Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Mg(s)+Cl2(g)⟶Mg2+(aq)+2Cl−(aq) Which substance gets oxidized? Mg2+ Cl− Mg Cl2 Which substance gets reduced? Cl2 Mg Cl− Mg2+ What is the oxidizing agent? Cl− Mg2+ Cl2 Mg What is the reducing agent? Mg2+ Cl2 Mg Cl−

Answer:

the concentration remains = 0.138 M

Explanation:

Let us calculate the charge passed:

The charge passed during the electrolysis = current X time(s)

charge passed = 2.40X30X60= 2160 C

we know that

96485 C = 1 F

Therefore 2160 C = 0.0224 F

according to Faraday's law of electrolysis if we pass one Faraday of charge through an electrolytic solution, 1 gram equivalent of metal will be deposited.

the gram equivalent of nickel deposited = 0.0224

The moles of Nickel deposited = 0.0224/2 = 0.0112 mol

the initial moles of Nickel ions present in solution is

[tex]moles=molarityXvolume(L)=0.250X0.1=0.0250mol[/tex]

The moles of nickel ion gets consumed = 0.0112

So moles of nickel ion left after electrolysis = 0.0138

[tex][Ni^{+2}]=\frac{moles}{volume(L)}\frac{0.0138}{0.1}=0.138M[/tex]

The concentration of Ni2+ ions remaining in the solution can be found by applying Faraday's first law of electrolysis and subtracting the moles of Ni2+ discharged from the initial moles present in the solution.

The electrolysis process involves a chemical reaction in which electricity is used to break down a substance into its constituent elements. In this scenario, we are determining the concentration of Ni2+ in the solution after the electrolysis process of NiSO4.

As per Faraday's first law of electrolysis, the mass of any element discharged during electrolysis is directly proportional to the quantity of electricity (current) passed. Thus, using Faraday's constant (96485  C/mol e-) and the given current and time, we can calculate how many moles of Ni2+ are plated out.

The number of moles of Ni2+ discharged would be equal to the number of moles initially present in the 100.0 mL of 0.250 M NiSO4 solution. By subtracting the moles of Ni2+ that were discharged from the total initial moles of Ni2+, we can obtain the number of moles of Ni2+ remaining in the solution. This value can be converted back into molarity by dividing the moles of Ni2+ remaining by the volume of the solution in liters.

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Answer:

70.3824 grams of carbondioxide is produced.

Explanation:

[tex]C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)[/tex]

Moles of glucose =  [tex]\frac{48.0 g}{180 g/mol}=0.2666 mol[/tex]

According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.

Then 0.2666 mol of gluocse will reactwith :

[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of oxygen

Mass of 1.5996 moles of oxygen gas:

[tex]1.5996 mol\times 32 g/mol = 51.1872 g[/tex]

51.1872 grams of oxygen are required to convert 48.0 grams of glucose to [tex]CO_2[/tex] and [tex]H_2O[/tex].

According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.

Then 0.2666 moles of glucose will give:

[tex]\frac{6}{1}\times 0.2666 mol =1.5996 mol[/tex] of carbon dioxide

Mass of 1.5996 moles of carbon dioxide gas:

[tex]1.5996 mol\times 44 g/mol = 70.3824 g[/tex]

70.3824 grams of carbondioxide is produced.

Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.

Let's consider the overall equation for the combustion of glucose in the human body.

C₆H₁₂O₆(aq) + 6 O₂(g) ⟶ 6 CO₂(g) + 6 H₂O(l)

We can calculate the grams of oxygen required to react with 48.0 g of glucose considering the following relations.

[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molO_2}{1molGlucose} \times \frac{32.00 gO_2}{1 molO_2} = 51.2 gO_2[/tex]

We can calculate the grams of carbon dioxide produced from 48.0 g of glucose considering the following relations.

[tex]48.0gGlucose \times \frac{1molGlucose}{180.16gGlucose} \times \frac{6molCO_2}{1molGlucose} \times \frac{44.01 gCO_2}{1 molCO_2} = 70.4 gCO_2[/tex]

Glucose reacts with 51.2 g of oxygen to produce 70.4 g of carbon dioxide and water.

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Answer:

Any one structure in the image can used to show the Lewis structure for [tex]N_2O[/tex].

Explanation:

Lewis structure is used to represent the arrangement  of the electrons around the atom in a molecule. The electrons are shown by dots and the bonding electrons are shown by line between the two atoms.

For [tex]N_2O[/tex],

The number of valence electrons = [tex](5\times 2)+6[/tex] = 16

The skeleton structure of the compound is in which one N molecule act as a central atom and is bonded to another N and O with one sigma bond each.

Number of valence electrons used in skeletal structure = 4

So,

Remaining valence electrons = 12

The central atom still needs 2 lone pairs and the nitrogen and oxygen attached to it, still needs 3 lone pairs each. So, total number of valence electrons needed to complete the octet of all the atoms = 16

But, we have 12 electrons and thus, 2 additional bonds are required in the structure which can be distributed in the 3 atoms in 3 ways.

[tex]\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}[/tex]

For Structure A:

Formal charge on central atom N:

[tex]\text{Formal charge}=\text{5}-\text{0}-\frac{\text{8}}{2}[/tex]

[tex]\text{Formal charge}= +1[/tex]

Formal charge on substituent atom N:

[tex]\text{Formal charge}=\text{5}-\text{4}-\frac{\text{4}}{2}[/tex]

[tex]\text{Formal charge}= -1[/tex]

Formal charge on substituent atom O:

[tex]\text{Formal charge}=\text{6}-\text{4}-\frac{\text{4}}{2}[/tex]

[tex]\text{Formal charge}= 0[/tex]

For Structure B:

Formal charge on central atom N:

[tex]\text{Formal charge}=\text{5}-\text{0}-\frac{\text{8}}{2}[/tex]

[tex]\text{Formal charge}= +1[/tex]

Formal charge on substituent atom N:

[tex]\text{Formal charge}=\text{5}-\text{6}-\frac{\text{2}}{2}[/tex]

[tex]\text{Formal charge}= -2[/tex]

Formal charge on substituent atom O:

[tex]\text{Formal charge}=\text{6}-\text{2}-\frac{\text{6}}{2}[/tex]

[tex]\text{Formal charge}= +1[/tex]

For Structure C:

Formal charge on central atom N:

[tex]\text{Formal charge}=\text{5}-\text{0}-\frac{\text{8}}{2}[/tex]

[tex]\text{Formal charge}= +1[/tex]

Formal charge on substituent atom N:

[tex]\text{Formal charge}=\text{5}-\text{4}-\frac{\text{2}}{2}[/tex]

[tex]\text{Formal charge}= 0[/tex]

Formal charge on substituent atom O:

[tex]\text{Formal charge}=\text{6}-\text{6}-\frac{\text{2}}{2}[/tex]

[tex]\text{Formal charge}= -1[/tex]

The Lewis structures in resonance of the compound is shown in the image.

A resonance structure of N2O can be drawn showing the nitrogen atom in the center, the formal charges of the atoms, and the lone pairs of electrons. The double bond between nitrogen and oxygen can be delocalized or moved to form a resonance hybrid.

One of the resonance structures for dinitrogen monoxide (N2O) can be drawn as follows:

In this structure, the nitrogen atom in the center has a formal charge of +1 and both nitrogen and oxygen atoms have lone pairs of electrons. The double bond between the nitrogen and oxygen atoms can be delocalized or moved to form a resonance hybrid, which is an average of the resonance structures.

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Answer:

Part A: 36 MBq; Part B: 18 MBq

Explanation:

The half-life is the time it takes for half the substance to disappear.

The activity decreases by half every half-life

A =Ao(½)^n, where n is the number of half-lives.

Part A

3.0 da = 1 half-life

A =  Ao(½) = ½ × 72 MBq = 36 MBq

Part B

6.0 da = 2 half-lives

A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq

Final answer:

Part A: The activity of 201Tl after 3.0 days is 36 MBq. Part B: The activity of 201Tl after 6.0 days is 18 MBq.

Explanation:

Part A: To calculate the activity of 201Tl after 3.0 days, we need to determine the number of half-lives that have passed. Since the half-life of 201Tl is 3.0 days, dividing 3.0 days by the half-life gives us 1 half-life. Each half-life reduces the activity of the isotope by half, so after 1 half-life, the activity would be half of the initial activity. Therefore, the activity after 3.0 days would be 72 MBq / 2 = 36 MBq.

Part B: To calculate the activity of 201Tl after 6.0 days, we can use the same method as in Part A. Since the half-life is 3.0 days, 6.0 days is equivalent to 2 half-lives. Each half-life reduces the activity by half, so after 2 half-lives, the activity would be 72 MBq / 2 / 2 = 18 MBq.

Answer:

[tex]P=5.6*10^{6} Pa[/tex]

Explanation:

Consider that, as the system is adiabatic, [tex]U_{1}= U_{2}[/tex] where U1 and U2 are the internal energies before the process and after that respectively.

Consider that: [tex]U=H-PV[/tex], and that the internal energy of the first state is the sum of the internal energy of each tank.

So, [tex]H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}[/tex]

Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: [tex]H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}[/tex]

So, [tex]P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}[/tex]

Isolating [tex]P_{2}[/tex],

[tex]P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}[/tex]

[tex]V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}[/tex]

So,

[tex]P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa[/tex]

Answer:

35.5 amu is the average atomic mass of chlorine

Explanation:

Fractional abundance of chlorine‑35 = 75.5%

Fractional abundance of chlorine‑37 = 24.5%

Average atomic mass is equal to summation of products of all isotopes masses into their fractional abundance.

Average atomic mass =  Σ(Mass of an isotope × fractional abundance)

Average atomic of chlorine :

[tex]35 amu\times 0.755+37 amu\times 0.245=35.49 amu\approx 35.5 amu[/tex]

35.5 amu is the average atomic mass of chlorine

Answer: B) [tex]1.60\times 10^{-6}[/tex] M, acidic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.

[tex]pH=-\log [H_3O^+][/tex]

[tex]pOH=-log{OH^-}[/tex]

[tex]pH+pOH=14[/tex]

Given : [tex][OH^-]=6.25\times 10^{-9}M[/tex]

[tex]pOH=-log[6.25\times 10^{-9}M][/tex]

[tex]pOH=8.20[/tex]

[tex]pH=14-8.20=5.8[/tex]

As pH is less than 7, the solution is acidic.

[tex]5.8=-log[H_3O^+][/tex]

[tex][H_3O^+]=1.60\times [10^{-6}M[/tex]

Thus solution is acidic and concentration of [tex]H_3O^+[/tex] is [tex]1.60\times [10^{-6}M[/tex]

Hey there!:

240 mL bottle of the oral solution , so 43.2% alcohol

Therefore:

240 mL *  43.2 / 100

= 2.4 * 43.2

=> 103.68 mL of  alchohol

Hope this helps!

Ritonavir oral solution contains, in addition to ritonavir, 43.2% alcohol and 26.57% propylene glycol. The quantity of alcohol in a 240 ml bottle of the oral solution is 103.68 ml of alcohol.

Alcohol is a drink that is made up of rotten fruits and vegetables, It is made by fermentation. They cause unconsciousness to the brain and the body. It creates hallucinations.

Ritonavir is taken with meals twice a day. Ask your doctor or pharmacist to explain any instructions on your prescription label that you are unsure about following.

Given the 43.2% alcohol and 26.57% propylene glycol. The bottle is of 240 ml.

240 mL x  43.2 / 100

2.4 x 43.2 = 103.68 ml of  alcohol.

Thus, the quantity of alcohol in a 240 ml bottle is 103.68 ml of alcohol.

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Answer:

The partial pressure of the [tex]CO_2[/tex] in the final mixture is 200 kPa.

Explanation:

Pressure of nitrogen gas when the two tanks are disconnected = 500 kPa

Pressure of the carbon-dioxide gas when the two tanks are disconnected = 200 kPa

Moles of nitrogen gas =[tex]n_1= 2 kmol[/tex]

Moles of carbon dioxide gas =[tex]n_2=8 kmol[/tex]

After connecting both the tanks:

The total pressure of the both gasses in the tank = p = 250 kPa

According to Dalton' law of partial pressure:

Total pressure is equal to sum of partial pressures of all the gases

Partial pressure of nitrogen =[tex]p_{N_2}^o[/tex]

Partial pressure of carbon dioxide=[tex]p_{CO_2}^o[/tex]

[tex]p_{N_2}^o=p\times \frac{n_1}{n_1+n_2}[/tex]

[tex]p_{N_2}^o=250 kPa\times \frac{0.2}{0.2+0.8}=50 kPa[/tex]

[tex]p_{CO_2}^o=p\times \frac{n_2}{n_1+n_2}[/tex]

[tex]p_{CO_2}^o=250 kPa\times \frac{0.8}{0.2+0.8}=200 kPa[/tex]

The partial pressure of the [tex]CO_2[/tex] in the final mixture is 200 kPa.

Answer: [tex]H^+[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Example:  For the given chemical equation:

[tex]H_3BO_3(aq.)+HS^-(aq.)\rightarrow H_2BO_3^-(aq.)+H_2S(aq.)[/tex]

Here, [tex]H_3BO_3[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]H_2BO_3^-[/tex] which is a conjugate base.

And, [tex]HS^-[/tex] is gaining a proton, thus it is considered as a base and after gaining a proton, it forms [tex]H_2S[/tex] which is a conjugate acid.

Thus the chemical symbol that is commonly used to represent a "proton" in the context of Bronsted-Lowry acids and bases is [tex]H^+[/tex]

Final answer:

In Brønsted-Lowry acid-base theory, a "proton" is symbolized by H₊ and is the key player in reactions where a proton donor (acid) transfers a proton to a proton acceptor (base).

Explanation:

The chemical symbol commonly used to represent a "proton" in the context of Brønsted-Lowry acids and bases is H₊. A proton essentially refers to a hydrogen ion (H₊) devoid of its electron, leaving just a single positive charge. In Brønsted-Lowry theory, a proton donor (acid) transfers a proton to a proton acceptor (base), thereby defining the roles of the acid and base in an acid-base reaction. This concept is crucial in understanding reactions such as the combination of ammonia and water, where NH₃ acts as the proton acceptor and water acts as the proton donor, forming NH₊₄ and OH₋.

Hey there!:

Ksp =(Mg⁺²)(2F⁻)²

Ksp = (1.18*10⁻³)(2*1.18*10⁻³)²

Ksp = 6.57*10⁻⁹

Hope that helps!

Answer:

Li

Explanation:

The group 1 elements are the alkali elements, and they are the ones that have one valence electron int heir outer shell, along wiht them Hydrogen is also an element that has one valence electron but is not considered group 1 because it is not a solid in temperature room, but a gas and has different properties, from the options the most different from the others is litium, because it can react with the same elements than the other elements in group one, plus nitrogen in a very violent way.

The group 1 element that exhibits slightly different chemistry from the others is: Lithium (Li).

The chemical elements found in group 1 of the periodic table are highly electro positive metals and have a single (1) valence electron in its outermost shell.

Some examples of these chemical elements (alkali metals) are;

Li is the symbol for the chemical element referred to as Lithium.

Lithium (Li) is an alkali metal and as such it is found in group 1 of the periodic table with a single (1) valence electron in its outermost shell.

Also, the electronic configuration of Lithium (Li) is written as;

As a result of the small size of Lithium (Li), it is exhibits slightly different chemistry from the others such as:

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Answer:

The final temperature of the water system is 51.5 °C.

Explanation:

A 50.0 gram sample of water initially at 100 °C

Mass of the water = 50 g

Initial temperature of the water = [tex]T_i=100[/tex]

Final temperature of the water after mixing = [tex]T_f=[/tex]

-Q' = heat lost by 50.0 g of water

[tex]-Q'=mc\Delta (T_f-T_i)[/tex]

A 100.0 gram sample of water initially at 27.32°C

Mass of the water ,m'= 10.00 g

Initial temperature of the water = [tex]T_i'=100[/tex]

Final temperature of the water after mixing = [tex]T_f=[/tex]

Q = heat gained by 100.0 g of water after mixing

[tex]Q=m'c\Delta (T_f-T_i')[/tex]

-Q'=Q (Energy remains conserved)

[tex]-(mc\Delta (T_f-T_i))=m'c\Delta (T_f-T_i')[/tex]

[tex]-(50 g\times 4.184J/g^oC(T_f-100^oC))=100 g\times 4.184J/g^oC(T_f-27.32^oC)[/tex]

[tex]T_f=51.54^oC\approx 51.5^oC[/tex]

The final temperature of the water system is 51.5 °C.

To determine the final temperature of the system, we consider the exchange of heat between the warm and cold water samples. We use the heat transfer formula and conservation of energy principle. Then, solving for 'T', we get the final temperature – in scientific notation.

This question revolves around the concept of specific heat and its application in heat transfer. When two substances with different initial temperatures are mixed, they will exchange heat until an equilibrium temperature is reached.

The equation to calculate heat exchanged is: Q = mcΔT, where 'm' is mass, 'c' is specific heat, and 'ΔT' is the change in temperature.

In this context, water with a temperature of 100 °C will lose heat (Qhot), whereas water at 27.32 °C will gain heat (Qcold). When the system reaches thermal equilibrium, Qhot = -Qcold due to the law of conservation of energy.

So, we will have two equations:

We then solve for T, which represents the final temperature of the system.

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Answer:

False

Explanation:

Atomic bonding can be of two types:

Ionic bonds: The bond formed between two opposite charged ions. The strong electrostatic force acting between the ions make them compounds of high stability . Example: NaCl

Covalent bonds: The bond formed by sharing of the electrons between the atoms. The balance between the attractive force (nucleus, electron) and repulsive force (electron, electron or nucleus, nucleus) makes the compound stable. Example: [tex]CH_4[/tex]

The bonds are either formed by transferring (losing and gaining) or sharing of electrons. Thus, compound forming these bonds share or transfer electrons to attain noble gas configuration. Thus, they lack free electrons.

Answer: The mass of water required will be 10.848 g.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]    ......(1)

Given mass of nitric acid = 75.9 g

Molar mass of nitric acid = 63 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of nitric acid}=\frac{75.9g}{63g/mol}=1.204mol[/tex]

For the given chemical equation:

[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]

By Stoichiometry of the reaction:

2 moles of nitric acid is produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced by = [tex]\frac{1}{2}\times 1.204=0.602mol[/tex] of water.

Now, calculating the amount of water, we use equation 1:

Moles of water = 0.602 mol

Molar mass of water = 18.02 g/mol

Putting values in equation 1, we get:

[tex]0.602=\frac{\text{Mass of water}}{18.02g/mol}\\\\\text{Mass of water}=10.848g[/tex]

Hence, the mass of water required will be 10.848 g.

To form 75.9 g of HNO3, 10.85 g of water is required.

To determine the number of grams of water required to form 75.9 g of HNO3, we need to use the given balanced equation and apply stoichiometry. From the equation, we can see that 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3. Using the molar masses of H2O and HNO3, we can calculate the amount of water needed.

First, convert the mass of HNO3 to moles using its molar mass:

HNO3: 75.9 g ÷ 63.02 g/mol = 1.204 mol HNO3

Next, use the stoichiometric ratio to find the moles of water:

H2O: (1.204 mol HNO3) × (1 mol H2O) ÷ (2 mol HNO3) = 0.602 mol H2O

Finally, convert the moles of water to grams using its molar mass:

0.602 mol H2O × 18.02 g/mol = 10.85 g H2O

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The question involves calculating the CO2 emissions from the annual production of calcium oxide in the U.S., based on the stoichiometry of the reaction converting CaCO3 to CaO, which releases CO2.

The question is about calculating the yearly release of CO2 to the atmosphere resulting from the annual production of calcium oxide (CaO) in the United States, given that the reaction for producing CaO from calcium carbonate (CaCO3) releases CO2. Since the reaction is CaCO3(s) → CaO(s) + CO2(g), for each mole of CaCO3 decomposed, one mole of CO2 is released. Calcium carbonate (mol weight = 100.09 g/mol) decomposes to give calcium oxide (mol weight = 56.08 g/mol) and carbon dioxide (mol weight = 44.01 g/mol). Given the annual production of CaO is 8.6 × 10¹° kg, we first convert this mass to moles (using CaO's molar mass), and then calculate the corresponding moles (and mass) of CO2 released.

hey there!

The combined gas equation is  :

P1V1/T1 = P2V2/T2  

Since Pressure is constant for this question it will cancel out leaving  

V1/T1 = V2/T2  

where  :

V1 = initial volume = 20.9 L  

T2 = initial temp (must be in Kelvin) = 286 K  

V2 = final volume = ? L  

T2 = final temp = 264 K  

Solve for V2  

V2 = V1T2 / T1  

= 20.9 L * 264 K / 286 K  

= 19.29 L  

= 19.3 L (3 sig digits)

Hope this helps!

Applying Charles' law, which establishes a direct relationship between the temperature and volume of a gas at constant pressure, the new volume of the helium gas in the balloon when the temperature falls from 13 degrees Celsius to -9 degrees Celsius, is approximately 19.3 litres.

The student's question revolves around temperature changes and its effects on the volume of a gas, which is regulated by Charles' law. Charles' law expresses that the volume of a gas is directly proportional to its temperature (in Kelvin), assuming pressure is kept constant.

Given the initial volume (V1) is 20.9L and the initial temperature (T1) is 13 degrees Celsius (or 286.15K since K = C + 273.15), when the temperature falls to -9 degrees Celsius (or 264.15K), the new volume (V2) can be calculated. By Charles' law: V1/T1 = V2/T2, we will then have: (20.9 L/286.15 K) * 264.15 K = V2. Thus, the new volume of the balloon, rounded to significant digits, is approximately 19.3 L

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Answer: [tex]10.17\times 10^{-23}g[/tex]

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

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

[tex]\text{Number of moles}=\frac{\text{Given atoms}}{\text{Avogadro's number}}=\frac{1}{6.023\times 10^{23}}=0.16\times 10^{-23}atoms[/tex]

1 mole of zinc weighs = 65.38 g

[tex]0.16\times 10^{-23}[/tex]  moles of Zinc weigh = [tex]\frac{63.58}{1}\times 0.16\times 10^{-23}=10.17\times 10^{-23}g[/tex]

Thus mass of one atom of Zinc weigh [tex]10.17\times 10^{-23}g[/tex]

Answer: e. 3.5 atm

Explanation:

[tex]\pi =CRT[/tex]

[tex]\pi[/tex] = osmotic pressure  = ?

C= concentration in Molarity

R= solution constant  = 0.0821 Latm/Kmol

T= temperature = [tex]36^0C=(36+273)K=309K[/tex]

For the given solution: 0.82 grams of [tex]NaCl[/tex] is dissolved in 100 g of solution.

[tex]{\text {volume of solution}}=\frac{\text {mass of solution}}{\text {Density of solution}}=\frac{100g}{1.0g/ml}=100ml[/tex]

[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}[/tex]

Putting in the values we get:

[tex]C_{NaCl}=\frac{0.82\times 1000}{58.5\times 100}=0.14M[/tex]

[tex]\pi=0.14mol/L\times 0.0821Latm/Kmol\times 309K[/tex]

[tex]\pi=3.5atm[/tex]

The osmotic pressure of a 0.82% NaCl by weight aqueous solution is 3.5 atm

The osmotic pressure of a 0.82% NaCl solution at 36°C is 34 atm.

The osmotic pressure of a solution can be calculated using the formula II = MRT, where II is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.

In this case, the molarity of the solution is 0.70 M NaCl. Since 1 mol of NaCl produces 2 mol of particles, the total concentration of dissolved particles is (2)(0.70 M) = 1.4 M.

Plugging in the values into the formula, II = (1.4 mol/L) [0.0821 (L· atm)/(K · mol)] (298 K) = 34 atm.

Answer : The value of [tex]K_c[/tex] for the following reaction will be, 1024

Explanation :

[tex]K_c[/tex] is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants

The given balanced equilibrium reaction is,

[tex]P_4O_{10}(s)\rightleftharpoons P_4(s)+5O_2(g)[/tex]

As we know that the concentrations of pure solids are constant that means they do not change. Thus, they are not included in the equilibrium expression.

The expression for equilibrium constant for this reaction will be,

[tex]K_c=[O_2]^5[/tex]

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

[tex]K_c=(4.000)^5[/tex]

[tex]K_c=1024[/tex]

Therefore, the value of [tex]K_c[/tex] for the following reaction will be, 1024

The value of the equilibrium constant for the given reaction is 1,024.

Equilibrium constant of any chemical reaction at the equilibrium state is calculated as:

Equilibrium constant Kc = Concentration of product / Concentration of reactant

Given chemical reaction is:

P₄O₁₀(s) → P₄(s) + 5O₂(g)

Equilibrium constant for the above reaction is written as:

Kc = [P₄][O₂]⁵ / [P₄O₁₀]

For this reaction only concentration of oxygen gas is taken into consideration, as all other quantities are present in solid form and their value at equilibrium is 1. So, value of Kc is calculated as:

Kc = [O₂]⁵

Given concentration of O₂ = 4M

Kc = (4)⁵ = 1,024

Hence, 1,024 is the value of Kc.

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Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex] Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

[tex]PV=nRT[/tex]

[tex]n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol[/tex]

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

[tex]\frac{2}{1}\times 12.1597 mol=24.3194 mol[/tex] of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

Answer:

28.12 g Na2CrO4

Explanation:

To solve this question, we have to work with stoichiometry:

Then, if we have 45.5 g of Silver Nitrate:

[tex]45,5 g AgNo3 x \frac{1 mol AgNo3}{169.87 g AgNo3} = 0.2678 mol AgNo3\\[/tex]

We know that for each Na2CrO4's mol, we need 2 AgNo3's moles, so, we can calculate the amount of AgNo3's moles as follows:

[tex]0.2678 mol AgNo3 * \frac{1 mol Na2CrO4}{2 Mol AgNO3} =0.134 mol Na2CrO4\\[/tex]

and, now we can calculate the grams of sodium chromate as:

[tex]0.134 mol Na2CrO4 *\frac{209.9714 g Na2CrO4}{ molNa2CrO4} =28.12 g Na2CrO4[/tex]

To react completely with 45.5 g of silver nitrate, you would need approximately 21.6 g of sodium chromate based on the stoichiometric ratio from the balanced chemical equation.

In order to determine how many grams of sodium chromate is needed to react completely with the silver nitrate, we first look at the balanced chemical equation given: 2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3. This equation tells us that it takes 2 moles of AgNO3 to react completely with 1 mole of Na2CrO4. Now, we need to know the molar mass of AgNO3 and Na2CrO4 which are approximately 170 g/mol and 162 g/mol respectively.

Then, calculate the number of moles of AgNO3 we have by dividing the mass by the molar mass: 45.5 g / 170 g/mol = 0.267 moles. According to the balanced chemical equation, 0.267 moles of AgNO3 will react completely with half the amount of Na2CrO4. So, the number of moles of Na2CrO4 = 0.267 moles / 2 = 0.1335 moles. To find how many grams this is, we multiply by the molar mass of Na2CrO4: 0.1335 moles * 162 g/mol = 21.63 g. Therefore, you would need approximately 21.6 g of sodium chromate to react completely with 45.5 g of silver nitrate.

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Hey there!:

Option: C) a solution that is 0.10 M HC₂H₃O₂and 0.10 M LiC₂H₃O₂

Reason : Buffer mixture means mixture of either weak acid and salt of weak acid or weak base and salt of weak base.

Here acetic acid is a weak acid and its Li salt

Hope that helps!

Hey there !

below is the answer to the attached question

Answer:The isoelctric point of phenylalanine is 5.91. Kindly refer the attachment for structure.

Explanation:

Isoeectronic point is the pH at which the amino acid do not migrate in presence of an electric field.Hence at isoelectric point the molecule exists as a neutral molecule.

So for a molecule to not migrate in electric field it must be neutral. At isoelectric point the net charge on the amino acid molecule is zero.

The amino acid at isoelectric point exists as a zwitter ion which has a equal positive and equal negative charge making the net charge of the amino acid to be zero.

since the amino acid molecule has two kinds of functional group present which are an amine group and a carboxylic acid group.

So to calculate the isoelectric point we must know the pKa of these functional groups.

The formula for calculating the isoelectic point (pI) is :

[tex]pI=[pKa1+pKa2]\times 1/2[/tex]

pI=Isoelectric point

pKa1=pKa of carboxy group

pKa2=pKa of amine group

we have the pKa1 and pKa2 given in the question so putting these values in above formula we can calculate the isoelectric point.

pKa1=2.58

pKa2=9.24

pI=1/2×[pKa1+pKa2]

pI=1/2×[2.58+9.24]

pI=5.91

The isoelectric point is 5.91 hence at a pH of 5.91 phenylalanine would exist as a zwitterion and would not migrate in presence of a electric field.

Answer:

79.0 g

Explanation:

1. Gather the information in one place.

MM:    148.89        253.81

           2NaI + Cl2 → I2 + 2NaCl

m/g:                         67.3

2. Moles of I2

n = 67.3 g × (1 mol/253.81 g) = 0.2652 mol I2

2. Moles of NaI needed

From the balanced equation, the molar ratio is 2 mol NaI: 1 mol I2

n = 0.028 76 mol I2× (2 mol NaI/1 mol I2) = 0.5303 mol NaI

3. Mass of NaI

m = 0.5303 mol × (148.89 g/1 mol) = 79.0 g NaI

It takes 79.0 g of NaI to produce 67.3 g of I2.

We need approximately 79.5 g of sodium iodide, NaI, to produce 67.3 g of iodine, I2, based on the stoichiometry of the reaction 2NaI(aq)+Cl2(g)⇾I2(s)+2NaCl(aq).

The question is about determining the amount of sodium iodide, NaI needed to produce a certain amount of iodine, I2, based on the equation 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq). To solve this, we invoke stoichiometry. From the balanced equation, we know that 2 moles of NaI yield 1 mole of I2. Therefore, if we want to find the mass of NaI needed to produce 67.3 g of I2, we need to convert the mass of I2 to moles using its molar mass (about 253.8 g/mol), then multiply that by the ratio of moles of NaI to moles of I2 (which is 2:1), and then convert that to grams using the molar mass of NaI (about 149.9 g/mol). So, the calculation would be as follows: (67.3 g I2)/(253.8 g/mol) * 2 (moles of NaI/mole of I2) * 149.9 g/mol = about 79.5 g of NaI.

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Based on Coulomb's Law of electrostatic attraction of oppositely-charged species, the chlorine-containing compound that is predicted to have the greatest (most exothermic) lattice energy is; MgCl₂

     The lattice energy is defined as the energy required to dissociate one mole of an ionic compound to its constituent gaseous ions.

       Thus; F =  (q₁  ×  q ₂)/r²

        Now, Lattice energy is inversely proportional to the size of the ions. This implies that as the size of the ions increases, then the lattice energy will decrease. The size of the ions is also called the atomic radii.

Thus, as atomic radii increases, lattice energy will decrease.

      From the formula above and definition, we can tell that Lattice energy decreases down a group. Also, lattice energy will increase as the charges increase.

Now, the compounds we are dealing with are;

CaCl₂ - Calcium Chloride

NaCl - Sodium Chloride

MgCl₂ - Magnesium Chloride

KCl - Potassium Chloride

CCl₄ - Carbon tetrachloride

The charges of the metals that form the chlorides above are;

Ca = +2

Na = +1

Mg = +2  

K =  +1

The highest charges are Ca and Mg.

Now, they both belong to group two of the periodic table with Ca below Mg in the periodic table and as such Mg will have a greater Lattice energy.

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From the list of chlorine-containing compounds,  MgCl₂ would have the greatest lattice energy. Mg²⁺ has a greater charge than the other cations, resulting in stronger electrostatic attraction and most exothermic energy. CCl₄ is a covalently bonded molecule not an ionic compound, so does not possess lattice energy.

Based on Coulomb's Law of electrostatic attraction, the strongest (most exothermic) lattice energy would be present in the compound with the highest positively charged ion and smallest ionic radii, which results in a significant increase in lattice energy. This comes from the idea that lattice energy increases with higher charges and decreases with larger ionic radii.

In this case, it would be MgCl₂, because Mg⁺ has a greater charge than the other cations listed (Ca²⁺, Na⁺, K⁺). Greater charge results in stronger electrostatic attraction and hence the most exothermic lattice energy.

It's worth noting that CCl₄ is a covalently bonded molecule and not an ionic compound, thus it does not have lattice energy in the context of ionic compounds.

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The initial partial pressures of NO and Cl₂ were both 230 torr. We determined this by using the equilibrium constant (Kp) expression for the reaction and solving for the unknown initial partial pressures while knowing the equilibrium partial pressure of NOCl.

To find the initial partial pressures of NO and Cl₂, we'll use the equilibrium constant expression for the given reaction:

2 NO(g) + Cl₂(g) → 2 NOCl(g), Kp = 0.27 at 700 K

Let the initial partial pressures of NO and Cl₂ be p. At equilibrium, the partial pressure of NOCl is 115 torr. The change in partial pressure for NO and Cl₂ will be equal to the partial pressure of NOCl due to stoichiometry, so we can write p - 115 torr for both of them. Now, plug these values into the equilibrium expression:

Kp =  [tex]\frac{{(PNOCl)^2}}{{(PNO)^2 (PCl_2)}}[/tex]

Substitute the values:

0.27 =[tex]\frac{{(115 torr)^2}}{{(p - 115 torr)^2}}[/tex]

Solve for p to find the initial partial pressures. Upon calculation, you'll find that p = 230 torr.

Answer:

Mole fraction of glycerol is 0.02666.

Volume of the mixture is 99.75 mL.

Explanation:

Volume of glycerol,V = 10 mL

Mass of glycerol = m

Density of glycerol =[tex]\rho = 1.25802 g/cm^3=1.25802 g/mL[/tex]

[tex]m=\rho \times V=1.25802 g/mL\times 10 mL=12.5802 g[/tex]

Volume of water,V' = 90 mL

Mass of water= m'

Density of water =[tex]\rho '= 0.99708 g/cm^3=0.99708 g/mL[/tex]

[tex]m'=\rho '\times V'=0.99708 g/mL\times 90 mL=89.7372 g[/tex]

Mole fraction of glycerol =[tex]\chi_g=\frac{n_g}{n_g+n_w}[/tex]

[tex]\chi_g=\frac{\frac{12.5802 g}{92.09 g/mol}}{\frac{12.5802 g}{92.09 g/mol}+\frac{89.7372 g}{18 g/mol}}=0.02666[/tex]

Volume of the solution: v

Mass of the solution = M = 12.5802 g + 89.7372 g =102.3174 g[/tex]

Density of the mixture = [tex]\rho _{mix}=1.02567 g/cm^3=1.02567 g/mL[/tex]

[tex]v=\frac{M}{\rho _{mix}}=\frac{102.3174 g}{1.02567 g/mL}=99.75 mL=99.75 cm^3[/tex]

Volume of the mixture is 99.75 mL.

Theoretically the volume of the solution should be 100 ml.But experimentally the volume of the solution is 99.75 ml which is less than the theoretical volume.

This is because water molecules and glycerol molecules are getting associated with each other due to hydrogen bonding which results in less volume of the mixture.

Answer:

the molecular formula of the compound is N2O4

Explanation:

- Find the empirical formula

mole of N present = mass of N divided by molar mass of N = 0.140/14 = 0.01 mole

mole of O present = mass of O divided by molar mass of O = 0.320/16 = 0.02 mole

Divide both by the smallest number of mole to determine the coefficient of each, the smallest number of mole is 0.01 thus:

quantity of N = 0.01/0.01 = 1

quantity of O = 0.02/0.01= 2

thus the empirical formula = NO2

- Now determine the molecular formula by finding the ratio of molecular formula and empirical formula

Molar mass of molecular formula = 92.02 amu = 92.02 g/mole

Molar mass of empirical formula NO2 = (14 + (16 x 2)) = 46 g/mole

the x factor = 92.02/46 = 2

Molecular formula = 2 x NO2 = N2O4

Final answer:

The molecular formula of the compound with 0.140 grams of N and 0.320 grams of O and a molecular mass of 92.02 amu is N₂O₄.

Explanation:

To determine the molecular formula of the compound, we first need to calculate the number of moles of nitrogen (N) and oxygen (O) in the sample. Using the atomic masses of N (14.0 amu) and O (16.0 amu), we divide the mass of each element in the sample by its respective atomic mass to find the moles:

Moles of O = 0.320 g / 16.0 g/mol = 0.02 mol

Next, we determine the simplest whole number ratio of the moles of each element, which gives us the empirical formula. In this case, the ratio of N to O is 1:2, so the empirical formula is NO₂.

Since we know the molecular mass of the compound is 92.02 amu, we can compare it to the mass of the empirical formula to find the molecular formula. The mass of NO₂ is 14.0 + (2 x 16.0) = 46.0 amu. Thus, the molecular mass of our compound (92.02 amu) is approximately twice that of NO₂, so the molecular formula is N₂O₄.