Provide a structure for the compound with molecular formula C9H12 and with the following spectroscopic data.1H NMR: 1.2δ (doublet, I=6H), 3.0δ (septet, I=1H), 7.1δ (multiplet, I=5H)

Answer:

a) [tex]S_{AgI} = 9.11 \cdot 10^{-9} mol/L[/tex]

b) Keq = 8.3x10⁴

c) [tex] S_{[ AgI]} = 0.05 M [/tex]  

Explanation:

a) To find the molar solubility of AgI in water we need to use the solubility product constant (Ksp) of the following reaction:

AgI(s)  ⇄  Ag⁺(aq) + I⁻(aq)

[tex] K_{sp} = [Ag^{+}][I^{-}] = 8.3\cdot 10^{-17} [/tex]  

Since [Ag⁺] = [I⁻]:

[tex]K_{sp} = [Ag^{+}]^{2} \rightarrow S = \sqrt{K_{sp}} = \sqrt{8.3\cdot 10^{-17}} = 9.11 \cdot 10^{-9} mol/L[/tex]      

Hence, the molar solubility of AgI in water is 9.11x10⁻⁹ mol/L.

b) The equilibrium constant of AgI in CN, first we need to evaluate the reactions involved:

AgI(s)  ⇄  Ag⁺(aq) + I⁻(aq)   (1)

[tex] K_{sp} = [Ag^{+}][I^{-}] [/tex]   (2)

Ag⁺(aq) + 2CN⁻(aq)  ⇄  Ag(CN)₂⁻(aq)   (3)

[tex] K_{f} = \frac{[Ag(CN)_{2}^{-}]}{[Ag^{+}][CN^{-}]^{2}} [/tex]   (4)

The net equation is given by the sum of the reactions (1) and (3):

AgI(s) + 2CN⁻(aq)  ⇄  Ag(CN)₂⁻(aq) + I⁻(aq)    (5)

Hence, the equilibrium constant of the reaction (5) is:

[tex] K_{eq} = \frac{[Ag(CN)_{2}^{-}][I^{-}]}{[CN^{-}]^{2}} [/tex]   (6)

From equation (2):

[tex] [I^{-}] = \frac{K_{sp}}{[Ag^{+}]} [/tex]   (7)

By entering equations (7) and (4) into equation (6) we have:

[tex] K_{eq} = \frac{[Ag(CN)_{2}^{-}]}{[CN^{-}]^{2}}*\frac{K_{sp}}{[Ag^{+}]} [/tex]  

[tex] K_{eq} = K_{f}*K_{sp} = 1.0 \cdot 10^{21}*8.3\cdot 10^{-17} = 8.3 \cdot 10^{4} [/tex]                                      

Therefore, the equilibrium constant for the reaction (5) is 8.3x10⁴.  

c) From reaction (5):

AgI(s) + 2CN⁻(aq)  ⇄  Ag(CN)₂⁻(aq) + I⁻(aq)    

             0.1 - 2x                x                  x

[tex] K_{eq} = \frac{[Ag(CN)_{2}^{-}][I^{-}]}{[CN^{-}]^{2}} = \frac{x*x}{(0.1 - 2x)^{2}} [/tex]

[tex] x^{2} - 8.3 \cdot 10^{4}*(0.1 - 2x)^{2} = 0 [/tex]

Solving the above equation for x:

[tex] x = 0.05 mol/L = S_{[ AgI]} [/tex]

Hence, the molar solubility of AgI in a NaCN solution is 0.05 M.

I hope it helps you!

Molar solubility of AgI in pure water and in a NaCN solution can be calculated using the given values of Ksp and Kf through the described steps leading to the calculations of concentration of various ions required.

To determine the molar solubility of AgI in pure water, we need to use the solubility product constant (Ksp) for AgI. The Ksp expression for AgI is [Ag+][I-] ; when AgI dissolves, it separates into Ag+ and I- ions in a 1:1 ratio. If we represent the molar solubility as 's', then at equilibrium, the concentrations of Ag+ and I- are both 's'. So the Ksp is (s)(s)=s².

Next, for the reaction of AgI with CN-, we need to use the formation constant (Kf) for Ag(CN)2− ; this reaction can be written as Ag+ + 2CN- ⇌ Ag(CN)2-. If we denote the concentration of CN- as 'y', and we know the molar solubility of AgI is 's', then the equilibrium concentrations of Ag+ and CN- are s and y and the concentration of Ag(CN)2- is s. Therefore, the equilibrium constant for the given reaction is K=kf/Ksp which is equal to [s]/{[s][y]^2).

Finally, in a 0.100M NaCN solution, the CN- ion (from NaCN) reacts with Ag+ ion (from AgI) to form Ag(CN)2- complex, thereby decreasing the concentration of Ag+ ion which in turn increases the solubility of AgI. This increase in solubility of AgI can be calculated by the formula: Solubility=s+y=s+[NaCN]=[Ag(CN)2-].

https://brainly.com/question/34262172

#SPJ11

Answer:

In order to ensure consistent membrane fluidity

Explanation:

Cold-blooded animals rely on the temperature of the surrounding environment to maintain its internal temperature, their blood is not cold.  Their body temperature fluctuates,  based on the external temperature of the environment. If it  is 30 °F outside, their body temperature  will eventually normalize to  30 °F, as well. If it eventually rises to 120 °F, their body temperature will follow the same pattern to 120 °F.

The cell membrane of cold-blooded animals contains cholesterol, which acts as a shield for the membrane. Membrane fluidity is enhanced by temperature, as the temperature increases membrane fluidity increases and it decreases when the temperature goes down.

Most cold-blooded animals adjust their feeding habit to contain  more unsaturated fats from plants. This helps them to maintain their motor coordination and membrane fluidity during the long winter

Cold-blooded animals modulate the fatty acid composition of their membrane to stabilize their membrane fluidity

Old-blooded animals modulate the fatty acid composition of their membranes as a function of temperature to ensure consistent membrane fluidity, and cholesterol plays a role in maintaining appropriate fluidity across a range of temperatures.

Old-blooded animals, also known as poikilothermic organisms, modulate the fatty acid composition of their membranes in response to temperature changes. This modulation helps them maintain consistent membrane fluidity, which is crucial for proper cell function. These animals increase the unsaturated fatty acid content of their cell membranes in colder temperatures and increase the saturated fatty acid content in higher temperatures. Additionally, cholesterol, which lies alongside phospholipids in the membrane, acts as a buffer to dampen the effects of temperature on the membrane, extending the range of temperature in which the membrane is appropriately fluid and functional. Cholesterol also serves other functions, such as organizing clusters of transmembrane proteins into lipid rafts.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The predominate specie is [tex]HCO_3^-[/tex]

Explanation:

From the question we can see that [tex]pk_{a1}(6.351) < pH(6.73) pk_{a2}(10.329)[/tex]

Hence looking at the question we can see that the predominant specie is [tex]HCO_3^-[/tex]  

Final answer:

At a pH of 6.73, which is between the pKa₁ (6.351) and pKa₂ (10.329) of carbonic acid, the predominant form of carbonic acid is HCO³⁻ (bicarbonate ion).

Explanation:

The question pertains to the acid dissociation of carbonic acid (H₂CO₃) and which form is predominant at a certain pH. Carbonic acid is a diprotic weak acid, which means it can donate two protons (H⁺), and has two dissociation steps each with different pKa values.

The pKa₁ of carbonic acid is 6.351, and below this pH, the predominant form of carbonic acid is H₂CO₃. The pKa₂ is 10.329, and above this pH, the carbonate ion (CO₃²⁻) is predominant. Between these pKa values, the bicarbonate ion (HCO³⁻) is the predominant species. Therefore, at a pH of 6.73, which lies between the two pKa values, the predominant form is HCO³⁻ (bicarbonate ion).

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

Answer:

1.38 M is the concentration of blue solution.

Explanation:

The equation of calibration curve:

[tex]A=\epsilon \times l\times c[/tex]

where,

A = absorbance of solution

C = concentration of solution

l = path length

[tex]\epsilon[/tex] = molar absorptivity coefficient

The graph between A and c will straight line with slope equal to prduct of [tex]\epsilon and l[/tex].

y = 0.194 x + 0.025.

A = 0.194 c+ 0.025

Given the absorbance of solution :

A = 0.293

So, concentration at this value of absorbance will be= c

[tex]0.293=0.194 c+0.025[/tex]

[tex]c=\frac{0.293-0.025}{0.194}=1.381\approx 1.38[/tex]

1.38 M is the concentration of blue solution.

The concentration of the Blue #1 solution can be calculated by rearranging the equation of the calibration curve to solve for 'x' when 'y' is the measured absorbance, 0.293. The calculated concentration to three decimal places is approximately 1.381.

The equation of the straight line formed by your calibration curve for Blue #1 at 635 nm is y = 0.194x + 0.025. In this equation, 'y' represents the measured absorbance and 'x' represents the concentration. If the measured absorbance, or 'y', is 0.293, you can calculate the concentration, or 'x', by rearranging the equation to solve for 'x'.

To isolate 'x', you would subtract 0.025 from each side of your equation.

0.293 - 0.025 = 0.194x, which simplifies to 0.268 = 0.194x. Finally, divide each side of the equation by 0.194 to solve for 'x':

0.268 / 0.194 = x

When you divide 0.268 by 0.194, you get approximately 1.381. Therefore the concentration of the Blue #1 solution to three decimal places is approximately 1.381.

https://brainly.com/question/29729322

#SPJ3

Answer:

The specific heat of the copper is 0.771 cal/ grams °C

Explanation:

Step 1: Data given

Mass of the piece of copper = 15.0 grams

The temperature of the wire changes from 12.0 °C to 79.0 °C

The amount of heat absorbed is 775 cal

Step 2: Calculate the specific heat of copper

Q = m*c*ΔT

⇒with Q = the heat absorbed = 775 cal

⇒with m = the mass of the copper = 15.0 grams

⇒with c = the specific heat of copper = TO BE DETERMINED

⇒with ΔT = The change in temperature = 79.0 °C - 12.0 °C = 67.0 °C

775 cal = 15.0 grams * c * 67.0 °C

c = 0.771 cal/gm °C

The specific heat of the copper is 0.771 cal/ grams °C

Answer:

In between H and Cl the bond will be covalent.

In between Mg and F the bond will be ionic.

In between Li and N the bond will be ionic .

In between N and S the bond will be polar covalent.

Explanation:

Ionic bonds are defined as the bonds which are formed by the complete transfer of electrons from cation (positively charged ions) to anion (negatively charged ions). For Example: NaCl, [tex]MgF_2[/tex] etc.

Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. For Example: HCl, [tex]CH_4[/tex] etc.

Its of two type:

In between H and Cl the bond will be covalent.

In between Mg and F the bond will be ionic.

In between Li and N the bond will be ionic .

In between N and S the bond will be polar covalent.

The type of bond that would be formed between each of the following pairs of atoms will be:

Ionic bonds are formed by the transfer of electrons from cation to anion. An example is NaCl.

Covalent bond is a bond that's formed by the sharing of electrons between atoms. e.g. HCl.

A polar covalent bond is formed when there is the presence of difference in electronegativity between the atoms.

Read related link on:

https://brainly.com/question/3759728

Answer:

razor and blade strategy

Explanation:

Razor and blade strategy -

It refers to the method of pricing , where the price of one of the item is reduced in order to increase the sale of another item , is referred to as razor and blade strategy .

It is a type of pricing tactics used by the company to indirectly earn profit for their goods and services .

Sometimes , fews goods are given free along with certain products , in order to earn profit .

Hence , from the given information of the question ,

The correct answer is razor and blade strategy .

Answer:see the picture attached

Explanation:

Final answer:

Resonance structures for benzoic acid highlight electron delocalization, particularly the interaction between the carboxyl group and the para position on the benzene ring. Valence electrons are shown to resonate, creating multiple valid forms which collectively define the resonance hybrid of the molecule.

Explanation:

Resonance Structures of Benzoic Acid

When drawing resonance structures for benzoic acid, especially illustrating the resonance interaction with the para position (position opposite to the carboxyl group), we focus on the delocalization of π-electrons within the benzene ring and the adjacent carboxyl group. For the carboxyl group, one of the oxygen atoms will have a double bond with the carbon to fulfill the octet rule. However, due to the equivalent nature of the oxygen atoms, the double bond can resonate between the two oxygens, creating two resonance forms. At the para position, another resonance form is created by the movement of π-electrons from the benzene ring towards the carboxyl group, thus extending the conjugation and delocalization of electrons throughout the molecule.

Each resonance structure will be carefully drawn ensuring that all atoms obey the octet rule and formal charges are correctly assigned. It's important to note that the actual molecule is better represented by a resonance hybrid, which is an average of all valid resonance forms. This concept signifies that electrons are not strictly localized in one structure, but are distributed across the molecule in a delocalized π-electron cloud.

Answer: The value of the equilibrium constant is 0.024

Explanation:

Initial moles of  [tex]NOBr[/tex] = 0.612 mole

Volume of container = 100 L

Initial concentration of [tex]NOBr=\frac{moles}{volume}=\frac{0.612moles}{100L}=6.12\times 10^{-3}M[/tex]  

equilibrium concentration of [tex]Br_2=2.12\times 10^{-3}M[/tex]

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

[tex]NOBr(g)\rightleftharpoons NO(g)+\frac{1}{2}Br_2(g)[/tex]

at t=0   [tex]6.12\times 10^{-3}M[/tex]     0       0

At eqm. conc. [tex](6.12\times 10^{-3}-x)M[/tex]      x       x/2     

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[NO]^2[Br_2]}{[NOBr]}[/tex]

[tex]K_c=\frac{(2x)^2\times x/2}{(6.12\times 10^{-3}-x)}[/tex]

we are given : x/2 =[tex]2.12\times 10^{-3}[/tex]

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

[tex]K_c=\frac{(2.12\times 10^{-3})^{\frac{1}{2}}\times (2.12\times 10^{-3})}{(6.12\times 10^{-3})-(2.12\times 10^{-3})}[/tex]

[tex]K_c=0.024[/tex]

Thus the value of the equilibrium constant is 0.024

Explanation:

(a)  When we place the bulb of thermometer below then there would be vapors due to high temperature. And, since you are measuring temperature of the vapors closer to when evaporated.

As a result, there will be inaccurate temperature reading (higher than the actual reading).

(b)   When we place the bulb above the condenser then we are not able to measure the vapors directly as they are escaping into the condenser before their temperature can be measured.

As a result, we will get inaccurate results of room temperature.

The total heat change for the process of ice melting at -6.5oC can be calculated by considering the temperature change and the phase change. The correct calculations are: temperature change is 337.25 J and phase change is 150.5 J. The total heat change is 487.75 J or 0.48775 kJ.

The total heat change for the process of ice melting at -6.5oC can be calculated by considering the temperature change from -6.5oC to 0.0oC and the phase change from solid to liquid.

For the temperature change, we use the equation q = m * C * ΔT, where q is the heat change, m is the mass of ice, C is the specific heat capacity of ice, and ΔT is the temperature change. Plugging in the values, we get q = 25.0 g * 2.09 J/g.°C * (0.0-(-6.5)°C) = 337.25 J. This calculation is incorrectly represented as option A.

For the phase change, we use the equation q = m * ΔH, where q is the heat change, m is the mass of ice, and ΔH is the enthalpy of fusion. Plugging in the values, we get q = 25.0 g * 6.02 kJ/mol = 150.5 J. This calculation is incorrectly represented as option B.

Since there are only two stages in this process (temperature change and phase change), option C is false.

Therefore, the correct calculations to determine the total heat change for this process are: the temperature change is 337.25 J and the phase change is 150.5 J. Adding these up, the total heat change is 487.75 J, which is equivalent to 0.48775 kJ. This calculation is incorrectly represented as option D but not as option E.

https://brainly.com/question/34039707

#SPJ11

Answer: The equilibrium concentration of NO after it is re-established is 0.55 M

Explanation:

For the given chemical equation:

[tex]N_2(g)+O_(g)\rightleftharpoons 2NO(g)[/tex]

The expression of [tex]K_{eq}[/tex] for above equation follows:

[tex]K_{eq}=\frac{[NO]^2}{[N_2][O_2]}[/tex]     .....(1)

We are given:

[tex][NO]_{eq}=0.400M[/tex]

[tex][N_2]_{eq}=0.200M[/tex]

[tex][O_2]_{eq}=0.200M[/tex]

Putting values in expression 1, we get:

[tex]K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4[/tex]

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

                           [tex]N_2(g)+O_(g)\rightleftharpoons 2NO(g)[/tex]

Initial:               0.200    0.200        0.700

At eqllm:      0.200+x   0.200+x     0.700-2x

Putting values in expression 1, we get:

[tex]4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075[/tex]

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

Explanation:

The given data is as follows.

        P = 1 bar,     T = 298 K

For [tex]CO_{2}[/tex];  [tex]V_{1C} = 0.7 m^{3}[/tex]

For [tex]N_{2}[/tex];    [tex]V_{1N} = 0.3 m^{3}[/tex]

Therefore, final volume will be as follows.

           [tex]V_{1C} + V_{1N} = 1 m^{3} = V_{2}[/tex]

Hence,            [tex]V_{2} = 1 m^{3}[/tex]

Formula for change in entropy is as follows.

          dS = [tex]nR ln (\frac{V_{2}}{V_{1}})[/tex]

For [tex]CO_{2}[/tex],  [tex]PV_{1C} = nRT[/tex]

or,         [tex]nR = \frac{PV_{1C}}{T}[/tex]            

                       = [tex]\frac{10^{5} \times 0.7}{298}[/tex]        

                       = 234.89

      [tex]dS_{CO_{2}} = 234.89 \times \frac{1}{0.7}[/tex]  

                       = 83.77 J/K

For [tex]N_{2}[/tex],       nR = [tex]\frac{P_{1}V_{1}N}{T}[/tex]

                  = [tex]\frac{10^{5} \times 0.3}{298}[/tex]

                  = 100.67

       [tex]dS_{N_{2}} = 100.67 ln (\frac{1}{0.3})[/tex]

                   = 121.20 J/K

Now, total change in the entropy is calculated as follows.

               dS = [tex]dS_{CO_{2}} + dS_{N_{2}}[/tex]

                    = (83.77 + 121.20) J/K

                    = 204.97 J/K

Thus, we can conclude that the change in entropy is 204.97 J/K.

Entropy change will be 204.97 J/K.

Firstly, let's write the given values:

Since, it's an ideal gas therefore;

Thus, the final volume will be V₂= [tex]0.7+0.3=1m^3[/tex]

What does change in entropy mean?

Entropy change can be defined as the change in the state of disorder of a thermodynamic system that is associated with the conversion of heat or enthalpy into work.

Entropy change can be calculated by using the given formula:

[tex]\text{dS}=n R ln \frac{V_2}{V_1}[/tex]

⇒For CO₂,

From ideal gas equation it is known that: pV=nRT

∴ nR=pV/T

[tex]nR=\frac{10^5 *0.7}{298} =234.89[/tex]

So, entropy change for CO₂ will be:

[tex]dS_{CO_2}=234.89ln (\frac{1}{0.7} )=83.77J/K[/tex]

⇒For N₂,

[tex]nR=\frac{10^5*0.3}{298} =100.67[/tex]

So, entropy change for N₂ will be:

[tex]dS_{N_2}=100.67 ln(\frac{1}{0.3}) =121.20 J/K[/tex]

Now, total entropy change can be calculated as:

[tex]dS_{Total}=dS_{CO_2}+dS_{N_2}\\\\=83.77+121.20\\\\=204.97 J/K[/tex]

Thus, entropy change will be 204.97 J/K.

Learn more:

brainly.com/question/23084720

Answer:

7.79 moles

Explanation:

Let the mass of helium gas = Mass of argon gas = x g

Moles of helium = [tex]\frac{x}{4}[/tex] moles

Moles of argon = [tex]\frac{x}{40}[/tex] moles

Total moles = [tex]\frac{x}{4}+\frac{x}{40}=\frac{11x}{40}\ moles[/tex]

Given that:

Temperature = 398 K

V = 80.0 L

Pressure = 3.50 atm

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

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

Applying the equation as:

[tex]3.50\times 80.0=\frac{11x}{40}\times 0.0821\times 398[/tex]

x=31.16 g

Moles of helium = 31.16 / 4 = 7.79 moles

The mixture contains approximately 7.79 moles of helium."

To solve this problem, we will use the Ideal Gas Law, which is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given:

- The total volume of the gas mixture (V) is 80.0 L.

- The temperature of the gas mixture (T) is 398 K.

- The pressure of the gas mixture (P) is 3.50 atm.

- The mass of helium gas (m_He) is equal to the mass of argon gas (m_Ar).

First, we need to find the total number of moles of gas in the mixture using the Ideal Gas Law:

[tex]\[ PV = nRT \][/tex]

We can rearrange this equation to solve for the total number of moles (n_total):

[tex]\[ n_{\text{total}} = \frac{PV}{RT} \][/tex]

The value of the ideal gas constant (R) is 0.0821 L·atm/(mol·K). Plugging in the given values:

[tex]\[ n_{\text{total}} = \frac{(3.50 \, \text{atm})(80.0 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K)})(398 \, \text{K})} \][/tex]

[tex]\[ n_{\text{total}} = \frac{280 \, \text{atm·L}}{32.6028 \, \text{atm·L/mol}} \][/tex]

[tex]\[ n_{\text{total}} \approx 8.59 \, \text{mol} \][/tex]

Since the mass of helium is equal to the mass of argon, and assuming no other gases are present, the mixture consists of two parts: one part helium and one part argon by mass. Therefore, the moles of helium (n_He) and the moles of argon (n_Ar) are in the same ratio as their molar masses. The molar mass of helium (M_He) is approximately 4.0026 g/mol and the molar mass of argon (M_Ar) is approximately 39.948 g/mol.

The ratio of the number of moles of helium to the total number of moles is the same as the ratio of the molar mass of argon to the sum of the molar masses of helium and argon:

[tex]\[ \frac{n_{\text{He}}}{n_{\text{total}}} = \frac{M_{\text{Ar}}}{M_{\text{He}} + M_{\text{Ar}}} \][/tex]

[tex]\[ \frac{n_{\text{He}}}{8.59 \, \text{mol}} = \frac{39.948 \, \text{g/mol}}{4.0026 \, \text{g/mol} + 39.948 \, \text{g/mol}} \][/tex]

[tex]\[ \frac{n_{\text{He}}}{8.59 \, \text{mol}} = \frac{39.948}{43.9506} \][/tex]

[tex]\[ \frac{n_{\text{He}}}{8.59 \, \text{mol}} \approx 0.9087 \][/tex]

[tex]\[ n_{\text{He}} \approx 8.59 \, \text{mol} \times 0.9087 \][/tex]

[tex]\[ n_{\text{He}} \approx 7.79 \, \text{mol} \][/tex]

Therefore, the mixture contains approximately 7.79 moles of helium."

Answer:

0.03 M

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

HCl + NaOH → NaCl + H₂O

0.2 L of 0.1 M NaOH were used. The moles of NaOH that reacted are:

0.2 L × 0.1 mol/L = 0.02 mol

The molar ratio of HCl to NaOH is 1:1. The moles of HCl that reacted are 0.02 mol.

0.02 moles of HCl are in 0.8 L of solution. The molarity of HCl is:

0.02 mol / 0.8 L = 0.03 M

Answer:

1

Explanation:

The principal quantum number, n, showsthe principal electron shell. which can inturn be describe as the most probable distance of the electrons from the nucleus, the larger the number n is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. n can be any positive integer starting at 1, as n=1 designates the first principal shell (the innermost shell). When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where n=2

As the energy of the electron increases, so does the principal quantum number, e.g., n = 3 indicates the third principal shell, n = 4 indicates the fourth principal shell, and so on. n=1,2,3,4…

On the other hand, the magnetic quantum number ml determines the number of orbitals and their orientation within a subshell. Consequently, its value depends on the orbital angular momentum quantum number l. Given a certain l, ml is an interval ranging from –l to +l, so it can be zero, a negative integer, or a positive integer. ml=−l,(−l+1),(−l+2),…,−2,−1,0,1,2,…(l–1),(l–2),+l

note also that: The orbital angular momentum quantum number l determines the shape of an orbital, and therefore the angular distribution. The number of angular nodes is equal to the value of the angular momentum quantum number l. Each value of l indicates a specific s, p, d, f subshell (each unique in shape.) The value of l is dependent on the principal quantum number n. Unlike n, the value of l can be zero. It can also be a positive integer, but it cannot be larger than one less than the principal quantum number (n-1): l=0,1,2,3,4…,(n−1).

So the answer is 1

Given the magnetic quantum number (m₁) of an electron is 0, the smallest possible value of the principal quantum number (n) under this scenario would be 1.

In quantum mechanics, the state of an electron in an atom is well defined by a set of quantum numbers: n, l, m₁, and ms. These numerical attributes express key principles regarding the electron's characteristics within the atom such as its energy and orientation.

In your question, the provided magnetic quantum number (m₁) is given as 0. The nature of the magnetic quantum number is that it ranges from -l to +l, including zero. So having m₁ = 0 implies that the lowest possible value of the angular momentum quantum number (l) should also be zero.

The principal quantum number (n), which indicates the main energy level of the electron, is always an integer greater than zero. Since l can take any value from 0 to n - 1, if l=0, the smallest possible value for n would also be 1. Therefore, the smallest possible value of the principal quantum number of the state with m₁=0 is 1.

https://brainly.com/question/27152536

#SPJ11

Answer:

The diagram of the Possibilities Frontier (PPF) is shown on the first uploaded image

A Possibilities Frontier (PPF)  can also be referred to as the production possibility curve and it can be defined as a curve that denotes the production of two types of goods in an economy at a given period of time in different proportion of combinations

When this curve shift to the right it means that there is technological advancement in that economy

If this curve a points lies within it then the resources are not fully utilized    

Explanation:

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

Based on the buffer's molarity and the amount of substances added, it appears that none of the additions would exceed the buffer's capacity to neutralize them. Yet, they would consume a significant portion of the buffering agents, thereby potentially affecting the buffer's effectiveness in further reactions.

The subject question is asking about the buffer capacity and if certain substances when added would exceed this capacity. The buffer in this case is a solution of HNO2 (nitrous acid) and KNO2 (potassium nitrite). The buffer can neutralize added acids or bases because HNO2 can donate a proton (H+) to neutralize OH- ions from a base, and KNO2 can donate OH- ions to neutralize excess H+ ions from an acid.

Now, for each addition, let's convert the mass of the substance to moles. For NaOH, 250.0 mg is about 0.00625 moles. For KOH, 350.0 mg is roughly 0.00624 moles. For HBr, 1.25 g is about 0.00736 moles. Finally, for HI, 1.35 g is quite close to 0.00736 moles.

Comparing these measures to the buffer's concentration in the solution, all of the added substances are less than 0.100 M or 0.150 M. Thus, these added amounts will not exceed the buffer's capacity to neutralize them, although it should be noted that they would extensively consume the buffering agents in the solution.

https://brainly.com/question/29458673

#SPJ3

Answer: the structures are shown in the image attached.

Explanation:

Before the invention of mordern spectroscopy, Korner's absolute method was commonly applied in determining whether a disubstituted benzene derivative was the ortho, meta, or para isomer. Korner's method depends on the addition of a third group, nitro groups are usually added. The number of isomers formed is then determined. The isomers are usually obtained only in small amounts and analysed to know the structure of the original compound.

Final answer:

The original compound is p-bromotoluene or 4-bromotoluene. The three nitrated derivatives are m-dinitrobenzene or 1,3-dinitrobenzene.

Explanation:

The turn-of-the-century chemist isolated a compound with the molecular formula C6H4Br2. After nitrating this compound, three isomers of formula C6H3Br2NO2 were obtained.

The original compound is called p-bromotoluene or 4-bromotoluene. It contains a methyl (CH3) group and the bromine atom is attached to the fourth carbon atom.

The three nitrated derivatives are called m-dinitrobenzene or 1,3-dinitrobenzene. These compounds have two nitro (NO2) groups attached to the benzene ring, at the first and third positions.

Answer:

Option d. 2615.0g

Explanation:

Let M1, M2, and M3 represent the masses of the three different samples

M1 = 0.1568934 kg = 156.8934g

M2 = 1.215mg = 1.215x10^-3 = 0.001215g

M3 = 2458.1g

Total Mass = M1 + M2 + M3

Total Mass = 156.8934 + 0.001215 + 2458.1

Total Mass = 2614.994651g

Total Mass = 2615.0g

To find the total mass, convert all weights to the same unit (grams), then add them. The sum needs to be rounded to the correct number of significant figures. The correct answer should be 2616.0 g, which isn't provided among the options.

The total mass of the samples is found by converting all the masses to the same unit (in this case, grams) and then adding them together. The weight in kilograms is 0.1568934 kilograms or 156.8934 grams, 1.215 milligrams is equal to 0.001215 grams and 2458.1 grams remains unchanged. Adding these, we get a total mass of 2615.994615 grams. However, in science, you often report to the correct number of significant figures, in this case should be the least number of decimal places we have in our data, which is 1 decimal place. Therefore 2615.994615 grams rounded is 2616.0 grams. So none of the options provided (a-e) are correct. The correct answer should be 2616.0 g.

Learn more about Mass Calculation & Significant Figures here:

https://brainly.com/question/34660333

#SPJ3

Answer:

B. Triazine-resistant weeds were more likely to survive and reproduce

Explanation:

According to darwin's theory of evolution, variation is already present in some members of a population and this variation lets them survive in the adverse condition and those who do not have that variation which helps in survival are lost with time.

So as before using triazine herbicide the major population of the weed were not resistant to this herbicide so in the first few years the nonresistant weeds were lost and only resistant weed which was very less in number survived.

So after several years these resistant weeds reproduced and transferred their gene to their offsprings and became predominant in the field. Therefore the correct answer is B.

Triazine-resistant weeds survived and reproduced due to natural selection, rendering the herbicide less effective over time.

Option (B) is correct.

Option B, the development of triazine-resistant weeds, is the most likely explanation for the increasing pigweed problem. Over time, the repeated use of the same herbicide, like triazine, exerts selective pressure on weed populations.

Some pigweed plants may carry genetic mutations that make them naturally resistant to the herbicide. When the herbicide kills most pigweed, these resistant individuals survive and pass on their resistant traits to their offspring. Eventually, the population becomes dominated by triazine-resistant pigweed, making the herbicide less effective.

This is a classic example of natural selection and the evolution of herbicide resistance in weed populations, a common issue in agriculture when the same herbicide is used repeatedly. The other options are less plausible or unrelated to herbicide resistance.

To learn more about triazine-resistant weeds  here

https://brainly.com/question/26346230

#SPJ12

Answer:

It would take 20 seconds

Explanation:

Let the final concentration of AB be C

Rate = k[C]^2 = change in concentration of AB/time

k is the rate constant = 0.2 L/mol.s

Initial concentration of AB = 1.5 M

Final concentration of AB = 1/3 × 1.5 = 0.5 M

Change in concentration of AB = 1.5 - 0.5 = 1 M

0.2 × 0.5^2 = 1/time

time = 1/0.05 = 20 s

Answer:

4.1

Explanation:

First, we will calculate the moles of each species.

The volume of the mixture is 0.225 L + 0.435 L = 0.660 L

The concentration of the species in the buffer are:

We can find the pH of the buffer using the Henderson-Hasselbach equation.

pH = pKa + log [lactate] / [lactic acid]

pH = -log 1.38 × 10⁻⁴ + log 0.45 M / 0.29 M

pH = 4.1

Answer :

(a) Reaction at anode (oxidation) : [tex]4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-[/tex]  

(b) Reaction at cathode (reduction) : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]  

(c) [tex]O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}[/tex]

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : [tex]Fe^{2+}\rightarrow Fe^{3+}+e^-[/tex]     [tex]E^0_{anode}=+0.771V[/tex]

Reaction at cathode (reduction) : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]     [tex]E^0_{cathode}=+1.23V[/tex]

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : [tex]4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-[/tex]     [tex]E^0_{anode}=+0.771V[/tex]

Part (b):

Reaction at cathode (reduction) : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]     [tex]E^0_{cathode}=+1.23V[/tex]

Part (c):

The balanced cell reaction will be,

[tex]O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}[/tex]

Part (d):

Now we have to calculate the standard electrode potential of the cell.

[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]

[tex]E^o=(1.23V)-(0.771V)=+0.459V[/tex]

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

Answer:

The answer is attached and other details about the answer is also attached.

Explanation:

The molecular formula (C9H10O2) indicates five

degrees of unsaturation (see Section 14.16), which is

strongly suggestive of an aromatic ring, as well as one

additional double bond or ring. The signal just above

3000 cm-1 in the IR spectrum confirms the aromatic ring,

as does the signal just above 1600 cm-1. The 1

H NMR

spectrum exhibits two doublets between 6.9 and 7.9

ppm, each with an integration of 2. This is the

characteristic pattern of a disubstituted aromatic ring, in

which the two substituents are different from each other:

The singlet at 3.9 ppm (with an integration of 3)

represents a methyl group. The chemical shift is

downfield from the expected benchmark value of 0.9

ppm for a methyl group, indicating that it is likely next to

an oxygen atom:

The singlet at 2.6 ppm (with an integration of 3)

represents an isolated methyl group. The chemical shift

of this signal suggests that the methyl group is

neighboring a carbonyl group:

The carbonyl group accounts for one degree of

unsaturation, and together with the aromatic ring, this

would account for all five degrees of unsaturation. The

presence of a carbonyl group is also confirmed by the

signal at 196.6 ppm in the 13C NMR spectrum.

We have uncovered three pieces, which can only be

connected in one way, as shown:

This structure is consistent with the 13C NMR data: four

signals for the sp2 hybridized carbon atoms of the

aromatic ring, and two signals for the sp3 hybridized

carbon atoms (one of which is above 50 ppm because it

is next to an oxygen atom).

Also notice that the carbonyl group is conjugated to the

aromatic ring, which explains why the signal for the

C=O bond in the IR spectrum appears at 1676 cm-1,

rather than 1720 cm-1.

The molarity of the chromium(III) acetate solution is 0.398 M. The concentration of chromium(III) cation is also 0.398 M, and the concentration of the acetate anion is 1.194 M.

First, we need to recall that the molar mass of chromium(III) acetate is approximately 229.13 g/mol. We can then find the moles of the compound by the equation: moles = mass (g) / molar mass (g/mol). So, moles of chromium(III) acetate = 22.8 g / 229.13 g/mol = 0.0995 mol. Then, using the formula for molarity, M = moles / volume (L). Volume must be in liters, so 250 mL is converted to 0.25 L. This gives us M = 0.0995 mol / 0.25 L = 0.398 M.

As for the concentrations of the chromium(III) cation and the acetate anion, we have to consider the formula of chromium(III) acetate, which is Cr(C2H3O2)3. This indicates that for every 1 mol of compound, there is 1 mol of Cr3+ and 3 mol of C2H3O2-. So the molar concentration of Cr3+ is the same as that of the substance, 0.398 M. The molar concentration of C2H3O2- is three times as much, 0.398 M x 3 = 1.194 M.

https://brainly.com/question/10720472

#SPJ2

Answer:

Light of wavelength 200 nm will have lowest frequency in while traveling through diamond.

Explanation:

Speed of the light in vacuum = c

Relation between speed of light , wavelength (λ) and frequency (ν);

[tex]\nu =\frac{c}{\lambda }[/tex]

Speed of light in a medium = c' = c/n

[tex]\nu =\frac{c'}{\lambda }=\frac{c}{n\times \lambda }[/tex]...[1]

Where : n = refractive index of a medium

So, the medium with greater value of refractive index lower the speed of light to greater extent.

From [1] , we can see that frequency of light is inversely proportional to the refractive index of the medium :

[tex]\nu \propto \frac{1}{n} [/tex]

This means that higher the value of refractive index lower will be the value of frequency of light in that medium or vice-versa.

According to question, light of wavelength 200 nm will have lowest frequency in while traveling through diamond because refractive index of diamond out of the given mediums is greatest.

Increasing order of refractive indices:

[tex]H_2<H_2O<CCL_4<\text{Silicon oil}< Diamond[/tex]

Decreasing order of frequency of light 210 nm in these medium :

[tex]H_2>H_2O>CCL_4>\text{Silicon oil}> Diamond[/tex]

To find the Ksp of copper(I) chloride at 25 ºC with a solubility of 1.00 x 10^-3 mol/L, the Ksp value is calculated to be 1.00 x 10^-6.

Ksp = [Ag+][Cl-] = (1.67 × 10^-5)² = 2.79 × 10^-10

Therefore, at 25°C, the Ksp value of copper(I) chloride would be 1.00 x 10^-6 as calculated from the given solubility information.

Remember the formula for Ksp and plug in the given values to obtain the accurate result for the given conditions.