A 17.0 g sample of quartz, which has a specific heat capacity of 0.730 Jg , is dropped into an insulated container containing 200.0 g of water at 85.0 °C and a constant pressure of 1 atm. The initial temperature of the quartz is 7.2 ℃. Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 3 significant digits. ec

Answers

Answer 1

Answer:

the final temperature T final = 83.863 °C

Explanation:

Assuming that no heat is absorbed from the container , then all the heat absorbed by the water Q water comes from the heat released by the quartz (-Q quatz), since

Q water + Q quatz = Q surroundings =0

denoting w as water and q as quartz then

Q water = mw * cpw * (Tfinal - T initial w)

Q quartz = mq * cpq * (Tfinal - T initial q)

where

m= mass

cp= specific heat capacity at constant pressure

T final = final temperature

T initial w and T initial q = initial temperature of water and quartz respectively

then

mw * cpw * (Tfinal - T initial w) + mq * cpq * (Tfinal - T initial q) = 0

mw * cpw * Tfinal + mq * cpq * Tfinal = mw * cpw *T initial w + mq * cpq * T initial q

Tfinal = (mw * cpw *T initial w+ mq * cpq * T initial q)/(mw * cpw +mq * cpq)

replacing values and assuming cpw= 1 cal/gr°C = 4.186 J/g°C

Tfinal = (200 g * 4.186 J/g°C * 85 °C + 17g * 0.730 J/g°C * 7.2 °C)/(200 g * 4.186 J/g°C + 17g * 0.730 J/g°C) = 83.863 °C

Tfinal = 83.863 °C


Related Questions

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

Answers

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]

Arrange the following aqueous solutions in terms of freezing point depression with the least depression first: 0.45m CaCl2, 0.45m glucose or 0.45m NaCl? Then determine the freezing point depression for each. For water, Kf=1.86 C/m

Answers

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]

What is the minimum % relative error for a 0.5M NaOH solution?

Assume the absolute error is +/- 0.1.

Answers

Answer: 20%

Explanation:

Relative error = Absolute error/actual value

= 0.1/0.5

= 0.2

Hence,

Minimum % Relative error = 0.2 ×100%

= 20%

Answer:20%

Explanation:

Let us take some important definitions of some terms in the question.

What is an ABSOLUTE ERROR?.

Absolute error is the difference between the expected value and the value you got from an experimental calculations or the approximations made. It can be represented mathematically in the equation (1) below;

Absolute error= expected value - Approximated/measured value. ---------------------------------------------------------------------(1).

What is A RELATIVE ERROR?.

A relative error is the the ratio of absolute value to that of the exact value. This can be illustrated mathematically by the equation (2) below;

Relative error= absolute error/exact value. --------------------------------------(3).

From equation (2) the solutio to the question is;

Relative error= absolute error/exact value.

Relative error=[ 0.1/0.5] = 0.2.

Relative error Percentage= 0.2×100= 20 Percent.

A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.11 g/mL, 1.81 g/mL, 1.95 g/mL, 1.75 g/mL. If the actual value for the density of the sugar solution is 1.75 g/mL, which statement below best describes her results

A) Her results are precise, but not accurate.
B) Her results are accurate, but not precise.
C) Her results are both precise and accurate
D) Her results are neither precise nor accurate.
E) It isn't possible to determine with the information given.

Answers

A) Her results are precise, but not accurate, as she only obtained the real density once from her experiment

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:

(A) Her results are precise, but not accurate.(B) Her results are accurate, but not precise.(C) Her results are both precise and accurate.(D) Her results are neither precise nor accurate.(E) It isn't possible to determine with the information given.

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.

A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium carbonate the chemist has added to the flask. Round your answer to 3 significant digits. mmol

Answers

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

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

[tex]\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}[/tex]     .....(1)

Molarity of [tex]BaCl_2[/tex] solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

[tex]1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol[/tex]

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

Final answer:

In order to calculate the millimoles of sodium carbonate added to the flask, one must rearrange the formula for molarity, convert the volume from milliliters to liters, substitute, solve, then, finally, convert the solution from moles to millimoles. This results in an approximate 255.6mmol of sodium carbonate.

Explanation:

The chemist added 180.0 mL of a 1.42M sodium carbonate (Na2CO3) solution. Molarity (M) is defined as the number of moles of solute (Na2CO3 in this case) per liter of solution. Therefore, to find the number of millimoles, you would use the equation:

M = moles/volume(L)

Rearrange the formula to solve for moles:

moles = M x volume (L)

Next, convert the volume from milliliters to liters: 180.0 mL = 0.180 L. Substituting into the equation gives:

moles = 1.42M x 0.180 L = 0.2556 moles

To convert moles to millimoles, you'll need to know that 1 mole = 1000 millimoles. Therefore, 0.2556 moles x 1000 = 255.6 millimoles rounded to three significant digits.

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When 4.31 g of a nonelectrolyte solute is dissolved in water to make 345 mL of solution at 25 °C, the solution exerts an osmotic pressure of 851 torr.

a. What is the molar concentration of the solution?
b. How many moles of solute are in the solution?c. What is the molar mass of the solute?

Answers

Answer:

a) 0.046 mol/L

b) 0.016 mol

c) 271.58 g/mol

Explanation:

A nonelectrolyte solute is a solute that, when dissolved in a solvent, will not make the solution a conductor, and so the electricity will not pass through it. The osmotic pressure is a colligative property, and it's the pressure difference needed to stop the flow of a solution across a semipermeable membrane. It can be calculated by:

π = MRT

Where π is the osmotic pressure, M is the molarity of the solute (mol/L), R is the ideal gas constant, and T is the temperature (in K). For a pressure in torr, R = 62.3637 torr.L/(mol.K).

a) π = MRT , T = 25°C = 298 K

851 = M*62.3637*298

18584.3826M = 851

M = 0.046 mol/L

b) The number of moles of the solute (n) is the molar concentration (molarity) multiplied by the volume. So, for a solution of 345 mL = 0.345 L,

n = 0.046 * 0.345

n = 0.016 mol

c) The molar mass (MM) is the mass divided by the number of moles:

MM = 4.31/0.016

MM = 271.58 g/mol

Neon has three naturally occuring isotopes. The abundance of 20Ne is 90.48% and 22Ne is 9.25%. What is the percent abundance of 21Ne?

Answers

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

For isotope 1 (Ne-20) :

Mass of isotope 1 = 20 amu

Percentage abundance of isotope 1 = 90.48 %

Fractional abundance of isotope 1 = 0.9048

For isotope 2 (Ne-21) :

Mass of isotope 2 = 21 amu

Fractional abundance of isotope 2 = x

For isotope 3 (Ne-22) :

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 %

Final answer:

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.

Explanation:

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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Carbon has four electrons in its outer electron shell, therefore: Select one: a. it has a filled outer electron shell. b. it can form four single covalent bonds c. it does not react with any other atoms. d. it has a positive charge.

Answers

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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If you were to use Lewis theory to predict the formula for the compound between rubidium and fluorine, it would most likely be:

A. Rb2F
B. RbF
C. RbF2
D. Rb3F

Answers

Answer: The formula of the compound formed between rubidium and fluorine is RbF

Explanation:

Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.

The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.

Rubidium is the 37th element of the periodic table having electronic configuration of [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1[/tex]

This will loose 1 electron to form [tex]Rb^+[/tex] ion

Fluoride is the 9th element of the periodic table having electronic configuration of [tex]1s^22s^22p^5[/tex]

This will gain 1 electron to form [tex]F^-[/tex] ion

To form [tex]RbF[/tex] compound, 1 rubidium ion is needed to neutralize the charge on fluoride ion

The formation of the given compounds is shown in the image below.

A water sample tested positive for lead with a concentration of 35 ppm. The density of the solution is 1.00 g/mL Which of the following statements is correct?

A. 100 g of the solution contains 35 g of lead
B. there are 35 mg of lead in 1.0 L of this solution
C. 100 g of the solution contains 35 mg of lead
D. the solution is 35% by mass of lead
E. Ethe molarity of the solution is 35 M

Answers

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

A student synthesized a nickel (II) ammonia complex with a molecular formula of [Ni(NH3)x(H2O)y]Clz. The amount of ammonia in the complex was analyzed using 0.2005M HCl. The student determined that the reaction required 20.02mL of HCl to react with 0.1550g of the nickel(II) ammonia complex. Answer the following questions.Atomic masses: Ni = 58.69g/mol; H= 1.00g/mol; O= 16.00g/mol; Cl = 35.45g/mol; N= 14.00g/mol(a) Determine the value for "z" in the compound. Briefly explain your reasoning. Hint: the complex salt must be electrically neutral (i.e. NO NET CHARGE)(b) Calculate the EXPERIMENTAL EQUIVALENT WEIGTH for the nickel(II) ammonia complex (i.e. grams of complex per mole of ammonia).(c) Based on the results from (a) and (b), propose a molecular formula for the neutral nickel(II) ammonia coordination complex. Show all your work. Hint: x+y≤6. Try different combinations of "x" and "y" to see which combination best matches the calculated experimental equivalent weight.

Answers

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.

A sample of an ideal gas occupies 2.78 x 10^3 mL at 25°C and 760 mm Hg.
What volume will it occupy at the same temperature and 475 mm Hg?

Answers

Answer: It will occupy [tex]4.45\times 10^3ml[/tex] at the same temperature and 475 mm Hg.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]P_1V_1=P_2V_2[/tex]    (At constant temperature and number of moles)

where,

[tex]P_1[/tex] = initial pressure of gas = 760 mm Hg

[tex]P_2[/tex] = final pressure of gas = 475 mm Hg

[tex]V_1[/tex] = initial volume of gas = [tex]2.78\times 10^3ml[/tex]

[tex]V_2[/tex]  = final volume of gas = ?

Putting in the values:

[tex]760mm Hg\times 2.78\times 10^3ml=475 mm Hg\times V_2[/tex]

[tex]V_2=4.45\times 10^3ml[/tex]

Thus it will occupy [tex]4.45\times 10^3ml[/tex] at the same temperature and 475 mm Hg

The volume that it will occupy at the same temperature is 4448mL

According to Boyle's law, the pressure of a given mass of gas is inversely proportional to the volume. It is expressed mathematically as:

[tex]P\alpha\frac{1}{V}\\P=\frac{k}{V}\\PV=k[/tex]

This can be expressed as [tex]P_1V_1=P_2V_2[/tex]

Given the following parameters

P₁ = 760mmHg

V₁ = 2.78 x 10³ mL

P₂ = 475mmHg

V₂ = ?

Substitute the given parameters into the formula

[tex]V_2=\frac{P_1V_1}{P_2}\\V_2=\frac{760\times 2780}{475}\\V_2= 4448mL[/tex]

Hence the volume that it will occupy at the same temperature is 4448mL

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The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm. What is the molecular formula of the compound

Answers

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].

Each of the following substrates can react with a nucleophile in a substitution reaction. Select the substrate that cannot undergo substitution via neighboring group participation (NGP). A B C D

Answers

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.

The following reaction between ethyl alcohol and ethyl bromide forms diethyl ether via a substitution reaction (followed by deprotonation).cannot explain figure

Answers

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.

Final answer:

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.

Explanation:

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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For each compound below, identify any polar covalent bonds and indicate the direction of the dipole moment using the symbols 84 and δ-.

(a) HBr
(b) HCI
(c) H2O
(d) CH40

Answers

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

Final answer:

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

Explanation:Polar Covalent Bonds and 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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A chemistry student needs 50.00g of acetic acid for an experiment. He has available 450. g of a 31.9%w solution of acetic acid in ethanol. Calculate the mass of solution the student should use

Answers

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

A compound contains 10.13% C and 89.87% Cl (by mass). Determine both the empirical formula and the molecular formula of the compound given that the molar mass is 237 g/mol.

CCl3
C2Cl
CCl

Answers

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

1. How to determine the empirical formula C = 10.13%Cl = 89.87%Empirical formula =?

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

2. How to determine the molecular formula Molar mass of compound = 237 g/mol.Empirical formula = CCl₃Molecular formula =?

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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How many electrons in an atom can have each of the following quantum number or sublevel designations? (a) 4p (b) n = 3, l = 1, m1 = +1 (c) n = 5, l = 3

Answers

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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Draw the product formed when the following diene is treated with one equivalent of HCl. Do not show stereochemistry in your answer.

Answers

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.

What is substitution reaction?

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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In the Minnesota Department of Health set a health risk limit for methanol in groundwater of 4.00 μg. Suppose an analytical chemist receives a sample of groundwater with a measured volume of 76.0 mL.
Calculate the maximum mass in milligrams of methanol which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits.

Answers

The question is incomplete, here is the complete question:

In the Minnesota Department of Health set a health risk limit for methanol in groundwater of 4.00 μg/L. Suppose an analytical chemist receives a sample of groundwater with a measured volume of 76.0 mL.

Calculate the maximum mass in milligrams of methanol which the chemist could measure in this sample and still certify that the groundwater from which it came met Minnesota Department of Health standards. Be sure your answer has the correct number of significant digits.

Answer: The risk limit of methanol in the given amount of methanol is [tex]3.04\times 10^{-4}mg[/tex]

Explanation:

We are given:

Risk limit for methanol in groundwater = 4.00 μg/L = 0.004 mg/L    (Conversion factor:  [tex]1mg=1000\mu g[/tex] )

Volume of groundwater that is to be measured = 76.0 mL

We know that:

1 L = 1000 mL

Applying unitary method:

In 1000 mL of groundwater, the risk limit of methanol is 0.004 mg

So, in 76.0 mL of groundwater, the risk limit of methanol will be = [tex]\frac{0.004}{1000}\times 76.0=3.04\times 10^{-4}mg[/tex]

Hence, the risk limit of methanol in the given amount of methanol is [tex]3.04\times 10^{-4}mg[/tex]

The metal zirconium becomes superconducting at temperatures below 546 mk. Calculate the temperature at which zirconium becomes superconducting in degrees Celsius. Round your answer to 3 decimal places.

Answers

Final answer:

The temperature at which zirconium becomes superconducting in degrees Celsius is approximately 272.85°C.

Explanation:

To convert the temperature from millikelvin (mk) to degrees Celsius (°C), we need to use the formula:

T(°C) = T(mk) - 273.15

Applying this formula to the given temperature of 546 mk:

T(°C) = 546 - 273.15 = 272.85°C

Therefore, the temperature at which zirconium becomes superconducting in degrees Celsius is approximately 272.85°C.

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Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.

Answers

Answer:

The mass of xenon in the compound is 2.950 grams

Explanation:

Step 1: Data given

Mass of XeF4 = 4.658 grams

Molar mass of XeF4 = 207.28 g/mol

Step 2: Calculate moles of XeF4

Moles XeF4 = mass XeF4 / molar mass XeF4

Moles XeF4 = 4.658 grams / 207.28 g/mol

Moles XeF4 = 0.02247 moles

Step 3: Calculate moles of xenon

XeF4 → Xe + 4F-

For 1 mol xenon tetrafluoride, we have 1 mol of xenon

For 0.02247 moles XeF4 we have 0.02247 moles Xe

Step 4: Calculate mass of xenon

Mass xenon = moles xenon * molar mass xenon

Mass xenon = 0.02247 moles * 131.29 g/mol

Mass xenon = 2.950 grams

The mass of xenon in the compound is 2.950 grams

Final answer:

To calculate the number of grams of xenon in 4.658 g of xenon tetrafluoride, we need to determine the molar mass of XeF4 and convert grams to moles. The molar mass of XeF4 is 207.282 g/mol. Using this molar mass, we find that there are 0.0225 moles of XeF4, which corresponds to 2.968 grams of xenon.

Explanation:

To calculate the number of grams of xenon in 4.658 g of xenon tetrafluoride, we need to first determine the molar mass of xenon tetrafluoride (XeF4) and then use it to convert grams to moles. The molar mass of XeF4 is calculated as follows:

Molar mass of Xe = 131.293 g/mol

Molar mass of F = 18.998 g/mol

Molar mass of XeF4 = (131.293 g/mol) + 4 * (18.998 g/mol) = 207.282 g/mol

Now, we can use the molar mass to convert grams to moles:

Moles of XeF4 = (4.658 g) / (207.282 g/mol) = 0.0225 mol

Finally, we can convert moles of XeF4 to grams of xenon:

Grams of xenon = (0.0225 mol) * (131.293 g/mol) = 2.968 g

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At this pressure, how many molecules of air would there be in a 20 ∘C experimental chamber with a volume of 0.020 m3 ?

Answers

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

Classify these molecules as polar or nonpolar. Drag each item to the appropriate bin.

Answers

Final answer:

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

Explanation:

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.

Water (H2O) - This molecule is polar because it has polar bonds and is asymmetrical in shape.Carbon Dioxide (CO2) - This molecule is nonpolar because it has polar bonds but is symmetrical in shape.Methane (CH4) - This molecule is nonpolar because it has nonpolar bonds and is symmetrical in shape.

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An example of a pure substance is ____.

a) an element
b) a compound
c) H2O
d) carbon dioxide
e) all of these

Answers

Answer: e) all of these

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: Copper

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: water  [tex](H_2O)[/tex], carbon dioxide [tex](CO_2)[/tex]

Mixtures are not pure substances as they consist of elements and compounds combined physically and not in any fixed ratio. Example: Air

Thus all given substances are pure substances.

Final answer:

The correct answer to the example of a pure substance is e) all of these, which includes elements, compounds, water (H₂O), and carbon dioxide (CO₂).

Explanation:

An example of a pure substance can be any of the following: a) an element, b) a compound, c) H₂O (which is water, a compound), and d) carbon dioxide (also a compound). Therefore, the correct answer to the question is e) all of these.

A pure substance has a constant composition and characteristic properties. Each element is made up of one type of atom, and cannot be broken down further by chemical means. Examples include carbon and oxygen. On the other hand, a compound is constituted by two or more elements chemically bonded in a fixed ratio, such as water (H₂O) and carbon dioxide (CO₂).

An aqueous NaBr solution has a mass of 270.0 g and contains 27.50 g NaBr. Calculate the mass percent NaBr .

Answers

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 %

Final answer:

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

Explanation:

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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Calculate the density, in grams per liter, of carbon dioxide (CO2) gas at STP.

Answers

Answer:

Density of CO₂ at STP = 1.96 g/L

Explanation:

1 mol of any gas at STP occupies 22.4L of volume.

The rule for the Ideal gases.

Assume 1 mol of CO₂, we know that has 44 grams of mass.

Density = mass / volume

44 g/ 22.4L = 1.96 g/L

Answer: 1.96g/L

Explanation:

1mole of CO2 contains 22.4L at stp.

1mole of CO2 = 12 + ( 2x16) = 12 + 32 = 44g

Density = Mass /volume

Density = 44g /22.4L

Density = 1.96g/L

The number of ____ is a defining characteristic of the two types of angiosperms. A. Rhizomes B.Cotyledons C. Petals D. Vascular bundles

Answers

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

When comparing certain amino acid sequences in the protein hemoglobin of humans to those of other animals, scientists found that baboons had seven different amino acids (than the human protein), dogs had ten different amino acids, gorillas had one different amino acid and lemurs had eight different amino acids. Which of these animals is most closely related to humans

Answers

Answer: helps them survive

Explanation: because they want to and need to

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