A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C. There are now ________ mol of gas in the flask.

Answer: 28.3 g/mol

Explanation:

According to the ideal gas equation:'

[tex]PV=nRT[/tex]

P= Pressure of the gas = 0.951 atm

V= Volume of the gas = 280 mL = 0.28 L     (1L=1000 ml)

T= Temperature of the gas = 25°C=(25+273)K=298 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm/K mol

[tex]n=\frac{PV}{RT}=\frac{0.951\times 0.28L}{0.0821 \times 298}=0.010moles[/tex]

To calculate the moles, we use the equation:

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

[tex]0.010=\frac{0.283}{\text {Molar mass}}[/tex]

[tex]{\text {Molar mass}}=28.3g/mol[/tex]

Thus the molar mass of the gas is 28.3 g/mol.

The molar mass of the gas is 25.73 g/mol. This was calculated using the ideal gas law and conversion of units, finally using the formula molar mass = mass in g/moles.

To calculate the molar mass of the gas, we can use the ideal gas law, which is PV = nRT. Here, n = number of moles, R = ideal gas constant, T = temperature in Kelvin, P = pressure and V = volume. First, convert the temperature from Celsius to Kelvin by adding 273.15: 25.0°C + 273.15 = 298.15 K. We need to find n, the number of moles. From the ideal gas law, we know that n = PV/RT. Substituting the given values: n = (0.951 atm * 280 mL) / (0.0821 L atm mol⁻¹K⁻¹ * 298.15 K) = 0.011 mol.

The mass given is 0.283 g. So, using the relation molar mass = mass in g/moles; molar mass of gas = 0.283g / 0.011 mol = 25.73 g/mol.

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Answer : The concentration of [tex]COF_2[/tex] remains at equilibrium will be, 0.37 M

Explanation :  Given,

Equilibrium constant = 4.90

Initial concentration of [tex]COF_2[/tex] = 2.00 M

The balanced equilibrium reaction is,

                       [tex]2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)[/tex]

Initial conc.    2 M                 0             0

At eqm.         (2-2x) M          x M         x M

The expression of equilibrium constant for the reaction will be:

[tex]K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}[/tex]

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

[tex]4.90=\frac{(x)\times (x)}{(2-2x)^2}[/tex]

By solving the term 'x' by quadratic equation, we get two value of 'x'.

[tex]x=1.291M\text{ and }0.815M[/tex]

Now put the values of 'x' in concentration of [tex]COF_2[/tex] remains at equilibrium.

Concentration of [tex]COF_2[/tex] remains at equilibrium = [tex](2-2x)M=[2-2(1.219)]M=-0.582M[/tex]

Concentration of [tex]COF_2[/tex] remains at equilibrium = [tex](2-2x)M=[2-2(0.815)]M=0.37M[/tex]

From this we conclude that, the amount of substance can not be negative at equilibrium. So, the value of 'x' which is equal to 1.291 M is not considered.

Therefore, the concentration of [tex]COF_2[/tex] remains at equilibrium will be, 0.37 M

First we need to calculate the number of moles of NaOH titrated.

molar concentration = number of moles / solution volume (liter)

number of moles = molar concentration × solution volume

number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles

Then we look at the chemical reaction:

CH[tex]_{3}[/tex]-COOH +  NaOH = CH[tex]_{3}[/tex]COONa + H[tex]_{2}[/tex]O

We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.

Now we can get the mass of acetic acid:

number of moles = mass (grams) / molecular mass (g/mol)

mass = number of moles × molecular mass

mass of acetic acid = 0,01125 × 60 = 0.675 g

We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:

concentration of acetic acid = (mass of acetic acid / mass of vinegar) × 100

concentration of acetic acid = (0.674 / 7) ×  100 = 9.6 %

The mass percent of acetic acid in the vinegar sample 9.6 %.

The reaction will be,

[tex]\rm \bold { CH_3-COOH + NaOH \leftrightharpoons CH_3COONa + H_2O}[/tex]

Means 1 mole  acetic acid is reacting with one mole of sodium hydroxide.

Number of moles of NaOH titrated,

Number of moles of  [tex]\rm \bold{ NaOH = 1.5 \times 0.0075 = 0.01125 moles}[/tex]

number of moles of [tex]\rm \bold { CH_3COOH}[/tex]  = 0.01125 moles

Mass of [tex]\rm \bold { CH_3COOH}[/tex]  = [tex]\rm \bold{ 0.01125 \times 60 = 0.675 g}[/tex]

Assume that the density of the vinegar is  1 g/mL

The mass percent of acetic acid can be calculated,

The mass percent of acetic acid = (mass of acetic acid / mass of vinegar) 100

The mass percent of acetic acid = [tex]\rm \bold{\frac{0.674}{7} \times 100 = 9.6 } \\[/tex]

Hence, we can conclude that  the mass percent of acetic acid in the vinegar sample 9.6 %.

To know more about mass percent, refer to the link:

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Substance X is the substrate for the initial enzyme in the metabolic pathway. The sequence represents a process known as feedback inhibition, where the end product (A) inhibits an enzyme early in the pathway, thus controlling the production rate.

In the provided sequence X → Y → Z → A, product A inhibits the enzyme that catalyzes the reaction converting X to Y. In this scenario, substance X is the substrate for the initial enzyme in this metabolic pathway. Enzymes, which are usually proteins, catalyze biochemical reactions by forming temporary and reversible complexes with their substrates at the enzyme's active site. They do this by lowering the activation energy for a chemical reaction to occur. Substrate X gets converted to Y, Y then converts to Z, and finally, Z gets converted to A.

This process is an example of feedback inhibition, a regulatory mechanism in cells. Feedback inhibition involves the end product of a metabolic pathway, in this case substance A, binding to an enzyme that acts early in the pathway (converting X into Y). This binding typically occurs at a position remote from the enzyme's active site (an allosteric site), and it inhibits the enzyme's activity, thereby reducing the speed of production when the levels of the end product (A) are high. This mechanism prevents the unnecessary accumulation of product A when it is already abundant.

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Answer:

For a: The amount of heat transferred for the given amount of methanol is 94.6736 kJ.

For b: The mass of methane gas produced will be 10.384 g.

Explanation:

For the given chemical reaction:

[tex]2CH_3OH(g)\rightarrow 2CH_4(g)+O_2(g);\Delta H=+252.8kJ[/tex]

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 methanol = 24.0 g

Molar mass of methanol = 32.04 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of methanol}=\frac{24.0g}{32.04g/mol}=0.749mol[/tex]

By Stoichiometry of the reaction:

For every 2 moles of methanol, the amount of heat transferred is +252.8 kJ.

So, for every 0.749 moles of methanol, the amount of heat transferred will be = [tex]\frac{252.8}{2}\times 0.749=94.6736kJ[/tex]

Hence, the amount of heat transferred for the given amount of methanol is 94.6736 kJ.

By Stoichiometry of the reaction:

252.8 kJ of energy is absorbed when 2 moles of methane gas is produced.

So, 82.1 kJ of energy will be absorbed when = [tex]\frac{2}{252.8}\times 82.1=0.649mol[/tex] of methane gas is produced.

Now, calculating the mass of methane gas from equation 1, we get:

Molar mass of methane gas = 16 g/mol

Moles of methane gas = 0.649 moles

Putting values in equation 1, we get:

[tex]0.649mol=\frac{\text{Mass of methane gas}}{16g/mol}\\\\\text{Mass of methane}=10.384g[/tex]

Hence, the mass of methane gas produced will be 10.384 g.

To calculate the amount of heat transferred, use the equation q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. To calculate the grams of methane gas produced, use the equation q = ΔH, where ΔH is the enthalpy change during the reaction.

To calculate the amount of heat transferred, we need to use the equation q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we are working at constant pressure, so we can use the equation q = ΔH. First, calculate the moles of CH3OH(g) using the molar mass. Then, use the molar ratio from the balanced equation to determine the moles of CH4(g). Finally, use the molar mass of CH4(g) to calculate the mass of CH4(g). Substitute the values into the equation q = ΔH to find the amount of heat transferred.

The given enthalpy change during the reaction is +82.1 kJ. Using the equation q = ΔH, we can calculate the moles of CH4(g). Then, use the molar mass of CH4(g) to calculate the mass of CH4(g).

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Answer:

Rate = 1.09*10^-3 (mol H2/L)/s

Explanation:

Given:

Initial concentration of H2, C1 = 0 M

Final concentration of H2, C2 = 0.101 M

Time taken, t = 93.0 s

To determine:

The rate of the given reaction

Calculation:

The decomposition of PH3 is represented by the following chemical reaction

[tex]4 PH3(g)\rightarrow P4(g) + 6 H2(g)[/tex]

Reaction rate in terms of the appearance of H2 is given as:

[tex]Rate = +\frac{1}{6}*\frac{\Delta [H2]]}{\Delta t}[/tex]

[tex]Rate = +\frac{1}{6}*\frac{C2[H2]-C1[H2]}{\Delta t}[/tex]

Here C1(H2) = 0 M and C2(H2) = 0.101 M

Δt = 93.0 s

[tex]Rate = \frac{(0.101-0.0)M}{93.0 s} =1.09*10^{-3} M/s[/tex]

Since molarity M = mole/L

rate = 1.09*10^-3 (mol H2/L)/s

Answer: c. oxidation - gain of electrons

Explanation:

1. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Example: [tex]Li_2O+CO_2\rightarrow Li_2CO_3[/tex]  

2. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: [tex]Li_2CO_3\rightarrow Li_2O+CO_2[/tex]

3. Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

[tex]M\rightarrow M^{n+}+ne^-[/tex]

4. Dehydration reaction is defined as the reaction in which water is lost as product.

[tex]CH_3CH_2OH\rightarrow CH_2=CH_2+H_2O[/tex]

5. Hydrolysis  reaction is defined as the reaction in which water is used for decomposition.

Example: [tex]CH_3COOCH_2CH_3+H_2O\rightarrow CH_3COOH+CH_3CH_2OH[/tex]

The mismatched pair in the list is 'oxidation - gain of electrons.' Oxidation is actually characterized by the loss of electrons, not the gain. All other pairs accurately depict the respective chemical reactions.

The question asked is trying to identify which of the provided pairs misrepresents a type of chemical reaction. For the majority of these reactions, the descriptions are accurate:

However, the provided definition of an oxidation reaction is incorrect. Oxidation is characterized by the loss of electrons, not the gain. The oxidation-reduction (redox) reactions involve an exchange of electrons between reactants, where the substance losing electrons is being oxidized, and the substance gaining electrons is being reduced.

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hey there!:

In the case of ammonia, the hydrogen bond is formed using the lone pair present in nitrogen and the hydrogen having δ+ charge (due to bonding with electronegative N) of another ammonia molecule. Thus the inter molecular attraction increases which in turn increases the boiling point .

Option B is the correct answer.

Hope this helps!

The intermolecular force or bond which  is responsible for the high boiling point of NH₃ relative to PH₃ and AsH₃ is covalent bonding.

Covalent bonding is defined as a type of bonding which is formed by the mutual sharing of electrons to form electron pairs between the two atoms.These electron pairs are called as bonding pairs or shared pair of electrons.

Due to the sharing of valence electrons , the atoms are able to achieve a stable electronic configuration . Covalent bonding involves many types of interactions like σ bonding,π bonding ,metal-to-metal bonding ,etc.

Sigma bonds are the strongest covalent bonds while the pi bonds are weaker covalent bonds .Covalent bonds are affected by electronegativities of the atoms present in the molecules.Compounds having covalent bonds have lower melting points as compared to those with ionic bonds.

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Answer:

[tex]H_2PO_3^-[/tex] is Bronsted Lowry base.

[tex]H_2O[/tex] is Bronsted Lowry acid.

Explanation:

According to the Bronsted Lowry conjugate acid-base theory:

[tex]H_2PO_3^-(aq)+H2O(l)\rightarrow H_3PO_3(aq)+OH^-(aq)[/tex]

[tex]H_2PO_3^-[/tex] is Bronsted Lowry base.It accepts protons and forms conjugate acid [tex]H_3PO_3[/tex]

[tex]H_2O[/tex] is Bronsted Lowry acid.It donates protons and forms conjugate base [tex]OH^-[/tex]

[tex](CH_3)_2NH(g)+BF_3(g)\rightarrow (CH_3)_2NHBF_3(s)[/tex]

There in no exchange of proton in an above reaction.Neither of the reactants and products are Bronsted Lowry acid or Bronsted Lowry base

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

[tex]\Delta T_f=k_f\times m[/tex]

or,

[tex]\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]k_f[/tex] = freezing point constant  (for benzene} =[tex]5.12^0Ckg/mol[/tex]

m = molality

Putting in the values we get:

[tex]0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}[/tex]

[tex]{\text{ Mass of solute in g}}=0.028g[/tex]

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.

Answer:

B. Carbon dioxide

Explanation:

A catalytic converter is designed and placed in the exhaust of most automobiles to remove harmful gases from escaping into the environment.

The converter uses palladium and platinum which are both catalyst to reduce harmful gases such as nitrogen oxide which causes acid rain, carbon monoxide which affects human haemoglobin and other hydrocarbons into less harmful ones.  

The catalytic converter helps to convert carbon monoxide, a product of incomplete combustion of hydrocarbons into less harmful carbondioxde. Carbon monoxide is a very toxic gas. It causes harm to both plants and animals as well.

Hey there!:

Aqueous ethanol contains water , sodium borohydride will react with water and decompose to give hydrogen gas :

NHB4 + 2 H2O => NaBO2 + 4 H2

thus a large excess of borohydride should be used used to take into account losses from the reaction with water , so that there is sufficient sodium borohydride left for the reduction of the chemical compound os interest .

Hope this helps!

The importance of using a large excess of sodium borohydride in a reduction in aqueous ethanol lies in its role as a reducing agent that donates hydrogen ions. However, in an aqueous solution, some of the sodium borohydride may react with water in a hydrolysis reaction, forming hydrogen gas and borate ions. Using an excess ensures an ample amount for the intended reaction.

It's important to use a large excess of sodium borohydride when doing a reduction in aqueous ethanol because sodium borohydride (NaBH₄) is an excellent reducing agent. It donates hydrogen ions (H⁻) which do not survive in an acidic medium. The H⁻ ions react initially, and then the acid is added to donates a proton to oxygen (O) in the second step.

However, the aqueous environment complicates this process because water can react with sodium borohydride in a hydrolysis reaction. Water acts as an incoming nucleophile, and in the presence of water, NaBH4 may end up reducing water molecules, forming hydrogen gas (H₂) and borate ions (BO3⁻). Using an excess of sodium borohydride ensures that there's enough of it to react with the intended substance, even when a portion of it reacts with water.

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Answer:

It will take 6.6 hours for 75% of the lead to decay.

Explanation:

The radioactive decay follows first order rate law

The half life and rate constant are related as

[tex]k=rate constant=\frac{0.693}{halflife}=\frac{0.693}{3.3}=0.21h^{-1}[/tex]

The rate law for first order reaction is

[tex]time=\frac{1}{k}(ln[\frac{A_{0}}{A_{t}}][/tex]

Where

A0 = initial concentration = 1 g

At= final concentration = 0.25 g (as 75% undergoes decay so 25% left]

[tex]time=\frac{1}{0.21}(ln(\frac{1}{0.25})=6.6hours[/tex]

Answer:

Joule-Thomson coefficient for an ideal gas:

[tex]\mu_{J.T} = 0[/tex]

Explanation:

Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.

Thus,

[tex]\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H[/tex]

Also,

[tex]H= H (T,P)[/tex]

[tex]Differentiating\ it,[/tex]

[tex]dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT[/tex]

Also, [tex] C_p[/tex] is defined  as:

[tex]C_p = \left [\frac{\partial H}{\partial T}\right ]_P[/tex]

[tex]So,[/tex]

[tex]dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT[/tex]

Acoording to defination, the ethalpy is constant which means [tex]dH = 0[/tex]

[tex]So,[/tex]

[tex]\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H[/tex]

[tex]Also,[/tex]

[tex]\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H[/tex]

[tex]So,[/tex]

[tex]\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p[/tex]

For an ideal gas,

[tex]\left [\frac{\partial H}{\partial P}\right ]_T = 0[/tex]

So,

[tex]0 =-\mu_{J.T}\times C_p[/tex]

Thus, [tex]C_p[/tex] ≠0. So,

[tex] \mu_{J.T} = 0[/tex]

The Joule-Thompson Coefficient is zero for an ideal gas because the temperature of the gas doesn't change during an adiabatic expansion or compression due to the lack of intermolecular forces, which is a characteristic as per the ideal gas laws.

The Joule-Thompson Coefficient measures the change in temperature of a gas when it is forced to expand or contract at constant enthalpy. In an ideal gas, the internal energy, which includes information about temperature, is only a function of temperature. Since an ideal gas follows the ideal gas law, when a gas expands or contracts with no heat exchange (an adiabatic process), its temperature remains constant due to an absence of intermolecular forces. This is because any work done on (or by) the gas doesn’t result in a change in temperature.

We know from the equation of internal energy (Eint = 3nRAT/2), that any change in internal energy is due to a change in temperature (ΔT), and if ΔT = 0 for an isothermal process, then the change in internal energy of the system (ΔEint) also equals zero, indicating no change in temperature. Hence, for an ideal gas, the Joule-Thompson Coefficient is zero because the temperature of the gas doesn't change during an adiabatic expansion or compression.

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Answer: False

Explanation:

Viscosity is defined as the resistance to the flow of a fluid. More are the inter molecular forces between the particles of a liquid, the more the viscosity of the liquid and thus it will flow slowly.

Example: Oil is more viscous than water and thus flows slowly than water.

Effect of temperature: Viscosity decreases with increase in temperature as the forces among particles decrease due to increase in kinetic energy and thus they offer less resistance to flow.

Thus viscosity of liquids will decrease with increasing temperature.

To find the equilibrium concentration of Cl2 in the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), we use the equilibrium constant and initial concentrations to set up and solve a quadratic equation. The resulting equilibrium concentration of Cl2 is 0.054 M.

To calculate the equilibrium concentration of Cl2 from the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), let's denote the change in concentration of Cl2 as 'x' when the system reaches equilibrium. Since Cl2 and PCl3 start at 0.0571 M and PCl5 starts at 0 M, the equilibrium concentrations will be 0.0571 - x for PCl3 and Cl2, and x for PCl5.

The equilibrium constant expression is Kc = [PCl5]/[PCl3][Cl2]. Plugging in the equilibrium concentrations and the given Kc value of 0.021, we get:
0.021 = x / (0.0571 - x)2. After solving this quadratic equation, we find that the equilibrium concentration of Cl2 rounded to the nearest thousandth is 0.054 M.

Answer:

4600 Liters NH₃(g)

Explanation:

Applying Avogadro's law to the reaction of hydrogen and nitrogen gas forming ammonia, we find that 6.9 m3 of hydrogen will form about 4.6 m3 of ammonia, assuming that temperature and pressure remain constant.

The subject in question relates to the application of Avogadro's law to chemical reactions involving gases. More specifically, we're investigating the reaction between hydrogen gas and nitrogen gas to produce ammonia gas. As per Avogadro's law, gases react in definite and simple proportions by volume, if all gas volumes are measured at the same temperature and pressure.

Focusing on the hydrogen to ammonia conversion, the reaction equation N₂(g) + 3H₂(g) turns into 2NH3(g). We can surmise that three volumes of hydrogen gas (H2) react to form two volumes of ammonia gas (NH3). Considering that 6.9 m3 of hydrogen gas is consumed, by the rule of three, we can infer that the reaction would result in approximately 4.6 m3 of ammonia gas.

Bear in mind, these calculations are assuming that the temperature and pressure remain constant during the reaction and that it goes to completion.

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hey there!:

H2S(aq) <=> H⁺(aq) + HS⁻(aq)

K'c = [H⁺][HS⁻]/[H₂S] = 9.5*10⁻⁸

HS⁻(aq) <=> H⁺(aq) + S²⁻(aq)

K"c = [H⁺][S²⁻]/[HS⁻] = 1.0*10⁻¹⁹

H₂S(aq) <=> 2 H⁺(aq) + S²⁻(aq)

Kc = [H⁺]²[S²⁻] / [H₂S]

= [H+][HS⁻] / [H₂S] * [H⁺][S²⁻]/[HS⁻]

= K'c *K"c

= ( 9.5*10⁻⁸ ) * ( 1.0 x 10⁻¹⁹ )

= 9.5*10⁻²⁷

Hope this helps!

The equilibrium constants the following reaction at the same temperature is 9.5*10⁻²⁷.

Equilibrium is the state that is in control, a balance state in which no changes occur.

The calculation of equilibrium constants is

[tex]H_2S(aq) < = > H^+(aq) + HS^-(aq)[/tex]

K'c = [H⁺] [HS⁻] / [H₂S] = 9.5 * 10⁻⁸

HS⁻ (aq) <=> H⁺(aq) + S²⁻(aq)

K"c = [H⁺] [S²⁻]/[HS⁻] = 1.0 * 10⁻¹⁹

H₂S (aq) <=> 2 H⁺(aq) + S²⁻(aq)

Kc = [H⁺]²[S²⁻] / [H₂S]

= [H+] [HS⁻] / [H₂S] * [H⁺] [S²⁻] / [HS⁻]

= K'c *K"c

= ( 9.5*10⁻⁸ ) * ( 1.0 x 10⁻¹⁹ )

= 9.5*10⁻²⁷

Thus, the equilibrium constants of the following reaction at the same temperature is 9.5*10⁻²⁷.

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Answer:

CuSO4 + Fe -> FeSO4 + Cu

Explanation:

This reaction is a classic example of a redox reaction. I won't go in too deep, but the basic thing is that electrons from the Fe atom go to the Cu2+ ion. Therefore, Fe becomes an ion, and Cu - an electroneutral atom:

Fe + Cu2+ -> Fe2+ + Cu.

Silver is not a very reactive metal and it does not give up its electrons to Cu.

Iron displaces copper in Copper(II) sulfate solution, forming Iron sulfate and Copper. The reaction with the silver rod does not occur as silver is less reactive than copper.

When the student placed an iron nail into the copper(II) sulfate solution, a displacement reaction occurred. Iron is higher on the reactivity series than copper, so it displaced copper from the copper(II) sulfate. The balanced chemical equation for this reaction is: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

On the other hand, when the silver rod was placed into the copper(II) sulfate solution, no visible reaction occurred. This is because Silver is less reactive than Copper and cannot displace it from its sulfate salt.

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In the redox reaction, Mg gets oxidized and Cl2 gets reduced. The oxidizing agent is Cl- and the reducing agent is Mg.

In the redox reaction: Mg(s) + Cl2(g) → Mg2+(aq) + 2Cl-(aq), the substance that gets oxidized is Mg(s) and the substance that gets reduced is Cl2(g). The oxidizing agent is Cl- and the reducing agent is Mg(s).

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Answer: The volume of liquid in liters is 2.5812 L.

Explanation:

Density of an object is defined as the ratio of its mass and volume. The chemical equation representing density of an object is:

[tex]\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}[/tex]

We are given:

Mass of liquid = 3.02 kg = 3020 g     (Conversion factor: 1kg = 1000 g)

Density of liquid = [tex]1.17 g/cm^3[/tex]

Putting values in above equation, we get:

[tex]1.17g/cm^3=\frac{3020g}{\text{Volume of liquid}}\\\\\text{Volume of liquid}=2581.2cm^3[/tex]

Converting this into liters, we use the conversion factor:

[tex]1L=1000cm^3[/tex]

So, [tex]\Rightarrow \frac{1L}{1000cm^3}\times 2581.2cm^3[/tex]

[tex]\Rightarrow 2.5812L[/tex]

Hence, the volume of the liquid is 2.5812 L.

The volume in liters of 3.02 kg of the liquid is 2.58 Liters

Density is the ratio of mass to volume of a substance. The density of a substance is given by:

Density = mass / volume

Given that a liquid has a density of 1.17 g/cm³ and a mass of 3.02 kg, the volume is:

Density = 1.17 g/cm³ = 1.17 kg/L

Density = mass/volume

1.17 = 3.02/volume

Volume = 3.02/1.17

Volume = 2.58 Liters

Hence the volume in liters of 3.02 kg of the liquid is 2.58 Liters

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Answer:

-56.4 kJ/mol

Explanation:

There are two heat flows in this experiment.

Heat released by reaction + heat absorbed by solution = 0

                     q1                   +                     q2                     = 0

                   nΔH                 +                  mCΔT                  = 0

1. Moles of water formed

                  KOHL + H2SO4 → K2SO4 + 2H2O

V/mL:          40.0        40.0

c/(mol/L)      1.00       0.500

   Moles of KOH = 40.0 mL × (1.00 mmoL/1 mL)     = 40.00 mmol KOH

Moles of H2SO4 = 40.0 mL × (0.500 mmoL/1 mL) = 20.00 mmol H2SO4

Moles of H2O from KOH

= 40.00 mL KOH × (2.00 mmol H2O/2 mmoL KOH)  = 40.00 mmol H2O

Moles H2O from of H2SO4

= 20.00 mL × (2 mmoL H2O/1 mmol H2SO4)             = 40.00 mmol H2O

KOH and H2SO4 are present in equimolar amounts.

They form 40.00 mmol = 0.040 00 mol of water.

2. Calculate q1

q1 = 0.04000 mol × ΔH = 0.040 00ΔH J

3. Calculate q2

Mass of water formed = 0.040 00 mol × (18.02 g/1 mol) = 0.7208 g

                  V(solution) = 40.00 + 40.00 = 80.00 mL

        Mass of solution = 80.00 mL × (1.02 g/1 mL) = 81.60 g

                 Total mass = 81.60 + 0.7208 = 82.32 g

                              ΔT = T2 - T1 = 27.85 - 21.00 = 6.85 °C

                              q2 = 82.32 × 4.00 × 6.85 = 2256 J

4. Calculate ΔH

0.040 00ΔH + 2256 = 0

            0.040 00ΔH = -2256

                           ΔH = -2256/0.040 00 = = -56 400 J/mol = -56.4 kJ/mol

The enthalpy of neutralization is -56.4 kJ/mol.

Answer:

The annual production of carbon dioxide is [tex]9.12\times 10^{10} kg[/tex].

Explanation:

Distance covered by each car = 5790 miles

Rate of consumption of gasoline =24. mile/gal

For every 24.1 mile 1 gallon of gasoline is used

Gasoline used by a single car by travelling 5790 miles =

[tex]\frac{1}{24.1}\times 5790 mile=240.24 gal[/tex]

Number of cars in the United states = 40.0 million = [tex]4\times 10^7[/tex]

Total gallons of gasoline consumed by 40 million cars = [tex]4\times 10^7\times 240.24 gal=9.60\times 10^9 gal[/tex]

1 gallon of gasoline produces = 9.50 kg of [tex]CO_2[/tex]

Then [tex]9.60\times 10^9 gal[/tex] of gasoline will produce:

[tex]9.60\times 10^9\times 9.50 kg =9.12\times 10^{10} kg[/tex] of [tex]CO_2[/tex]

The annual production of carbon dioxide is [tex]9.12\times 10^{10} kg[/tex].

Final answer:

Intensive properties are constant regardless of the amount of substance, while extensive properties depend on the amount. Temperature and density are examples of intensive properties, whereas mass and volume are extensive properties.

Explanation:

The difference between an intensive property and an extensive property is fundamental in chemistry and relates to how properties of matter are defined in relation to the amount of substance present. Extensive properties are those that depend on the amount of matter, such as mass and volume. For instance, a larger amount of substance will have a greater mass and volume than a smaller amount. In contrast, intensive properties do not depend on the amount of matter. These properties, such as temperature and density, remain consistent regardless of how much substance you have.

For example, whether you have a gallon or a cup of milk at 20 °C, the temperature is the same because temperature is an intensive property. Similarly, the density of a material is an intensive property because it is defined as the mass per unit volume, a ratio of two extensive properties which effectively 'cancels out' the dependence on quantity.

Answer:

6.9

Explanation:

A weak acid dissociates in an equilibrium reaction, thus, it is in equilibrium with its conjugate base:

HA ⇄ H⁺ + A⁻

The equilibrium constant (Ka) can be calculated, where Ka = [H⁺]*[A⁻]/[HA]. Using the -log form, we can also have pKa = -logKa. By the Handerson-Halsebach (HH) equation, the relation between pH and pKa is:

pH = pKa + log[A⁻]/[HA]

So, when pH = 5.7, for this acid, the ratio of [acid]/[base] ([HA]/[A-]) is:

5.7 = 6.3 + log[A⁻]/[HA]

log[A⁻]/[HA] = -0.6

log[HA]/[A⁻] = 0.6

[HA]/[A⁻] = [tex]10^{0.6}[/tex]

[HA]/[A⁻] = 3.98 = 4.0

If the ratio of acid and base is 4 to 1, it means that 80%(4/5) of the acid is protonated after the addition of the HCl.

The initial number of moles of the weak acid was: 100 mL * 0.10 M = 10 mmol, so after the addition of HCl, 8 mmol is in the acid form (80% of 10). It was added 6.0 mmol of HCl (6.0 mL*1.0M). Thus, 6.0 mmol of H+ was added and reacted with the conjugate base of the weak acid.

For the mass conservation, the initial amount of the protonated weak acid must be 2.0 mmol (8 - 6), and the number of moles of the conjugate base was 8.0 mmol. Using the HH equation:

pH = 6.3 + log(8/2)

pH = 6.3 + 0.6

pH = 6.9

The original pH of the weak acid solution was 6.3 which can be determined using the pKa value and the Henderson-Hasselbalch equation. The addition of HCl does not affect this original pH.

A weak acid's initial pH can be calculated using the pKa of the acid and its initial concentration (0.10 M in this case) through the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). However, a weak acid exists in equilibrium with its conjugate base and there is no added base here initially, so the pH equals the pKa, making the original pH 6.3. The addition of HCl doesn't matter in this context because the question asks for the original pH, not the final pH.

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Answer: The [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of [tex]N_2O[/tex] is:

[tex]N_2(g)+\frac{1}{2}O_2(g)\rightarrow N_2O(g)[/tex]    [tex]\Delta H^o_{formation}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(l)[/tex]    [tex]\Delta H_1=-1010kJ[/tex]    ( ÷  3)

(2) [tex]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)    [tex]\Delta H_2=1531kJ[/tex] ( ÷  6)

Reversing Equation 1 and then adding both the equations, we get the enthalpy change for the chemical reaction.

[tex]\Delta H^o_{formation}=[\frac{\Delta H_1}{3}]+[\frac{\Delta H_2}{6}][/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{formation}=[\frac{1010}{3}]+[\frac{1531}{6}]\\\\\Delta H^o_{formation}=591.9kJ[/tex]

Hence, the [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.

Hess's law is defined as the sum of amount of heat absorbed or released in the given chemical equation remains constant, irrespective of the steps involved in the reaction. The [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex] is for the given reaction.

Given that,

Now, the given chemical equations are:

The intermediate reactions between the above equation are:

Reversing the equation and then adding both the values, the enthalpy change becomes:

Therefore, the enthalpy change of the reaction is [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex].

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Answer:

Explanation:

0 degrees kelvin is mighty cold. It should be 0 degrees Celsius.

The answer is false.

Answer:

For a: The molality and molarity of the given solution is 2.45m and 2.28 M respectively.

For b: The molarity of the solution when more water is added is 0.912 M

Explanation:

To calculate the molality of solution, we use the equation:

[tex]Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](C_3H_8O_3)[/tex] = 42.0 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](C_3H_8O_3)[/tex] = 92.093 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 186 g

Putting values in above equation, we get:

[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 186}\\\\\text{Molality of }C_3H_8O_3=2.45m[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]    .....(1)

We are given:

Molarity of solution = ?

Molar mass of [tex](C_3H_8O_3)[/tex] = 92.093 g/mol

Volume of solution = 200 mL

Mass of [tex](C_3H_8O_3)[/tex] = 42 g

Putting values in above equation, we get:

[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 200}\\\\\text{Molality of }C_3H_8O_3=2.28M[/tex]

Hence, the molality and molarity of the given solution is 2.45m and 2.28 M respectively.

Now, the 300 mL water is added to the solution. So, the total volume of the solution becomes (200 + 300) = 500 mL

Using equation 1 to calculate the molarity of solution, we get:

Molar mass of [tex](C_3H_8O_3)[/tex] = 92.093 g/mol

Volume of solution = 500 mL

Mass of [tex](C_3H_8O_3)[/tex] = 42 g

Putting values in equation 1, we get:

[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 500}\\\\\text{Molality of }C_3H_8O_3=0.912M[/tex]

Hence, the molarity of the solution when more water is added is 0.912 M

Answer : The mass of the water in two significant figures is, [tex]3.0\times 10^1g[/tex]

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of iron metal = [tex]0.45J/g^oC[/tex]

[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of iron metal = 32.3 g

[tex]m_2[/tex] = mass of water = ?

[tex]T_f[/tex] = final temperature of mixture = [tex]59.2^oC[/tex]

[tex]T_1[/tex] = initial temperature of iron metal = [tex]21.9^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]63.5^oC[/tex]

Now put all the given values in the above formula, we get

[tex]32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC[/tex]

[tex]m_2=30.16g\approx 3.0\times 10^1g[/tex]

Therefore, the mass of the water in two significant figures is, [tex]3.0\times 10^1g[/tex]

The mass of the water in the insulated container is 30.04 g

Heat loss = Heat gain

MᵥᵥC(Tᵥᵥ – Tₑ) = MᵢC(Tₑ – Mᵢ)

Mᵥᵥ × 4.184 (63.5 – 59.2) = 32.2 × 0.45(59.2 – 21.9)

Mᵥᵥ × 4.184(4.3) = 14.49(37.3)

Clear bracket

Mᵥᵥ × 17.9912 = 540.477

Divide both side by 17.9912

Mᵥᵥ = 540.477 / 17.9912

Mᵥᵥ = 30.04 g

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