The state of system is changed isobarically until itjustbecomes saturated. What is thetemperature of the system?

Answer:

The empirical formula is = [tex]CCl_3[/tex]

The molecular formula = [tex]C_2Cl_6[/tex]

Explanation:

[tex]Moles =\frac {Given\ mass}{Molar\ mass}[/tex]

% of C = 10.13

Molar mass of C = 12.0107 g/mol

% moles of C = 10.13 / 12.0107 = 0.8434

% of Cl = 89.87

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 89.87 / 35.453 = 2.5349

Taking the simplest ratio for C and Cl as:

0.8434 : 2.5349

= 1 : 3

The empirical formula is = [tex]CCl_3[/tex]

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12*1 + 3*35.5 = 118.5 g/mol

Molar mass = 237 g/mol

So,  

Molecular mass = n × Empirical mass

237 = n × 118.5

⇒ n ≅ 2

The molecular formula = [tex]C_2Cl_6[/tex]

1. The empirical formula of the compound containing 10.13% C and 89.87% Cl is CCl₃

2. The molecular formula of the compound is C₂Cl₆

Divide by their molar mass

C = 10.13 / 12 = 0.844

Cl = 89.87 / 35.5 = 2.532

Divide by the smallest

C = 0.844 / 0.844 = 1

Cl = 2.532 / 0.844 = 3

Thus, the empirical formula of the compound is CCl₃

Molecular formula = n × empirical = molar mass

[CCl₃]n = 237

[12 + (3×35.5)]n = 237

118.5n = 237

Divide both side by 118.5

n = 237 / 118.5

n = 2

Molecular formula = [CCl₃]n

Molecular formula = [CCl₃]₂

Molecular formula = C₂Cl₆

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Answer:

[Na₂S₂O₃] = 0.83 M

[Na₂S₂O₃] = 0.86 m

Mole fraction  = 0.015

Explanation:

Na₂S₂O₃ 12 % by mass. This data means, that 12 g of solute are contained in 100 g of solution.

Let's find out the volume of solution, with density to determine molarity.

Solution density = Solution mass / Solution volume

1.1003 g/cm³ = 100 g / Solution volume

100 g / 1.1003 g/cm³ = Solution volume → 90.88 mL (1cm³ = 1mL)

Now, that we have volume, we can calculate molarity

Molarity is mol/L

90.88 mL = 0.09088 L

12 g / 158.12 g/mol = 0.0759 moles

0.0759 moles / 0.09088 L = 0.83 M

Total mass of solution = 100 g

12 g + Solvent mass = 100 g

Solvent mass = 100 g - 12 g → 88 g

Molality = moles of solute /1kgof solvent

88 g = 0.088 kg

0.0759 moles / 0.088 kg = 0.86 m

As solvent mass is 88 g, let's determine solvent's moles for mole fraction

88 g / 18 g/mol = 4.89 moles

Mole fraction = moles of solute / moles of solutes + moles of solvent

Mole fraction = 0.0759 mol / 0.0759 mol + 4.89 moles = 0.015

Answer : The concentration of [tex]Cl_2O_5[/tex] in the vessel 0.820 seconds later is, 0.16 M

Explanation :

The given reaction is:

[tex]2Cl_2O_5(g)\rightarrow 2Cl_2(g)+5O_2(g)[/tex]

The rate law expression is:

[tex]rate=(6.48M^{-1}s^{-1})[Cl_2O_5]^2[/tex]

The expression used for second order kinetics is:

[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]

where,

k = rate constant = [tex]6.48M^{-1}s^{-1}[/tex]

t = time = 0.820 s

[tex][A_t][/tex] = final concentration = ?

[tex][A_o][/tex] = initial concentration = 1.16 M

Now put all the given values in the above expression, we get:

[tex]6.48\times 0.820=\frac{1}{[A_t]}-\frac{1}{1.16}[/tex]

[tex][A_t]=0.16M[/tex]

Therefore, the concentration of [tex]Cl_2O_5[/tex] in the vessel 0.820 seconds later is, 0.16 M

Answer:

In the final solution, the concentration of sucrose is 0.126 M

Explanation:

Hi there!

The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:

Ci · Vi = Cf · Vf

Where:

Ci = concentration of the original solution

Vi = volume of the solution taken to prepare the more diluted solution.

Cf = concentration of the more diluted solution.

Vf = volume of the more diluted solution.

For the first dillution:

26.6 ml · 2.50 M = 50.0 ml · Cf

Cf = 26.6 ml · 2.50 M / 50.0 ml

Cf = 1.33 M

For the second dilution:

16.0 ml · 1.33 M = 45.0 ml · Cf

Cf = 16.0 ml · 1.33 M / 45.0 ml

Cf = 0.473 M

For the third dilution:

20.0 ml · 0.473 M = 75.0 ml · Cf

Cf = 20.0 ml · 0.473 M / 75.0 ml

Cf = 0.126 M

In the final solution, the concentration of sucrose is 0.126 M

Answer:

  259.497 mg,   58.84%

Explanation:

BaSO₄ → Ba²⁺ + SO₄²⁻

to calculate the mole of BaSO₄

mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol

comparing the mole ratio

1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺

403 mg BaSO₄  yields     ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺

441 mg BaSO₄  will yield   ( 1.7268 × 137.327  × 441 mg ) / 403 mg = 259 .497 mg

mas percentage of the Barium compound = 259 .497 mg / 441 mg × 100 = 58.84%

Answer:

[tex]V=591.748 L[/tex]

Explanation:

Assumption:

Ideal Vapors/Ideal gas

Formula for ideal Gas:

[tex]PV=nR_uT[/tex]

Where:

P is the pressure

V is the Volume

n is the number of moles = m/M

R_u is Universal Gas Constant=0.08314 L*bar/(K*mol)

T is the temperature in Kelvin

Calculating Number of moles n:

n=Mass/Molar Mass

[tex]Mass=\rho_L*Volume\\Mass=0.692*(4000 cm^3)........... (4 liter * 1000cm^3/Liters =4000 cm^3)\\Mass=2768 g[/tex]

Molar Mass of gasoline=114g/mol

[tex]n=\frac{2768}{114} \\n=24.2807 moles[/tex]

Now:

[tex]PV=nR_uT[/tex]

[tex]V=\frac{nR_uT}{P}\\V=\frac{24.2807*0.08314*293}{1 bar}\\V=591.748 L[/tex]

Answer:

33.7 kg

Explanation:

Let's consider calcium phosphate Ca₃(PO₄)₂.

The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol and the molar mass of P is 30.97 g/mol. In 1 mole of Ca₃(PO₄)₂ (310.18 g) there are 2 × 30.97 g = 61.94 g of P. The mass of Ca₃(PO₄)₂ that contains 3.57 kg (3.57 × 10³ g) of P is:

3.57 × 10³ g × (310.18 g Ca₃(PO₄)₂/61.94 g P) = 1.79 × 10⁴ g Ca₃(PO₄)₂

A particular ore contains 53.1% calcium phosphate. The mass of the ore that contains 1.79 × 10⁴ g of Ca₃(PO₄)₂ is:

1.79 × 10⁴ g Ca₃(PO₄)₂ × (100 g Ore/ 53.1 g Ca₃(PO₄)₂) = 3.37 × 10⁴ g Ore = 33.7 kg Ore

Answer:

4-chloro-4-methyl-cyclohexene.

Explanation:

Hello,

On the attached picture you will find the chemical reaction forming the required product, 4-chloro-4-methyl-cyclohexene. In this case, according to the Markovnicov’s rule, it is more likely for the chlorine to be substituted at the carbon containing the methyl radical in addition to the hydrogen to the next carbon to break the double bond and yield the presented product.

Best regards.

The dienes are the compounds comprised of two double bonds in the structure. The reaction of sodium phenoxide with HCl produced phenol and the salt.

The substitution reaction is defined as the reaction in which the element is substituted or replaced by the reactive other element.

In diene molecules, the double bonds add the partial charge to the compound, and  the HCl or the other reactant in the product is diverted to join at the double bonds in the diene.

The image attached is the reaction of sodium phenoxide with the HCl. The reaction carried the process of substitution of H at the place of  Na, with the formation of phenol.

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Explanation:

Relation between [tex]k_{f}[/tex], molality and temperature is as follows.

                 T = [tex]K_{f} \times m \times i[/tex]

It is also known as depression between freezing point where, i is the Van't Hoff factor.

Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.

          i for [tex]CaCl_{2}[/tex] = 3

           i for glucose = 1

           i for NaCl = 2

Depression in freezing point will have a negative sign. Therefore, d

depression in freezing point for the given species is as follows.

        [tex]T_{CaCl_{2}} = -1.86 \times 0.45 \times 3[/tex]

                  = [tex]-2.511^{o}C[/tex]

        [tex]T_{glucose} = 1.86 \times 0.45 \times 1[/tex]

                   = [tex]-0.837^{o}C[/tex]

       [tex]T_{NaCl} = -1.86 \times 0.45 \times 2[/tex]

                    = [tex]-1.674^{o}C[/tex]

Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.

                     Glucose < NaCl < [tex]CaCl_{2}[/tex]

Answer: option B. there are 35 mg of lead in 1.0 L of this solution

Explanation:

1ppm = 1mg/L

Therefore 35ppm = 35mg/L

So, the solution contains 35mg of lead in 1L of the solution

Answer:

see explanation below

Explanation:

The question is incomplete. The missing parts are a) determine the electrophylic site. b) determine the nucleophylic site.

In order to do this, we need to write the reaction and do the mechanism. The nucleophylic site will be the site where the nucleophyle attacks to form the product. In this case the site is the carbon next to the bromine. In this place the Oxigen which is the nucleophyle goes. The electrophyle is the site where one atom substract to complete it's charges. In this case, the electrophyle is usually the hydrogen, so the site will be next to the oxygen after the nucleophyle attack.

You can see it better in the attached picture.

The reaction between ethyl alcohol and ethyl bromide forms diethyl ether via a substitution reaction, where a hydrogen atom of the alcohol is replaced with ethyl from the bromide. After this, deprotonation occurs to remove excess hydrogen and create the stable diethyl ether.

The question refers to the reaction between ethyl alcohol and ethyl bromide to produce diethyl ether, a type of ether. This reaction happens via a substitution reaction, during which a hydrogen atom of the ethyl alcohol is replaced by an ethyl group from the ethyl bromide. After the substitution, a deprotonation process occurs to form the final product, diethyl ether.

In chemical terms, the substitution reaction occurs when an alcohol, like ethyl alcohol, reacts with a halogenoalkane, like ethyl bromide, in the presence of sulphuric acid (H2SO4). The result is the formation of an ether, in this case, diethyl ether. A deprotonation process takes place to remove an excess hydrogen proton to form the final stable product.

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The question is incomplete, here is the complete question:

Suppose 2.19 g of barium acetate is dissolved in 150 mL of a 0.10M of aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer: The final molarity of acetate ion in the solution is 0.12 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex]     .....(1)

Molarity of sodium chromate solution = 0.10 M

Volume of solution = 150 mL

Putting values in equation 1, we get:

[tex]0.10M=\frac{\text{Moles of sodium chromate}\times 1000}{150}\\\\\text{Moles of sodium chromate}=\frac{(0.10\times 150)}{1000}=0.015mol[/tex]

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

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

Given mass of barium acetate = 2.19 g

Molar mass of barium acetate = 255.43 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of barium acetate}=\frac{2.19g}{255.43g/mol}=0.0086mol[/tex]

The chemical equation for the reaction of barium acetate and sodium chromate follows:

[tex]Ba(CH_3CO_2)_2+Na_2CrO_4\rightarrow BaCrO_4(s)+2Na^+(aq.)+2CH_3CO_2^-(aq.)[/tex]

By stoichiometry of the reaction:

1 mole of barium acetate reacts with 1 mole of sodium chromate

So, 0.0086 moles of barium acetate will react with = [tex]\frac{1}{1}\times 0.0086mol[/tex] of sodium chromate

As, given amount of sodium chromate is more than the required amount. So, it is considered as an excess reagent.

Thus, barium acetate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of barium acetate produces 2 moles of acetate ions

So, 0.0086 moles of barium acetate will produce = [tex]\frac{2}{1}\times 0.0086=0.0172mol[/tex] of acetate ion

Now, calculating the molarity of acetate ions in the solution by using equation 1:

Moles of acetate ion = 0.0172 moles

Volume of solution = 150 mL

Putting values in equation 1, we get:

[tex]\text{Molarity of acetate ions}=\frac{0.0172\times 1000}{150}\\\\\text{Molarity of acetate ions}=0.12M[/tex]

Hence, the final molarity of acetate ion in the solution is 0.12 M

Answer: helps them survive

Explanation: because they want to and need to

Answer:

Substrate D

Explanation:

In substitution reactions the tertiary substrates cannot undergo substitution via neighboring group participation (NGP) due to the steric impediment, this means that the volume occupied by the substituents is very large and makes it impossible to attack the nucleophile to the substrate carbon.

Answer:

a) Six electrons

b) Two electrons

c) Fourteen electrons

Explanation:

n is the principal quantum number and defines the energy level of orbital. The shape of the orbital is described by azimuthal quantum number (l) and it also determine the angular momentum. It values give the following information

l = 0, define s orbital (single orbital)

l = 1, define p orbitals (three orbitals)

l = 2, define d orbitals (five orbitals)

l = 3, define f orbitals (seven orbitals)

These are further specified by magnetic quantum number (ml) which gives the orientation of the orbital. Its value ranges from +1 to -1, for example ml value of five d orbitals are +2, +1, 0, -1, -2. From this information we can predict the number of electrons that will have the given sub-level designations

a) n = 4 and orbital is p, there are three p orbitals as the ml is not defined, so six electrons will have this quantum number

b) In this part, the orbital is defined i.e. ml = +1. A single orbital can have only two electrons, so these electrons will have the given quantum number.

c) l = 3, is for f orbital, which have seven orbitals. The total number of electrons in it is fourteen. All of these electrons will have this quantum number.

(a) An atom with the quantum number of 4p will have 6 electrons.

(b) An atom with the quantum number of n=3, i = 1, m1 = +1 will have 2 electrons.

(c) An atom with the quantum number of n=5, i = 3, will have 14 electrons.

The number of electrons an atom in the given quantum number can have is calculated as follows;

(a) 4p --- p-orbital has 3 sub-shells and the atom will have maximum of 6 electrons.

(b) n = 3, l = 1, m1 = +1 --- this corresponds to 3p - orbital and each atom will have maximum of 2 electrons.

(c) l = 3, corresponds to f - orbital

f-orbitals have 7 sub-shell and the atom will have maximum of 14 electrons.

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Answer:

B

Explanation:

Carbon is an element with an atomic number 6, the electron configuration is 2,4. This means it has four electrons in its outer-most shell or valence shell. The valence shell needs 8 electrons to be filled. Thus the outer electron shell of carbon is unfilled.

It reacts with other elements covalently and forms organic and in-organic compounds.

The 4 electrons in the valance shell of carbon makes it a metalliod, but commonly its considered as a non-metal.

However carbon can form single and double covalent bonds. It forms single covalent bond when it combines with 4 atoms of Hydrogen in the methane molecule

Carbon's possession of four valence electrons allows it to form up to four single covalent bonds with other atoms, making it highly versatile and reactive in the formation of various compounds. It does not have a filled outer electron shell, and it is not positively charged.

The correct answer is option B.

Carbon has four electrons in its outer electron shell, which is also known as the valence shell. This electron configuration plays a crucial role in carbon's chemical behavior and its ability to form a wide variety of compounds. Let's examine the given options:

a. it has a filled outer electron shell: This statement is incorrect. A filled outer electron shell for carbon would require eight electrons in its valence shell, following the octet rule. Carbon only has four valence electrons, so its outer shell is not filled.

b. it can form four single covalent bonds: This statement is correct. Carbon has four valence electrons, which allows it to form up to four single covalent bonds with other atoms. This ability to form multiple bonds is a fundamental characteristic of carbon and is the basis for the vast diversity of organic compounds found in nature.

c. it does not react with any other atoms: This statement is incorrect. Carbon is highly reactive and readily forms compounds through covalent bonding with other elements, especially hydrogen, oxygen, nitrogen, and other carbon atoms. Its reactivity is a key feature that makes carbon the foundation of organic chemistry.

d. it has a positive charge: This statement is incorrect. Carbon, like all neutral atoms, has a balanced number of protons and electrons, resulting in no net electrical charge. It is neither positively charged (cation) nor negatively charged (anion).

Therefore, from the given options the correct one is B.

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Answer:

Explanation: The lowest pressure in a laboratory is 4.0×10^-11Pa

Using Ideal gas equation

PV = nRT

P= 4.0×10^-11Pa

V= 0.020m^3

T= 20+273= 293k

n=number of moles = m/A

Where m is the number of molecules and A is the Avogradro's number=6.02×10²³/mol

R=8.314J/(mol × K)

PV= m/A(RT)

4.0×10^-11 ×0.020 = m/6.02×10²³(8.314×293)

m = 4.0×10^-11×0.020×6.02×10^23 / (8.314×293)

m = 1.98×10^8 molecules

Therefore,the number of molecules is 1.98×10^8

The student's results are best described as accurate but not precise, as one measurement matches the actual value, but the other measurements are inconsistent with each other and the actual value.

You are trying to determine the density of a sugar solution and have obtained results of 1.11 g/mL, 1.81 g/mL, 1.95 g/mL, and 1.75 g/mL, with the actual density being 1.75 g/mL.

When comparing these results to the actual value, we can see that one of the results matches the actual value, indicating accuracy for that particular measurement.

However, the other measurements are quite different from the actual value and from each other, indicating a lack of precision. Precision refers to how close multiple measurements are to each other, regardless of whether they are close to the actual value (which is accuracy).

So, the statement that best describes the results is:

The most appropriate choice is (B), Her results are accurate, but not precise, because one measurement matches the actual value while the other measurements do not consistently match each other.

Answer:

Enthalpy of gas is greater than that of liquid

Explanation

Ethyl chloride (C2H5Cl) boils at 12 degrees Celsius. When liquid C2H5Cl under pressure is sprayed on a room-temperature (25 degrees Celsius) surface in the air, the surface is cooled considerably.

this preamble could precede the question

Enthalpy

it can defined as the ability of a substance to change at constant pressure, enthalpy tells how much heat and work was added or removed from the substance. Enthalpy is similar to energy, but not the same. When a substance grows or shrinks, energy is used up or released.

The total heat content of a system is known as Enthalp.it is a thermodynamic property. It is the internal energy of the system plus the product of pressure and volume.

H=U+PV........................1

Answer: check the picture

Explanation:

Answer:

In the oxidation of Pyruvate to Acetyl CoA  one carbon atom is released as CO2, However, the oxidation of the remaining two carbon atoms in acetate to CO2, requires a complex, eight step pathway, the citric-acid cycle.

Explanation:

Overall, one turn of the citric acid cycle releases two carbon dioxide molecules and produces three {NADH} one {FADH2} and one ATP or GTP. The Citric acid cycle goes around twice for each molecule of glucose that enters cellular respiration because there are two pyruvate and thus, two acetyl {CoA}s are made per glucose.

Answer:

a. 27g/mol

b. 1.85 x 10^5 moles

Explanation:Please see attachment for explanation

To classify molecules as polar or nonpolar, we need to consider molecular geometry and polarity of bonds.

In order to determine if a molecule is polar or nonpolar, we need to look at the molecular geometry and the polarity of the individual bonds within the molecule. If the molecule has polar bonds and is asymmetrical in shape, it will be polar. If the molecule has either nonpolar bonds or is symmetrical in shape, it will be nonpolar.

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Answer: The percent abundance of Ne-21 isotope is 0.27 %

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]   .....(1)

Let the fractional abundance of Ne-21 isotope be x

Mass of isotope 1 = 20 amu

Percentage abundance of isotope 1 = 90.48 %

Fractional abundance of isotope 1 = 0.9048

Mass of isotope 2 = 21 amu

Fractional abundance of isotope 2 = x

Mass of isotope 3 = 22 amu

Percentage abundance of isotope 3 = 9.25 %

Fractional abundance of isotope 3 = 0.0925

Average atomic mass of neon = 20.18 amu

Putting values in equation 1, we get:

[tex]20.18=[(20\times 0.9048)+(21\times x)+(22\times 0.0925)][/tex]

x = 0.0027

Percentage abundance of Ne-21 isotope = [tex](0.0027\times 100)=0.27\%[/tex]

Hence, the percent abundance of Ne-21 isotope is 0.27 %

The abundance of the isotope 21Ne of Neon can be found by subtracting the sum of the abundances of the other two isotopes, 20Ne and 22Ne, from 100%. This calculation yields an abundance of approximately 0.27% for 21Ne.

Neon has three naturally occurring isotopes: 20Ne, 21Ne, and 22Ne. Given the percentage abundances of 20Ne and 22Ne, we can find the abundance of 21Ne by subtracting the sum of these percentages from 100%. Specifically,

Abundance of 21Ne = 100% - (90.48% + 9.25%)

So, the abundance of 21Ne is approximately 0.27%.

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Answer:

Answers explained below

Explanation:

(a)Given,

Molecular formula of the complex = [Ni(NH3)x(H2O)y]Clz

(i) Ni is in +2 oxidation state in the complex.

(ii) NH3 and H2O are the neutral ligands but Cl  is the negatively charged ligand.

(iii) complex is neutral

So, to make the nickel complex in +2 oxidation with neutral charge, we requires 2 Cl-.

Hence, form the above statements, we can say that here in the complex

z=2

(b) Molarity of HCl = 0.2005M

Volume of HCl used = 20.02mL = 20.02*10-3 L

Weight of the nickel(II) ammonia complex = 0.1550g

Reaction of HCl with Ammonia,

HCl (aq) + NH3 (aq) -> NH4Cl (aq)

HCl reacts with ammonia in 1:1 ratio to form ammonium salt (NH4Cl). That means 1 mol of HCl reacts with 1 mol of NH3.

So, we have to find number of moles of HCl used.

No. of moles of HCl used = Molarity of HCl * Volume of HCl used (L)

= 0.2005M * 20.02*10-3 L = 4.014*10-3 moles

Hence no. of moles of ammonia in the complex = No. of moles of HCl used = 4.014*10-3 mol

So, Experimental Equivalent Weight = Weight of the nickel(II) ammonia complex/ No. of moles of NH3

= 0.1550g / 4.014*10-3 mol

= 38.615 g/mol

Hence, Experimental Equivalent Weight = 38.615 g/mol

(c) Given,

x+y=< 6

Molar mass of [Ni(NH3)x(H2O)y]Clz = 58.69 + x(14.00+3*1) + y(16+2*1)+z(35.45)

= (58.69 + 17x + 18y + 35.45z) g/mol

Case1 x=6, y=0 and z=2

Molar mass of [Ni(NH3)6(H2O)0]Cl2   = (58.69 + 17*6 + 18*0 + 35.45*2) g/mol

= 58.69+102+70.90 = 231.59 g/mol

Experimental Equivalent Weight = 233.59/6 = 38.598 g/mol

So, This experimental equivalent weight is equal to the calculated experimental equivalent weight.

Hence the molecular formula of the complex is [Ni(NH3)6]Cl2 where x=6, y=0 and z=2.

Note: You can try other combination but in every case you will find lower or higher calculated experimental equivalent weight.

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           [tex]N_2[/tex]                         [tex]7.81\times 10^{-1}[/tex]         [tex]6.70\times 10^{-4}[/tex]

           [tex]O_2[/tex]                         [tex]2.10\times 10^{-1}[/tex]        [tex]1.30\times 10^{-3}[/tex]

           Ar                          [tex]9.34\times 10^{-3}[/tex]        [tex]1.40\times 10^{-3}[/tex]

          [tex]CO_2[/tex]                        [tex]3.33\times 10^{-4}[/tex]        [tex]3.50\times 10^{-2}[/tex]

          [tex]CH_4[/tex]                       [tex]2.00\times 10^{-6}[/tex]         [tex]1.40\times 10^{-3}[/tex]

          [tex]H_2[/tex]                          [tex]5.00\times 10^{-7}[/tex]         [tex]7.80\times 10^{-4}[/tex]

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is [tex]1.48\times 10^{-10}M[/tex]

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

[tex]p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}[/tex]

where,

[tex]p_A[/tex] = partial pressure of hydrogen gas = ?

[tex]p_T[/tex] = total pressure = 0.380 atm

[tex]\chi_A[/tex] = mole fraction of hydrogen gas = [tex]5.00\times 10^{-7}[/tex]

Putting values in above equation, we get:

[tex]p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm[/tex]

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{H_2}=K_H\times p_{H_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]7.80\times 10^{-4}mol/L.atm[/tex]

[tex]p_{H_2}[/tex] = partial pressure of hydrogen gas = [tex]1.9\times 10^{-7}atm[/tex]

Putting values in above equation, we get:

[tex]C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M[/tex]

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is [tex]1.48\times 10^{-10}M[/tex]

Answer:

B:COTYLEDONS

Explanation:

Cotyledon is defined as;it is the part of embryo present whithin the seed of a plant and is often referred as "seed leaf"

The number of cotyledons is a defining characteristic of angiosperms.Cotyledons are present in the embryo of the angiosperms.On the basis of the number of cotyledons ;angiosperms are divided or distinguished into two classes which are termed as;Monocotyledonae (Monocots) and Dicotyledonae (Dicots).

MONOCOTS

The species which have one cotyledon in their seeds are monocots.corn,wheat,barley,rice.

DICOTS

The species having two cotyledons are called dicots.peas,beans,peanuts.

Yup. It's B, Cotyledons. I got it right on edge 2020

Answer:

The molecular formula = [tex]C_{6}H_{6}[/tex]

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,  

[tex]n=\frac{m}{M}[/tex]

Using ideal gas equation as:

[tex]PV=\frac{m}{M}RT[/tex]

where,  

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

[tex]0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15[/tex]

[tex]M=78.31\ g/mol[/tex]

The empirical formula is = [tex]CH[/tex]

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,  

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = [tex]C_{6}H_{6}[/tex]

The molecular formula of the compound CH is [tex]C_6H_6[/tex].

Given:

P = 0.74 atm

V = 97.2 mL = 0.0972 L (since 1 mL = 0.001 L)

T = 200°C + 273.15 = 473.15 K (to convert from Celsius to Kelvin)

R = 0.0821 L·atm/(mol·K)

First, we solve for n, the number of moles:

[tex]\[ n = \frac{PV}{RT} \][/tex]

[tex]\[ n = \frac{(0.74 \text{ atm})(0.0972 \text{ L})}{(0.0821 \text{ L·atm/(mol·K)})(473.15 \text{ K})} \][/tex]

[tex]\[ n \approx \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]

[tex]\[ n \approx \frac{0.072048}{38.84015} \][/tex]

[tex]\[ n \approx 0.001855 \text{ mol} \][/tex]

Next, we calculate the molar mass (M) using the given mass (m) of the compound: [tex]\[ M = \frac{m}{n} \][/tex]

[tex]\[ M = \frac{0.145 \text{ g}}{0.001855 \text{ mol}} \][/tex]

[tex]\[ M \approx \frac{0.145}{0.001855} \][/tex]

[tex]\[ M \approx 78.17 \text{ g/mol} \][/tex]

The empirical formula mass of CH is:

[tex]\[ (1 \times 12.01 \text{ g/mol}) + (1 \times 1.008 \text{ g/mol}) = 13.018 \text{ g/mol} \][/tex]

To find the molecular formula, we divide the molar mass by the empirical formula mass:

[tex]\[ \text{Molecular formula mass} = n \times \text{Empirical formula mass} \][/tex]

[tex]\[ \text{Molecular formula mass} = n \times \text{Empirical formula mass} \][/tex]

[tex]\[ n = \frac{78.17}{13.018} \][/tex]

[tex]\[ n \approx 6 \][/tex]

Therefore, the molecular formula is 6 times the empirical formula, which is (CH)†. However, since the empirical formula, CH already represents one carbon and one hydrogen atom, the molecular formula is simply C†H†.

Upon reviewing the problem, it appears there was an error in the calculation of n, the number of moles. Let's correct this:

[tex]\[ n = \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]

[tex]\[ n = \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]

[tex]\[ n = \frac{(0.74)(0.0972)}{(0.0821)(473.15)} \][/tex]

This value of n is correct, and the subsequent calculations are also correct. However, the final molecular formula should be CH, not C†H†, because the correct calculation for n (the multiplier) is:

[tex]\[ n = \frac{78.17}{13.018} \][/tex]

[tex]\[ n = \frac{78.17}{13.018} \][/tex]

Since the empirical formula, CH is CH‚ multiplying by 6 gives us C†H†. However, we must consider that the subscripts in the empirical formula are the smallest whole-number ratio of atoms in the compound. Therefore, the correct molecular formula is obtained by multiplying the subscripts in the empirical formula by the same number, n, which is 6. Thus, the molecular formula is C†H†, which simplifies to CH, since both subscripts can be divided by 3.

Therefore, the correct molecular formula of the compound CH is [tex]C_6H_6[/tex].

Answer:

H+  ----- Br-

H+  ----- Cl-

O₋₋ -----2H++

CH3 ----- O-   ------ H+

Explanation:

Dipole moment occurs when there is bonding between a very strong electronegative element and hydrogen atom.

Electronegative elements are the element which attract electrons towards themselves, (that is they have strong affinity for electrons).

Generally, group 7 elements (Fluorine, Chlorine, Bromine, Iodine) of the periodic table are highly electronegative

In HBr, HCI, and H2O, there are polar covalent bonds resulting in dipole moments

(a) HBr: H-Br bond is polar covalent. The dipole moment points from Br (delta negative) to H (delta positive).

(b) HCI: H-Cl bond is polar covalent. The dipole moment points from Cl (delta negative) to H (delta positive).

(c) H2O: O-H bonds are polar covalent. The dipole moments point from O (delta negative) to H (delta positive).

(d) CH40: No polar covalent bonds. The dipole moments cancel each other out due to the symmetrical arrangement of the atoms.

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Answer: The mass percent of NaBr in the solution is 10.18 %

Explanation:

To calculate the mass percentage of calcium ions in milk, we use the equation:

[tex]\text{Mass percent of NaBr}=\frac{\text{Mass of NaBr}}{\text{Mass of solution}}\times 100[/tex]

We are given:

Mass of solution = 270.0 g

Mass of NaBr = 27.50 g

Putting values in above equation, we get:

[tex]\text{Mass percent of NaBr}=\frac{27.50g}{270.0g}\times 100=10.18\%[/tex]

Hence, the mass percent of NaBr in the solution is 10.18 %

The mass percent of NaBr in the aqueous solution is 10.19%.

The mass percent of a component in a solution is calculated by taking the mass of the component divided by the total mass of the solution, then multiplied by 100. I

To calculate the mass percent of NaBr in an aqueous NaBr solution, we need to divide the mass of NaBr by the mass of the solution and multiply by 100. In this case, the mass of NaBr is 27.50 g and the mass of the solution is 270.0 g. So the mass percent of NaBr can be calculated as:

Mass percent NaBr = (mass of NaBr / mass of solution) x 100

=(27.50 g / 270.0 g) x 100

= 10.19%

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Answer:

The student should use 156.7 g of solution

Explanation:

31.9 %w solution of acetic acid in ethanol means 100 g of solution contains 31.9 g of acetic acid.

So, 450. g of solution contains [tex](\frac{31.9\times 450.}{100})g[/tex] of acetic acid or 143.55 g of acetic acid

Alternatively, 143.55 g of acetic acid is present in 450. g of solution

So, 50.00 g of acetic acid is present in [tex](\frac{450.\times 50.00}{143.55})g[/tex] solution or 156.7 g of solution

Hence the student should use 156.7 g of solution