A second order reaction is 50% complete in 15 min. How long after the start of the 10 reaction will it be 85% complete?

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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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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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:

Disorder

Explanation:

Entropy is a measure of disorder.

Entropy is a measure of disorder.

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!:

Pentane is a 5 member hydrocarbon

1 chloropentane can be on either side and will still have thesame name because of the IUPAC rules .

To determine the relative amounts of products obtained from radical chlorination of an alkane, both probability (the number of hydrogens that can be abstracted that will lead to the formation of the particular product) and reactivity (the relative rate at which a particular hydrogen is abstracted) must be taken into account.

The precise ratios differ at different temperatures.

For 3- chloro pentane , There are two hydrogen only and the reactivity  is same for all  2° hydrogen

The 2°  hydrogen % should = 100-22 = 78% ( all remaining are 2° hydrogen only)

since all are equally recative 2° hydrogen  , proportionate distribution will be there.

% Yield  =   {( Total hydrogen which can give 3 choloropentane) /  (all hydrogen of 2° ) } * 78

             =  (2/6)* 78 =  26%  

Hope this helps!

In the chlorination of pentane, if 1-chloropentane is 22% of the mixture and secondary hydrogens are equally reactive as in monochlorination, the percentage of 3-chloropentane in the mixture is approximately 15%. The percentages are derived taking into consideration that the ratio of monochlorination among the different positions in pentane (primary, secondary, tertiary) is 3:2:0.

In organic chemistry, different isomers of a compound can be formed via a process of substitution like chlorination. In the chlorination of pentane, we have a mixture of isomers, including 1-chloropentane and 3-chloropentane, among others. The question stated that the given percentage of 1-chloropentane is 22%. If the reactivity of secondary hydrogens in pentane is equal to that of monochlorination, then it implies the formation of 2-chloropentane and 3-chloropentane have the same ratio because both have their Cl bound to a secondary carbon. Thus, the distribution of monochlorination among 3 positions (primary, secondary, and tertiary) is in a ratio 3:2:0 for 1,2 and 3-chloropentane respectively.

To derive the individual rates of formation, we'd have to divide each by the total, e.g., for 1-chloropentane (primary position), it would be 3/5 = 0.60 or 60%. However, the question gives us a value of 22% for the 1-chloropentane, which suggests our chlorination is only 22/60 = 0.37 or 37% effective. Applying this scale factor to 2,3 chloropentane (secondary position), we get 0.37*40 = 14.8%. Therefore, 3-chloropentane will constitute around 15% of the mixture.

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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: True

Explanation: Intensive property does not depend on the amount of matter. Specific internal energy is an intensive property.

Hey there!:

Intensive property , does not depend upon the amount  of matter , specific internal energy is an intensive property , due to word specific  . TRUE

Hope this helps!

Answer:

1. [tex]60.0 mi/h=88 feet/seconds[/tex]

2. [tex]15 Ib/inch=267.86 kg/m[/tex]

3. [tex] 6.20 cm/hr=17,222.22 nm/s[/tex]

Explanation:

1. 60.0 mi/h to ft/s

1 mile = 5280 feet

1 hour = 3600 seconds

So, [tex]60.0 mi/h=\frac{60\times 5280 feet}{1\times 3600 s}=88 feet/seconds[/tex]

2. 15 Ib/in to kg/m

1 lb = 0.453592 kg

1 inch = 0.0254 m

So,[tex]15 Ib/inch=\frac{15\times 0.453592 kg}{1\times 0.0254 m}=267.86 kg/m[/tex]

3. 6.20 cm/hr to nm/s

1 cm = [tex]10^7 nm[/tex]

1 hour = 3600 seconds

So,[tex] 6.20 cm/hr=\frac{6.20\times 10^7 nm}{1\times 3600 s}=17,222.22 nm/s[/tex]

Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.  

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:

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:

see explanation...

Explanation:

The central nitrogen atom in the azide ion has a formal charge of +1. The other nitrogen atoms have a formal charge of -1.

The resonance structure of azide ion and hydrazoic acid is shown in the image attached to this answer. We can see that the central nitrogen atom in the azide ion has a formal charge of +1. The other nitrogen atoms have a formal charge of -1.

In hydrazoic acid, hydrogen is attached to one of the end nitrogen atoms while the other nitrogen atoms have a formal of +1 and -1 respectively.

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Answer:

The student conclude that the sample of benzoic acid is impure.

Explanation:

The observed melting point of benzoic acid when a student melts his/her sample is low than the actual value. The reason for this might be:

(a) The most probable reason is that the sample is impure. Impurities in the sample leads to lowering the value of the melting point. The reason for the phenomenon is that when impurity is present in the compound, the pattern of the crystal lattice disturbs and thus it less amount of heat is require to break the lattice.

(b) There may be some experimental errors like:

The student conclude that the sample is impure.

Final answer:

The student's benzoic acid sample is likely impure, as indicated by a melting point range significantly lower than the expected 121-122 degrees Celsius for a pure sample.

Explanation:

If a good sample of benzoic acid typically melts at 121-122 degrees Celsius, but a student's sample melted over a range of 105-115 degrees Celsius, it could indicate that their benzoic acid sample is impure. The presence of impurities in a compound will generally lower the melting point and cause the substance to melt over a broader temperature range. This can be seen in a MelTemp apparatus where samples with varying amounts of impurities, such as those containing acetanilide, melt at lower temperatures and have a wider melting range compared to pure benzoic acid.

Hey there !

below is the answer to the attached question

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]

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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Answer : The entropy change of the system is, 361.83 J/K

Solution :

Formula used :

[tex]\Delta S=\frac{n\times \Delta H_{vap}}{T_b}[/tex]

where,

[tex]\Delta S[/tex] = entropy change of the system = ?

[tex]\Delta H[/tex] = enthalpy of vaporization = 34.6 kJ/mole

n = number of moles of diethyl ether = 4.20 mole

[tex]T_b[/tex] = normal boiling point = [tex]26.5^oC=273+26.5=307.6K[/tex]

Now put all the given values in the above formula, we get the entropy change of the system.

[tex]\Delta S=\frac{4.20mole\times (34.6KJ/mole)}{307.6K}=0.36183kJ/K=0.36183\times 1000=361.83J/K[/tex]

Therefore, the entropy change of the system is, 361.83 J/K

Hey there!:

convert mass of H2O to mol  , use given chemical equation to calculate moles of H2 produced , use ideal gas equation to calculate volume  

convert mass of H2O to mol  :

mol of H2O = mass / molar mass

=  15.0 / 18.0

= 0.833 mol

use given chemical equation to calculate moles of H2 produced

from given equation, 4 mol of H2 is formed from 4 mol of H2O

So,

moles of H2 = moles of H2O

= 0.833 mol

use ideal gas equation to calculate volume

use:

P*V = n*R*T

(745/760) atm * V = 0.833 mol * 0.0821 atm.L/mol.K * (20+273) K

0.9803 * V = 0.833 * 0.0821*293

V = 20.4 L

Answer: 20.4 L

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:

molar mass of nicotine will be 162.16g/mol

Explanation:

The mass of nicotine taken = 0.60g

The volume of solution = 12mL

the osmotic pressure of solution = 7.55 atm

Temperature in kelvin =298.15K (25+ 273.15)

The formula which relates osmotic pressure and concentration (moles per L) is:

π = MRT

Where

π = osmotic pressure (unit atm) = 7.55 atm

M = molarity (mol /L)

T= temperature = (K) = 298.15 K

R = gas constant = 0.0821 L atm /mol K

Putting values

[tex]7.55=MX0.0821X298.15[/tex]

Therefore

[tex]M=\frac{7.55}{0.0821X298.15}=0.308\frac{mol}{L}[/tex]

Molarity is moles of solute dissolve per litre of solution

The volume of solution in litre = 0.012 L

[tex]molarity=\frac{moles}{V}[/tex]

[tex]moles=molarityXvolume=0.308X0.012=0.0037mol[/tex]

we know that

[tex]moles=\frac{mass}{ymolarmass}[/tex]

molar mass = [tex]\frac{mass}{moles}=\frac{0.60}{0.0037}=162.16\frac{g}{mol}[/tex]

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₋.

Answer:

Percent ionization of HCOOH is 3.69%

Explanation:

To calculate percent ionization of HCOOH, we have to construct an ICE table to determine changes in concentrations at equilibrium.

              [tex]HCOOH\rightleftharpoons HCOO^{-}+H^{+}[/tex]

    I:                0.125                                       0             0

    C:                -x                                           +x            +x

    E:              0.125-x                                     x              x

species inside third bracket represent equilibrium concentrations

So,      [tex]\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}=K_{a}(HCOOCH)[/tex]

   or,  [tex]\frac{x^{2}}{0.125-x}= 1.77\times 10^{-4}[/tex]

   or, [tex]x^{2}+0.000177x-0.0000221 = 0[/tex]

[tex]x=\frac{-0.000177+\sqrt{(0.000177)^{2}+(4\times 0.0000221)}}{2}[/tex]

or, [tex]x=4.61\times 10^{-3}[/tex] M

So, [tex][H^{+}]=4.61\times 10^{-3}M[/tex]

Percent ionization of formic acid  = [tex]\frac{[H^{+}]}{initial concentration of HCOOH}\times 100[/tex] = [tex]\frac{0.00461}{0.125}\times 100[/tex] = 3.69%

The percent ionization of the acid had been the concentration of hydrogen ion with respect to the acid concentration. The percent ionization of formic acid is 3.69%.

The acid dissociation constant has been the concentration of the acid dissociated into the constituent ions.

The dissociation of formic acid is given as:

[tex]\rm HCOOH\;\rightleftharpoons\;H^+\;HCOO^-[/tex]

The acid dissociation constant (Ka) for formic acid is given as:

[tex]\rm Ka=\dfrac{[H^+][HCOO^-]}{[HCOOH]}[/tex]

Substituting the concentration of the ions and the acid from the ICE table attached.

[tex]\rm 1.77\;\times\;10^{-4}=\dfrac{[x][x]}{[0.125-x]} \\\\x=4.61\;\times\;10^-3[/tex]

The hydrogen ion concentration in the solution has been 0.00461 M.

Substituting the concentration for the percent ionization of the formic acid:

[tex]\rm Percent\;ionization=\dfrac{[H^+]}{[HCOOH]}\;\times\;100\\\\ Percent\;ionization=\dfrac{0.00461}{0.125}\;\times\;100\\\\ Percent\;ionization=3.69\;\%[/tex]

The percent ionization of formic acid is 3.69%.

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Answer:

Is a dietary essential, especially for athletes shuttles .

Explanation:

Carnitine is a dietary essential, especially for athletes shuttles .

Carnitine shuttles fatty acids from the cytosol into the mitochondria, where they are broken down for energy in a process called oxidation. It is a nutrient associated with lipid metabolism and is involved in the Carnitine Shuttle, but it is not an essential dietary element for everyone, particularly those with already sufficient levels.

Carnitine is a nutrient involved in lipid metabolism within the mitochondria, especially for the transportation of fatty acids. In cellular metabolism, carnitine is not a dietary essential, but it shuttles fatty acids from the cytosol into the mitochondria, enabling their breakdown for energy production. This process, known as the Carnitine Shuttle, involves the conversion of fatty acyl-CoA molecules into acyl-carnitine by the enzyme carnitine palmitoyltransferase I, which allows the molecule to be transported across the inner mitochondrial membrane. Once within the matrix, carnitine acyltransferase II converts acyl-carnitine back into acyl-CoA for oxidation.

Despite being marketed as a supplement for athletes and individuals looking to enhance fat catabolism, carnitine supplementation has not been universally agreed upon as effective for those with sufficient carnitine levels. The role of carnitine in facilitating fatty acid oxidation is crucial, but it does not involve the shuttling of oxaloacetate from the mitochondria to the cytosol.

Answer:

weight percent of Iodine in the solution is 13.34%

Explanation:

Weight percentage -

weight percentage of A is given as , the weight of the substance A by weight of the total solution multiplied by 100.

i.e.

weight % A = weight of A / weight of solution * 100

From the question ,

weight of Iodine = 7.50 g

weight of carbon tetrachloride = 48.7 g

iodine is the solute and carbon tetrachloride is the solvent ,

Since,

solution = solute + solvent.

Hence,

weight of solution = weight of solute ( iodine ) + weight of solvent ( carbon tetrachloride)

weight of solution =  7.50 g + 48.7 g = 56.20 g

now,

weight of Iodine is calculated as,

weight % iodine = weight of iodine / weight of solution * 100

weight % iodine = 7.50 g / 56.20 g * 100

weight % iodine = 13.34 %

Answer : The correct pairs of elements likely to form ionic compounds are, potassium and sulfur, magnesium and chlorine.

Explanation :

Ionic compound : These are the compounds in which the atoms are bonded through the ionic bond. The ionic bond are formed by the complete transfer of the electrons.

That means, the atom which looses the electron is considered as an electropositive atom and the atom which gains the electron is considered as an electronegative atom.

And these bonds are formed between one metal and one non-metal.

From the given pairs of elements, potassium and sulfur, magnesium and chlorine are form ionic compound because potassium and magnesium are the metals and sulfur and chlorine are the non-metals. So, they can easily form ionic compound by the complete transfer of electrons.

While the other pairs, sodium and potassium are the metals, nitrogen and iodine, chlorine and bromine, helium and oxygen are the non-metals. They do not form ionic bond.

Hence, the correct pairs of elements likely to form ionic compounds are, potassium and sulfur, magnesium and chlorine.

Sodium and potassium, potassium and sulfur, and magnesium and chlorine are likely to form ionic compounds due to large differences in electronegativity.

Ionic compounds are formed when one element transfers electrons to another element, resulting in the formation of positive and negative ions. Elements with different electronegativities are more likely to form ionic compounds. Sodium and potassium, potassium and sulfur, and magnesium and chlorine are likely to form ionic compounds because of the large difference in electronegativity between them. In contrast, nitrogen and iodine, chlorine and bromine, helium and oxygen have similar electronegativities and are more likely to form covalent compounds.

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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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The answer is in the photo

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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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.