What is the pH of a 75.0 mL solution that is 0.047 M in weak base and 0.057 M in the conjugate weak acid ( K a = 7.2 × 10 − 8 ) ?

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

                                                                              Mn                         O    

% composition                                                      72.1                      27.9    

Divide by mass number                                       54.94                     16  

                                                                               1.31                      1.74    

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

Final answer:

The balanced half-reaction for the oxidation of gaseous nitrogen dioxide to nitrate ion in acidic aqueous solution is 4NO2(g) + O2(g) + 4H+(aq) -> 4NO3-(aq).

Explanation:

The balanced half-reaction for the oxidation of gaseous nitrogen dioxide to nitrate ion in acidic aqueous solution is:

4NO2(g) + O2(g) + 4H+(aq) → 4NO3-(aq)

In this reaction, nitrogen dioxide (NO2) is oxidized to nitrate ion (NO3-), and oxygen gas (O2) is reduced to water (H2O). The reaction takes place in acidic surroundings, indicated by the presence of hydrogen ions (H+).

Explanation:

A)

We know, each mole contains [tex]N_A=[/tex] [tex]6.023 \times 10^{23}[/tex] atoms.

It is given that mass of one oxygen atom is m= [tex]2.66\times 10^{-23}\ g[/tex].

Therefore, mass of one mole of oxygen, [tex]M=m\times N_A[/tex].

Putting value of n and [tex]N_A[/tex],

[tex]M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm[/tex]

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen[tex]=\dfrac{50}{16}=3.1 \ moles.[/tex]

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Avogadro's number

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The mass of 1 mole of oxygen atoms is 16.00 grams. There are 3125 moles of oxygen atoms with a mass equal to that of a glass of water.

To find the mass of 1 mole of oxygen atoms, we need to use the molar mass of oxygen. The molar mass is the mass of one mole of a substance, and it is equal to the atomic mass in grams. The atomic mass of oxygen is 16.00 g/mol, so the mass of 1 mole of oxygen atoms is 16.00 grams.

To determine how many moles of oxygen atoms have a mass equal to the mass of a glass of water, we need to use the given mass of the water and the molar mass of oxygen. The molar mass of oxygen is 16.00 g/mol, and the mass of a glass of water is 0.050 kg (or 50,000 grams). Using the molar mass, we can set up a proportion to find the number of moles of oxygen atoms.

Moles of oxygen atoms = (Mass of water / Molar mass of oxygen) = (50,000 g / 16.00 g/mol) = 3125 moles of oxygen atoms.

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Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             [tex]Q_{1} = mC_{1} \Delta T_{1}[/tex]

Putting the given values into the above equation as follows.

           [tex]Q_{1} = mC_{1} \Delta T_{1}[/tex]

                      = [tex]27.3 g \times 1.70 J/g K \times 41[/tex]

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           [tex]Q_{2}[/tex] = energy required = [tex]mL_{v}[/tex]

    [tex]L_{v}[/tex] = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             [tex]Q_{2}[/tex] = [tex]mL_{v}[/tex]

                         = [tex]27.3 \times 444[/tex]

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            [tex]Q_{3} = mC_{2} \Delta T_{2}[/tex]

Value of [tex]C_{2}[/tex] = 1.06 J/g,    [tex]\Delta T_{2}[/tex] = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             [tex]Q_{3} = mC_{2} \Delta T_{2}[/tex]

                       = [tex]27.3 g \times 1.06 J/g \times 34 K[/tex]

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = [tex]Q_{1} + Q_{2} + Q_{3}[/tex]

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

Answer:

When the neuron is at rest, what is primarily responsible for moving potassium ions OUT of the cell?​

The answer is "a concentration gradient"

Explanation:

A neuron is the main component of nervous tissue and it transmits information by electro-chemical signalling. For the nervous system to function, neurons must be able to send and receive signals.

A neuron at rest is negatively charged. The negative charge within the cell is created by the cell membrane being more permeable to potassium ion movement than sodium ion movement. At rest, there is a high concentration of potassium ions (K+) inside the cell compared to the extracellular fluid due to a net movement with the concentration gradient.

A concentration gradient acts on K+ (potassium ions). High number of potassium ions reside inside the cell, a chemical gradient occurs and pushes potassium out of the cell. The neuron membrane is more permeable to potassium ions than to other ions allowing it to selectively move out of the cell taking a positive charge with it down its concentration gradient.

Answer:

Basic

Explanation:

[tex]Sodium\ Fluoride\ +\ Water = Hydrogen\ Fluoride +\ Sodium\ Hydroxide[/tex]  

[tex]NaF + H_{2}O\rightarrow HF + NaOH[/tex]

Sodium fluoride, NaF, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, Na+, and fluoride anions, F-

[tex]NaF\rightarrow Na^{+} +F^{-}[/tex]

and when it dissolve in water the pH of the solution becomes greater than seven thereby becoming basic.

Final answer:

After sodium fluoride is dissolved in pure water, the solution becomes slightly basic due to the hydrolysis of fluoride ions which generates hydroxide ions.

Explanation:

When sodium fluoride is added to pure water and stirred until it dissolves, fluoride ions (F⁻) are released into the solution. These ions are capable of reacting, to a small extent, with water in a process known as hydrolysis. During this reaction, fluoride ions accept a proton from the water molecules, resulting in the formation of hydrofluoric acid (HF) and hydroxide ions (OH⁻). Since hydroxide ions increase the pH level of the solution, the new solution becomes slightly basic compared to pure water. Therefore, the corrected statement is: Sodium fluoride is added to pure water and stirred to dissolve. Compared to pure water, the new solution is slightly basic.

Answer: A. contains 18 electrons

Explanation:

Atomic number is defined as the number of protons or number of electrons that are present in an atom. It is characteristic of an element.

Atomic number of phosphorous is 15.

The electronic configuration of phosphorous (P) will be,

[tex]P:15:1s^22s^22p^63s^23p^3[/tex]

Atomic number = Number of electrons = Number of protons = 15

As the phosphorous atom has gained 3 electrons, it will have 15+3= 18 electrons , the phosphorous anion will be having a charge of -3.

The electronic configuration of [tex]P^{3-}[/tex] will be,

[tex]P^{3-}:18:1s^22s^22p^63s^23p^6[/tex]

Thus the correct statement is this ion contains 18 electrons

The phosphide ion contains 18 electrons. The correct answer is Option A.

The ion formed by phosphorus, called phosphide, has the formula P3−. To determine its properties, we can look at its electron configuration. Phosphorus has an atomic number of 15, meaning it has 15 electrons in its neutral state. When phosphorus forms the phosphide ion (P3−), it gains three extra electrons to achieve a stable electron configuration. So, the phosphide ion contains a total of 18 electrons.

Therefore, the correct answer is A. The ion contains 18 electrons.

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Answer:

The unknown gas is H2

Explanation:

Step 1: Data given

A mixture contains:

5.4 grams of Ar

5.4 grams of Ne

5.4 grams of X2

Molar mass of Ar = 39.95 g/mol

Molar mass of Ne = 20.18 g/mol

Step 2: Calculate moles of gases

.Moles of Ne = 4.5grams /20.18 g/mol = 0.223 moles

Moles of Ar= 4.5 grams /39.95 g/mol = 0.113 moles

Step 3: Calculate volume of gases

Volume of Ne =22.4 * 0.223 = 5

Volume of Ar =22.4 * 0.1525 = 2.53 L

Volume of unknown gas = 69.03 - 5 - 2.53 = 61.5 L

Step 4: Calculate moles of unknown gas

Moles of unkown gas =61.5/22.4 = 2.75 moles

Step 5: Calculate molar mass of unknown gas

Molar mass = mass / moles

Molar mass = 5.4 grams /2.75 moles ≈ 2 g/mol

The unknown gas is H2

Answer: sulfur pentafluoride

Explanation:

The rules for naming of binary molecular compound :

In the given formula, the lower group number element is written first in the name and keep its element name and the higher group number is written second.

First element i.e. sulphur in the formula is named first and keep its element name.

1) Gets a prefix if there is a subscript on it such as mono for 1, di for 2, tri for 3 and so on.

Second element i.e. fluorine is named second.

1) Use the root of the element name, if it is an anion then use suffix (-ide).

2) Always use a prefix on the second element such as mono for 1, di for 2, tri for 3 and so on.

Therefore, the chemical name of compound [tex]SF_5[/tex] is sulfur pentafluoride

The name of the molecular compound SF₅ is sulfur pentafluoride. Therefore, option A is correct.

A molecular compound is a compound composed of two or more nonmetallic elements. In molecular compounds, atoms are joined together by covalent bonds, which involve the sharing of electrons between atoms.

Examples of molecular compounds include water (H₂O), carbon dioxide (CO₂), methane (CH₄), and ammonia (NH₃). Molecular compounds often have specific naming conventions based on the elements present and their respective ratios, such as using prefixes to indicate the number of atoms for each element.

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Answer:

Hi

Williamson's ether reactions imply that an alkoxide reacts with a primary haloalkane. Alkoxides consisting of the conjugate base of an alcohol and are formed by a group R attached to an oxygen atom. They are often written as RO–, where R is the organic substituent (Step 1).

Sn2 reactions are characterized by the reversal of stereochemistry at the site of the leaving group. Williamson's synthesis is no exception and the reaction is initiated by the subsequent attack of the nucleophile. This requires that the nucleophile and electrophile be in anti-configuration (Step 2).

As an example (figure 3).

In the attached file are each of the steps of Williamson's synthesis.

Explanation:

The Williamson ether synthesis involves the reaction of an alkyl halide and an alcohol to form an ether. In the specific case of methyl t-butyl ether, methanol and t-butyl bromide are used. The reaction proceeds via the formation of an alkoxide ion which attacks the alkyl halide.

The Williamson ether synthesis is a method used to synthesize ethers. In this reaction, an alkyl halide (in this case, an alkyl bromide) reacts with an alcohol in the presence of a strong base to form an ether. The general mechanism involves the formation of an alkoxide ion, which then attacks the alkyl halide to form the desired ether.

In the specific case of methyl t-butyl ether, the alkyl bromide used is t-butyl bromide and the alcohol used is methanol. The reaction is as follows:

CH3OH + (CH3)3CBr → [(CH3)3CO]– + CH3Br
CH3Br + [(CH3)3CO]– → (CH3)3COCH3 + Br–

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Answer: the new volume will be 0.245L

Explanation:Please see attachment for explanation

Answer:

The volume will be 238 mL or 0.238 L

Explanation:

Step 1: Data given

Pressure = constant = 5.80 atm

The initial moles of helium = 0.04 moles

Temperature = 240.00 K

Volume = 0.14L

The number of moles increases to 0.07 moles

Step 2: Calculate the new volume

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles = 0.07 moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 240.00 K

⇒ with p = the pressure = 5.80 atm

V = (0.07 * 0.08206 * 240.00) / 5.80

V = 0.238 L = 238 mL

The volume will be 238 mL or 0.238 L

The question is incomplete, here is the complete question:

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.

Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer: The mass of copper (II) fluoride is 0.13 mg

Explanation:

We are given:

Millimolarity of copper (II) fluoride = 0.0013 mM

This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution

Converting millimoles into moles, we use the conversion factor:

1 moles = 1000 millimoles

So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]

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

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

We are given:

Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]

Hence, the mass of copper (II) fluoride is 0.13 mg

Answer:

The mass of copper(II) fluoride is 45.54 micrograms

Explanation:

A chemist adds 345.0 mL  of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.

Step 1: Data given

Molarity of the copper(II) fluoride solution = 0.0013 mM =

Volume of the solution 345.0 mL = 0.345 L

Molar mass copper(II) fluoride = 101.54 g/mol

Step 2: Calculate moles of copper(II) fluoride

Moles CuF2 = molarity * volume

Moles CuF2 = 0.0000013 M * 0.345 L

Moles CuF2 = 0.0000004485 moles

Step 3: Calculate mass of  CuF2

Mass CuF2 = moles * Molar mass

Mass CuF2 = 0.0000004485 moles * 101.54 g/mol

Mass CuF2 = 0.00004554 grams = 0.04554 miligrams = 45.54 micrograms

The mass of copper(II) fluoride is 45.54 micrograms

Answer:

0,034 moles

Explanation:

Answer:  The work energy lost by the system is -6078 Joules

Explanation:

According to first law of thermodynamics:

[tex]\Delta E=q+w[/tex]

[tex]\Delta E[/tex]=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=[tex]-P\Delta V[/tex]  {Work is done by the system as the final volume is greater than initial volume and is negative}

where P = pressure = 3 atm

[tex]\Delta V[/tex] = change in volume = (55-35) L = 20 L

w =[tex]-3atm\times (20)L=-60Latm=-6078Joules[/tex]  

{1Latm=101.3J}

Thus work energy lost by the system is -6078 Joules

Answer:

Atomic mass of 35.5 g/mol is of chlorine.

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for [tex]YCl_3[/tex].

1.835 grams of [tex]YCl_3[/tex]can be prepared.

Explanation:

[tex]2M+3X_2\rightarrow 2MX_3[/tex]

Moles of [tex]X_2[/tex] =n

Number of moleules of [tex]X_2=8.92\times 10^{20} molecules[/tex]

1 mole = [tex]6.022\times 10^{23} molecules[/tex]

[tex]n=\frac{8.92\times 10^{20} molecules}{6.022\times 10^{23} molecules}[/tex]

n = 0.001481 mole

Mass of [tex]X_2=0.105 g[/tex]

Molar mass of [tex]X_2=m[/tex]

[tex]n=\frac{Mass}{\text{Molar mass}}[/tex]

[tex]0.001481 mol=\frac{0.105 g}{m}[/tex]

m = 71 g/mol

Atomic mass of X = [tex]\frac{71 g/mol}{2}=35.5 g/mol[/tex]

Atomic mass of 35.5 g/mol is of chlorine.

The compound MX3 consists of 54.47% X by mass:

Molar mass of compound = M'

Percentage of element in compound :

[tex]=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100[/tex]

X:

[tex]54.47\%=\frac{3\times 35.5 g/mol}{M'}\times 100[/tex]

M' = 195.52 g/mol

Molar mass of compound = M'

M' = 1 × (atomic mass of M)+ 3 × (atomic mass of X)

195.52 g/mol = atomic mass of M + 3 × (35.5 g/mol)

Atomic mass of M = 89.02 g/mol

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for [tex]YCl_3[/tex].

[tex]2Y+3Cl_2\rightarrow 2YCl_3[/tex]

Moles of Yttrium = [tex]\frac{1g }{89.02 g/mol}=0.01123 mol[/tex]

Moles of chlorine gas= [tex]\frac{1 g}{71 g/mol}=0.01408 mol[/tex]

According to reaction, 3 moles of chlorine reacts with 2 moles of Y.

Then 0.01408 moles of chlorine gas will :

[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of Y.

This means that chlorine is in limiting amount., So, amount of yttrium (III) chloride will depend upon amount of chlorine gas.

According to reaction , 3 moles of chlorine gives 2 moles of [tex]YCl_3[/tex]

Then 0.01408 moles of chlorine will give :

[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of [tex]YCl_3[/tex]

Mass of 0.009387 moles of [tex]YCl_3[/tex]:

0.009387 mol × 195.52 g/mol = 1.835 g

1.835 grams of [tex]YCl_3[/tex]can be prepared.

To identify the elements in MX3, the calculations show that M is Iron (Fe) and X is Chlorine (Cl), forming Iron(III) Chloride (FeCl3). Given 1.00 g of each reactant, approximately 2.90 g of FeCl3 can be prepared. The limiting reagent in this process is Iron (Fe).

To determine the identities of M and X in the compound MX3 and the mass of MX3 that can be prepared, follow these steps:

Determine the molar mass of X2: The given data states that 0.105 g of X2 contains 8.92 × 1020 molecules. Using Avogadro's number (6.022 × 1023 molecules/mol), we can find the molar mass (MX2) of X2:

Calculate the atomic mass of X: Since X2 is diatomic, we divide the molar mass by 2:

Identify the element X - The atomic mass of 35.45 g/mol suggests that X is Chlorine (Cl).

Determine the molar mass and identity of M - Given that MX3 consists of 54.47% X by mass:

Solve for MM:

Identify the element M - The atomic mass of 55.85 g/mol suggests that M is Iron (Fe).

Name of MX3: Since M is Iron (Fe) and X is Chlorine (Cl), MX3 is Iron(III) Chloride (FeCl3).

Final Mass Calculation

Mole calculation for 1.00 g of M:

Limiting reagent: Each mole of X2 provides 2 moles of Cl, so: 0.0141 mol Cl2 × 2 = 0.0282 mol Cl (excess)

Iron (Fe) is the limiting reagent with 0.0179 mol producing 0.0179 mol of FeCl3

Calculate the mass of FeCl3:

Molar mass of FeCl3 = 55.85 g/mol + 3 × 35.45 g/mol ≈ 162.20 g/mol

Mass of FeCl3 = 0.0179 mol × 162.20 g/mol ≈ 2.90 g

Answer:

The absolute error = 3.1m

The relative error = 0.31

Explanation:Please see attachment for explanation

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

[tex]500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2[/tex]

[tex]42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl[/tex]

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The amount of hydrogen chloride yield in the second reaction is 1065.7 g.

The two given reactions

The amount of hydrogen gas yield in the first reaction is calculated as follows;

[tex]\frac{500 \ g\ H_2O}{18 \ g \ H_2O} \times (2 \ mol\ H_2) \times 0.753= 41.83 \ g \ H_2[/tex]

The amount of hydrogen chloride yield in the second reaction is calculated as follows;

[tex]41.83 \ g \times (36.5 \ HCl) \times 0.698 = 1065.7 \ g \ HCl[/tex]

Thus, the amount of hydrogen chloride yield in the second reaction is 1065.7 g.

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Answer:

The reaction rate is increased.

Explanation:

The pressure is the force the gas molecules do when hitting each other and the walls of the container they're. The partial pressure is the pressure of a substance in a mixture would have if it was alone at the same conditions.

Thus, when the partial pressure increases, it means that the molecules collide more often. The reaction happens when the molecules collide in the right way, so, if the collisions are happening more often, the rate must be higher.

Answer: 6.63

Explanation:Please see attachment for explanation

Final answer:

The pKa value for the weak acid HA can be calculated using the equation pKa = -log(Ka). Given the equilibrium concentrations, we can use the equation Ka = ([H+][A-])/[HA] to solve for Ka. After finding Ka, we can calculate pKa using the formula pKa = -log(Ka).

Explanation:

The pKa value for the weak acid HA can be calculated using the equation:

pKa = -log(Ka)

Given the equilibrium concentrations of [HA] = 0.170 M, [H+] = 2.00 × 10^-4 M, and [A-] = 2.00 × 10^-4 M, we can use the equation:

Ka = ([H+][A-])/[HA]

Plugging in the values gives us:

Ka = (2.00 × 10^-4)(2.00 × 10^-4)/(0.170)

After solving for Ka, we can calculate pKa using the formula pKa = -log(Ka).

Answer:

Xenon

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by [tex]N_0[/tex].

Avogadro constant:-

[tex]N_a=6.023\times 10^{23}[/tex]

Let the molar mass of the element is x g/mol

So,

[tex]6.023\times 10^{23}[/tex] atoms have a mass of x g

Also,

[tex]3.89\times 10^{24}[/tex] atoms have a mass of [tex]\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}[/tex] g

This mass is equal to 848 g

So,

[tex]\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}=848[/tex]

x= 131.3 g/mol

This mass correspond to xenon.

Answer: option B and option C

Explanation:

Answer: The Cu-63 isotope has the highest percent natural abundance

Explanation:

Average atomic mass 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]

We are given:

Two isotopes of Copper, which are Cu-63 and Cu-65

Average atomic mass of copper = 63.54 amu

As, the average atomic mass of copper is closer to the mass of Cu-63 isotope. This means that the relative abundance of this isotope is the highest as compared to the other isotope.

Percentage abundance of Cu-63 isotope = 69.2%

Percentage abundance of Cu-65 isotope = 30.8 %

Hence, the Cu-63 isotope has the highest percent natural abundance

Answer:

The answer will be that if there is a linear relationship exist between CO2 and the global temperature, then if one variable change, the other also will change correspondingly.

Answer: No crystals of potassium sulfate will be seen at 0°C for the given amount.

Explanation:

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be [tex]\frac{7.4}{100}\times 130=9.62g[/tex]

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

The question is incomplete, the complete question is:

For a certain chemical reaction, the standard Gibbs free energy of reaction is 144. kJ. Calculate the temperature at which the equilibrium constant K = [tex]5.9\times 10^{-26}[/tex] .

Round your answer to the nearest degree.

Answer:

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Explanation:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy = 144.0 kJ=144,000 J  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314 J/K mol[/tex]

T = temperature at which reaction is occurring = ?

K = Equilibrium constant of the reaction =[tex]5.9\times 10^{-26}[/tex]

Putting values in above equation, we get:

[tex]144,000 J/mol=-(8.3145J/Kmol)\times T\times \ln [5.9\times 10^{-26}][/tex]

[tex]T=\frac{144,000 J/mol}{-(8.314 J/Kmol)\times \ln [5.9\times 10^{-26}]}[/tex]

T = 298.15 K

T = 298.15 - 273 °C = 25°C

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Using the equation ΔG° = -RTlnK, which relates the Gibbs free energy change to the equilibrium constant, you can solve for temperature by rearranging the formula. Ensure that the units for Gibbs energy and the gas constant match. Plugging the given values into the rearranged formula will provide the temperature.

The student's question asked how to calculate the temperature at which the equilibrium constant K equals 5.9 x 10 with a standard Gibbs free energy of reaction of 144 kJ.

The relationship between the Gibbs free energy change and the equilibrium constant is defined by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/K mol), T is the absolute temperature in Kelvin, and K is the equilibrium constant.

To solve for temperature, rearrange the equation as T = -ΔG / (R * lnK). But first, you must ensure that the units for Gibbs energy and the gas constant match. If ΔG is given in kJ, it should be converted to J (1 kJ = 1000 J).

So, T = -(144,000 J) / (8.314 J/K mol * ln(5.9 x 10)), which should give you the answer.

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Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by  

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of  

hydration must be included.

1 Manganes  52.94 g = 63.55 g  

1 Sulphur  32.07 g =  

32.07 g 2 Hydrogen is  = 2.02 g

4 Oygen       =  

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol  18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

[tex]\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} \times 100 \%[/tex]

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

[tex]\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} \times 100 \%\\\\\text{Percent Mass} = 10.6 \%}[/tex]

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

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Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.

The balance chemical reaction will be:

[tex]2N_2(g)+O_2(g)\rightarrow 2N_2O(g)[/tex]

By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.

First we have to determine the limiting reagent.

From the reaction we conclude that,

As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas

So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas

It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.

Now we have to determine the moles of dinitrogen monoxide.

As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide

So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide

Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Final answer:

One part nitrogen gas (N₂) combines with one part oxygen gas (O₂) to produce two parts of nitric oxide (NO).

Explanation:

When nitrogen gas (N₂) combines with oxygen gas (O₂), they can form several different nitrogen oxides, depending on the conditions and proportions in which they react. For the specific formation of nitric oxide (NO), we need to understand that one volume of nitrogen gas will combine with one volume of oxygen gas to form two volumes of nitric oxide. This is evident from the balanced chemical equation for this reaction:

N₂(g) + O₂(g) → 2NO(g)

This means 1 part of nitrogen gas combines with 1 part of oxygen gas to form 2 parts of nitric oxide (NO). The balancing of the equation indicates that two molecules of NO are formed from one molecule of nitrogen and one molecule of oxygen, due to the conservation of mass and volume in chemical reactions according to the Law of Combining Volumes.

Answer:

A) 0.20 cm³

B) 49.7 m²

C) 99.99%

D) 17.7 mg

Explanation:

A) The density of a material represents the mass that it occupies in a "piece" of volume. Thus, the density (d) is the mass (m) divided by the volume (v):

d =m/v

If the mass is 40.0 mg = 0.04 g, and the density is 0.20 g/cm³, the volume is:

0.20 = 0.04/v

v = 0.04/0.20

v = 0.20 cm³

B) The surface area (S) is the are that is presented in each gram of the material, so, it's the area (a) divided by the mass (m):

S = a/m

If the mass is 40.0 mg = 0.04 g, and the surface area is 1242 m²/g, so:

1242 = a/0.04

a = 49.7 m²

C) The percent of mercury removed is the mass removed divided by the initial mass, this multiplied by 100%. The mass removed is the initial mass (m0) less the final mass (m), so:

%removed = [(7.748 - 0.001)/7.748] *00%

%removed = 99.99%

D) The final mass of the spongy material is it mass (10 mg) plus the mass removed of the mercury (7.748 - 0.001 = 7.747 mg), so:

m = 10 + 7.747

m = 17.747 mg

m = 17.7 mg

Final answer:

This detailed answer explains how to calculate the volume, surface area, percentage of mercury removed, and final mass of a new material used to remove mercury from water.

Explanation:

A) To calculate the volume of a 40.0-mg sample of the material, you can use the formula: Volume = Mass / Density. Therefore, Volume = 40.0 mg / 0.20 g/cm³ = 200 cm³.

B) The surface area for a 40.0-mg sample can be calculated by multiplying the specific surface area by the mass: 1242 m²/g × 40.0 mg = 49680 m².

C) The percentage of mercury removed from the water is: ((Initial mercury mass - Final mercury mass) / Initial mercury mass) × 100 = ((7.748 mg - 0.001 mg) / 7.748 mg) × 100 = 99.9874%.

D) The final mass of the spongy material after exposure to mercury is the initial mass minus the mass used: 40.0 mg - 10.0 mg = 30.0 mg.

Answer: No hydrogen bond cannot occur in this alpha helix structure.

Explanation: For hydrogen bond to form, the electronegativity difference should be more than 1.7. carbon has an electronegativity of 2.5 whereas hydrogen has 2.1 so their electronegativity difference is 0.3. So in this alpha helix structure cannot occur.

Molality  is the measure of concentration of solute in 1 kg of solution. The molality of the solution is 3.05 mol/kg.

10% of HCl (by mass) means 10 g of HCl and in 90 g of water.

Molar mass of HCl = 36.5 g/mol

Molality:

It is the measure of concentration of solute in 1 kg of solution. It can be calculated by the formula.

[tex]\bold {m = \dfrac {n }{w}\times 1000}[/tex]

Where,

m- molality

n - number of moles

w - weight of solvent in grams

Number of moles of HCl

[tex]\bold {n = \dfrac w{m} = \dfrac {10}{36.5} = 0.274 g}[/tex]

put the value in molality formula,

[tex]\bold {m = \dfrac {0.274 }{90}\times 1000}\\\\\bold {m = 3.05\ g/mol}[/tex]

Therefore, the molality of the solution is 3.05 mol/kg.

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Final answer:

To calculate the molarity of a 10.0% HCl solution, convert 10 grams of HCl to moles, and then divide by the volume of the solution in liters. This results in a molarity of 2.74 M.

Explanation:

To calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid (HCl), start by understanding that a 10.0% solution means there are 10 grams of HCl in 100 grams of the solution. First, since the solution is aqueous, we can assume the density is close to that of water, which is approximately 1 g/mL, so 100 grams of the solution is roughly equivalent to 100 mL (0.1 L) of solution.

Next, convert the mass of HCl to moles. The molar mass of HCl is about 36.46 g/mol:

10 grams HCl × (1 mol HCl / 36.46 grams) = 0.274 moles HCl

Then, divide the moles of HCl by the volume of the solution in liters to find the molarity:

Molarity = Moles of solute / Volume of solution in liters

Molarity = 0.274 moles HCl / 0.1 L = 2.74 M

The molarity of the 10.0% HCl solution is therefore 2.74 M.