hydroxide (Ba(OH)₂) dissociates into Ba²⁺ and OH⁻ ions.

In a neutralization reaction, the H+ ions from the acid combine with the OH⁻ ions from the base to form water (H₂O), meaning they are not spectator ions. Therefore, the spectator ions would be the ions that remain unchanged, which are ClO⁴⁻ and Ba²⁺. The complete ionic equation for this reaction would show all the ions separated, but the net ionic equation would exclude these spectators, focusing only on the ions that participate directly in forming the product, water.

The molecular equation for the reaction between hydroiodic acid (HI) and potassium hydroxide (KOH) is:

HI(aq) + KOH(aq) ⇒ KI(aq) + H₂O(l).

The molecular equation represents the chemical reaction between hydroiodic acid (HI) and potassium hydroxide (KOH). In the aqueous phase (aq), hydroiodic acid dissociates into hydrogen ions (H+) and iodide ions (I-), while potassium hydroxide dissociates into potassium ions (K+) and hydroxide ions (OH-).

During the reaction, the hydrogen ions from hydroiodic acid react with the hydroxide ions from potassium hydroxide, forming water (H₂O) in liquid (l) state. Additionally, the remaining potassium ions from potassium hydroxide and iodide ions from hydroiodic acid combine to form potassium iodide (KI) in aqueous state. This balanced equation illustrates the rearrangement of ions and atoms during the chemical reaction,

HI(aq) + KOH(aq) ⇒ KI(aq) + H₂O(l).

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The reaction between hydroiodic acid and potassium hydroxide forms potassium iodide and water. The balanced molecular equation is HI(aq) + KOH(s) → KI(aq) + H₂O(l). All the physical states are indicated for each compound.

In this reaction, hydroiodic acid (HI) is added to solid potassium hydroxide (KOH).

The reaction can be summarized as follows:

Correct question is: Write the overall molecular equation for the reaction of hydroiodic acid ( HI ) and potassium hydroxide. Include physical states. Enter the formula for water as H₂O .

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

Answer: 8.38 seconds

Explanation:

Integrated rate law for second order kinetics is given by:

[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]

[tex]a_0[/tex] = initial concentartion = 0.860 M

a= concentration left after time t = 0.230 M

k = rate constant =[tex]0.380M^{-1}s^{-1}[/tex]

[tex]\frac{1}{0.860}=0.380\times t+\frac{1}{0.230 }[/tex]

[tex]t=8.38s[/tex]

Thus it will take 8.38 seconds for the concentration of  A to decrease from 0.860 M to 0.230 M .

Final answer:

To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M in a second-order reaction, we can use the integrated rate law.

Explanation:

The reaction in question is second order and the rate constant is 0.380 M⁻¹⋅s⁻¹ at 300 ∘C. To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M, we can use the integrated rate law for a second-order reaction:

t = 1 / (k * [A])

Substituting the given values:

t = 1 / (0.380 M⁻¹⋅s⁻¹ * 0.860 M)

t ≈ 2.80 s

The mass of magnesium oxide is greater than that of the Magnesium ribbon because the Magnesium ribbon reacts with oxygen in the air to form magnesium oxide. The extra mass is from the oxygen atoms.

When comparing the mass of the Mg ribbon and the magnesium oxide, you may notice that the mass of the magnesium oxide seems to be greater. This apparent increase in mass can be attributed to a chemical reaction. In this case, when the Mg ribbon is exposed to air, it reacts with the oxygen present in the air to create magnesium oxide (MgO).

The increase in mass is due to the additional weight of the oxygen atoms that are now part of the compound. Moreover, smaller pieces of magnesium metal will react more rapidly than larger pieces because there is more reactive surface available. The increase in mass is thus directly related to the magnesium oxide formation with the interaction of magnesium and oxygen.

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The increase in mass of Magnesium Oxide compared to Magnesium ribbon is due to the incorporation of Oxygen during the oxidation process which forms Magnesium Oxide.

In comparing the mass of a Magnesium (Mg) ribbon and that of Magnesium Oxide (MgO), it's observed that the mass of the MgO is greater. This apparent increase in mass can be attributed to the chemical reaction that takes place when Magnesium reacts with Oxygen in the air to form Magnesium Oxide.

Magnesium + Oxygen -> Magnesium Oxide

Magnesium (Mg) atoms combine with Oxygen (O) molecules from the air. As a result, the atoms bond to form Magnesium Oxide (MgO) which results in an increase in mass since the mass of the Oxygen is now being taken into consideration.

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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]