Answer: The coefficients are 2, 7, 4 and 6.
Explanation:
Every balanced chemical equation follows Law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This means that total mass on the reactant side is equal to the total mass on the product side.
This also means that the total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side.
For the given chemical reaction, the balanced equation follows:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO2(g)+6H_2O(g)[/tex]
On reactant side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
On product side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
Hence, the coefficients are 2, 7, 4 and 6.
To balance the combustion of ethane (C2H6), the correct coefficients are 2 for C2H6, 7 for O2, 4 for CO2, and 6 for H2O resulting in the equation 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
Explanation:The chemical equation for the combustion of ethane to form carbon dioxide and water is C₂H₆ + O₂ → CO₂ + H₂O. After balancing the number of atoms of each element on both sides of the equation, we arrive at the balanced chemical equation: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Thus, the resulting coefficients are 2, 7, 4, and 6.
Firstly, the carbon (C) atoms are balanced. As we have 2 carbon atoms on the left side, we can put the coefficient 2 in front of CO₂ on the right side.Next, we balance the hydrogen (H) atoms. Since we have 6 hydrogen atoms on the left side, we put the coefficient 3 in front of H₂O on the right side.Finally, we balance the oxygen (O) atoms. Counting the oxygen atoms on both sides, we see that we have 7 oxygen atoms on the right (4 from CO₂ and 3 from H₂O), meaning we need 7 O₂ molecules on the left side.Learn more about Chemical reaction balancing here:https://brainly.com/question/32834309
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The heat capacity of solid sodiu capacity of solid sodium carbonate (Na2CO3) as estimated by Kopp's rule is A. 2.7 kJ/mol°C B. 0.121 KJ/mol°C C. 1.9 kJ/mol°C D. 0.11 KJ/mol°C
Final answer:
Without specific numeric values for the individual atomic heat capacities of the elements in sodium carbonate, directly answering the heat capacity multiple-choice question using Kopp's rule is not feasible. Additional data is required to estimate the heat capacity of solid sodium carbonate.
Explanation:
The question regarding the heat capacity of solid sodium carbonate (Na2CO3) can be answered by applying Kopp's rule, which is a method used to estimate the heat capacities of solids. According to Kopp's rule, the heat capacity of a compound in the solid state is the sum of the atomic heat capacities of the individual elements that compose the compound. Atomic heat capacities can be estimated using the Law of Dulong and Petit, which suggests that at room temperature, the molar heat capacity of many solid elements roughly equates to 3R, where R is the gas constant with a value approximately equal to 8.314 J/(mol·K).
Applying Kopp's rule would involve calculating the total molar heat capacity for sodium, carbon, and oxygen in sodium carbonate, based on their atomic weights and expected contributions. However, since specific numeric values are not provided in the question or reference information to apply Kopp's rule directly, directly answering the multiple-choice question is not feasible without additional data on the individual atomic heat capacities of the elements involved in sodium carbonate.
Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction 2COF2(g)⇌CO2(g)+CF4(g), Kc=4.90 If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
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
The student decided to do another experiment with his leftover copper(II) sulfate (CuSO4) solution. He divided the solution up into two beakers. He dropped an iron nail (Fe(s)) into one beaker. He dropped a silver rod (Ag(s)) into the other. Write the chemical equation of any reactions he saw.
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.
Explanation: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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technology is hurting education or enhancing education
The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hours of electricity required to produce 3.00kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 4.50V.
Explanation:
According to Faraday's law, the amount of a substance deposited or liberated in electrolysis process is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.
Formula to calculate the mass of substance liberated according to Faraday's law is as follows.
m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]
where, m = mass of substance liberated at electrode
Q = electric charge passing through the substance
F = Faraday constant = 96,487 C [tex]mol^{-1}[/tex]
M = molar mass of the substance
Z = valency number of ions of the substance
Since, it is given that mass is 3 kg or 3000 g (as 1 kg = 1000 g), molar mass of Al is 27, Z is 3.
Therefore, putting the values in the above formula as follows.
m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]
3000 g = [tex](\frac{Q}{96,487 C mol^{-1}})(\frac{27}{3})[/tex]
Q = 32162333.33 C
As it is given that V = 4.50 Volt. Also, it is known that
Energy = [tex]V \times Q[/tex]
Therefore, calculate the energy as follows.
Energy = [tex]V \times Q[/tex]
= [tex]4.50 V \times 32162333.33 C[/tex]
= 144730500 J
As it is known that [tex]3.6 \times 10^{6}[/tex] J = 1 KW Hr
So, convert 144730500 J into KW Hr as follows.
[tex]\frac{144730500 J \times 1 KW Hr}{3600000 J}[/tex]
= 40.202 KW Hr
Thus, we can conclude that the number of kilowatt-hours of electricity required to produce 3.00 kg of aluminum from electrolysis of compounds from bauxite is 40.202 KW Hr when the applied emf is 4.50V.
The production of aluminum from bauxite involves several chemical reactions and an electrolysis process in a Hall-Héroult cell. Aluminum oxide is reduced to aluminum metal during electrolysis. The exact amount of electricity required cannot be stated without additional industrial data.
Explanation:The process of making aluminum from bauxite involves several stages of chemical reactions. Initially, bauxite, AlO(OH), reacts with hot sodium hydroxide to form soluble sodium aluminate, leaving behind impurities. Aluminum hydroxide is then precipitated out and heated to form aluminum oxide, Al2O3, which is dissolved in a molten mixture of cryolite and calcium fluoride. An electrolytic cell, specifically called the Hall-Héroult cell, is used for the electrolysis process. During electrolysis, the reduction of aluminum ions to aluminum metal takes place at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode. The amount of electricity required to produce 3 kg of aluminum using a 4.5V emf cannot be directly calculated without additional data such as the exact chemical conversion efficiencies and energy losses in the process, which are typically proprietary information for aluminum production companies.
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The following equilibrium constants have been determined for hydrosulfuric acid at 25ºC:
H2S(aq) ⇌ H+(aq) + HS–(aq) K′c = 9.5 × 10^–8
HS–(aq) ⇌ H+(aq) + S2–(aq) K″c = 1.0 × 10^–19
Calculate the equilibrium constant for the following reaction at the same temperature: H2S(aq) ⇌ 2H+(aq) + S2–(aq).
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.
What is equilibrium?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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4. Why is it important to use a large excess of sodium borohydride when doing a reduction in aqueous ethanol? (Hint: Consider what reaction might occur between water and sodium borohydride.)
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.
Explanation: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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Hydrogen gas and nitrogen gas react to form ammonia gas. What volume of ammonia would be produced by this reaction if 6.9 m3 of hydrogen were consumed? Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.
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.
Explanation: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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Why is an intensive property different from an extensive property
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.
For the gas phase decomposition of phosphine at 120 °C, the rate of the reaction is determined by measuring the appearance of H2. 4 PH3(g)P4(g) + 6 H2(g) At the beginning of the reaction, the concentration of H2 is 0 M. After 93.0 s the concentration has increased to 0.101 M. What is the rate of the reaction? (mol H2/L) /s
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
A 280 mL bubble contains 0.283 g of a gas at 0.951 atm and 25.0°C What is the molar mass of this gas?
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.
Explanation: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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The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . How many grams of DDT, C14H9Cl5 (354.5 g/mol), must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C ?
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.
A liquid in the lab has a density of 1.17 g/cm3. What is the volume in liters of 3.02 kg of the liquid?
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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The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(water) = 0.512 °C/m In a laboratory experiment, students synthesized a new compound and found that when 13.62 grams of the compound were dissolved in 217.5 grams of water, the solution began to boil at 100.094 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
Answer:
341.1 g/molExplanation:
1) Data:
a) Tb₁ = 100.000°C
b) Kb = 0.512 °C/m
c) mass of solute = 13.62 g
d) mass of solvent = 217.5 g
e) Tb₂ = 100.094°C
f) Solute: nonvolatile and nonelectrolyte
g) MM = ?
2) Chemical principles and formulae:
a) The boiling point elevation of a non-volatile solute is a colligative property, which follows this equation:
ΔTb = Kb × m × iWhere:
ΔTb is the elevation of the boiling point = Tb₂ - Tb₁,Kb is the molal boiling constant of the solvent,m is the molality of the solution,i is the Van't Hoof constant, and is equal to 1 for non-electrolyte solutes.b) Molality, m:
m = number of moles of solute / kg of solventc) Molar mass, MM:
MM = mass in grams / number of moles3) Solution:
i) ΔTb = Tb₂ - Tb₁ = 100.094°C - 100.000°C = 0.094°C
ii) ΔTb = Kb × m ⇒ m = ΔTb / Kb = 0.094°C / (0.512°C/m) = 0.1836 m
iii) m = number of moles of solute / kg of solvent ⇒
number of moles of solute = m × kg of solvent = 0.1836 m × 0.2175 kg
number of moles of solute = 0.03993 mol
iv) MM = mass in grams / number of moles = 13.62 g / 0.03993 mol = 341.1 g/mol
Final answer:
The molecular weight of the synthesized compound is calculated using the change in boiling point and the molal boiling point elevation constant to first determine molality and then the number of moles of the compound. The molecular weight is then found by dividing the mass of the compound by the moles of the compound, yielding a value of 341.1 g/mol.
Explanation:
To calculate the molecular weight of the synthesized compound, we apply the boiling point elevation formula ΔT = Kb × m, where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant of water, and m is the molality of the solution. Given Kb for water is 0.512°C/m and ΔT is 0.094°C (100.094°C - 100.000°C), we can calculate the molality:
m = ΔT / Kb = 0.094°C / 0.512°C/m = 0.1836 m
Next, we calculate the moles of solute using molality and the mass of the solvent (water):
moles of solute = molality × mass of solvent in kg = 0.1836 m × 0.2175 kg = 0.0399 mol
The molecular weight (MW) is then found by dividing the mass of the compound by the moles of the compound:
MW = mass of compound / moles of compound = 13.62 g / 0.0399 mol = 341.1 g/mol
Show that the Joule-Thompson Coefficient is zero for ideal gas.
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.
Explanation: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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2 N H 3 ( g ) ⟷ N 2 ( g ) + 3 H 2 ( g ) K p = 0.83 Consider your answers above, if the initial pressures for all three species is 1 atm what is the equilibrium pressure of H2? (Hint: Your quadratic will have two solutions, which one is impossible?)
The equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.
Equilibrium pressure of H2The equilibrium pressure of H2 is calculated by creating ICE table as follows;
2 N H3 ( g ) ⟷ N2( g ) + 3H2
I: 1 1 1
C: -2x x 3x
E: 1 - 2x 1 + x 1 + 3x
[tex]KP = \frac{(N_2)(H_2)^3}{(NH_3)^2} \\\\0.83 = \frac{(1 + x)(1 + 3x)^3}{(1 - 2x)^2}[/tex]
0.83(1 - 2x)² = (1 + x)(1 + 3x)³
0.83(1 - 4x + 4x²) = (1 + x)((1 + 3x)³)
0.83 - 3.32x + 3.32x² = (1 + x)((1 + 3x)³)
0.83 - 3.32x + 3.32x² = 1 + 10x + 36x² + 54x³ + 27x⁴
27x⁴ + 54x³ + 32.68x² + 13.32x + 0.17 = 0
x = -1.379 or - 0.013
Partial pressure of H2 = 1 + 3xH2 = 1 + 3(-1.379)
H2 = -3.13 atm
H2 = 1 + 3(-0.013)
H2 = 0.96 atm
Thus, the equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.
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Consider the two reactions. 2NH3(g)+3N2O(g)4NH3(g)+3O2(g)⟶4N2(g)+3H2O(l)⟶2N2(g)+6H2O(l) Δ????∘=−1010 kJΔ????∘=1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N2(g)+12O2(g)⟶N2O(g)
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,
[tex]\Delta \text H^0_{1}&=-1010 \text{kJ}[/tex][tex]\Delta \text H^0_{2}&=-1531 \text{kJ}[/tex][tex]\Delta \text H^0_{\text{formation}}&=? \text{kJ}[/tex]Now, the given chemical equations are:
[tex]\text{N}_2_{\text (g)} + \dfrac{1}{2}\text O_2_{(\text g)} \rightarrow \text N_2\text O\;\;\;\;\Delta \text H^0_{\text{formation}}&=? \text{kJ}[/tex]The intermediate reactions between the above equation are:
[tex]\text {2 NH}_3_\text{(g)} + 3\text N_2\text O \rightarrow 4\text N_2 +3\text H_2\text O\;\;\;\;\;\Delta \text H^0_{1}&=-1010 \text{kJ}[/tex][tex]\text {4 NH}_3_\text{(g)} + \text O_2 \rightarrow 2\text N_2 +6\text H_2\text O\;\;\;\;\;\Delta \text H^0_{1}&=-1531\text{kJ}[/tex]Reversing the equation and then adding both the values, the enthalpy change becomes:
[tex]\Delta \text H^0_{\text{formation}}&=\dfrac{\Delta \text {H}_1}{3}+\dfrac{\Delta \text {H}_2}{6}\\\\\Delta \text H^0_{\text{formation}}&=\dfrac{1010}{3}+\dfrac{1531}{6}\\\\\Delta \text H^0_{\text{formation}}&= 591.9 \text{kJ}[/tex]Therefore, the enthalpy change of the reaction is [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex].
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A syringe containing 1.75 mL of oxygen gas is cooled from 97.8 ∘C to 0.7 ∘C. What is the final volume Vf of oxygen gas? (Assume that the pressure is constant.)
Answer : The final volume of oxygen gas will be, 1.292 ml
Explanation :
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of oxygen gas = 1.75 ml
[tex]V_2[/tex] = final volume of oxygen gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]97.8^oC=273+97.8=370.8K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]0.7^oC=273+0.7=273.7K[/tex]
Now put all the given values in the above formula, we get the final volume of the oxygen gas.
[tex]\frac{1.75ml}{V_2}=\frac{370.8K}{273.7K}[/tex]
[tex]V_2=1.292ml[/tex]
Therefore, the final volume of oxygen gas will be, 1.292 ml
Using the Br?nsted-Lowry concept of acids and bases, identify the Br?nsted-Lowry acid and base in each of the following reactions:
H2PO3?(aq)+H2O(l)?H3PO3(aq)+OH?(aq)
(CH3)2NH(g)+BF3(g)?(CH3)2NHBF3(s)
Drag the appropriate items to their respective bins.
H2PO3- H2O BF3 (CH3)2NH
Bronsted Lowry Acid Bronsted Lowry Base Neither
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:
An acid is defined as a substance which donates protons and form conjugate baseA base is defined as a substance which accepts protons and forms conjugate acid.[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
Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 5.0 kg of limestone.
Answer: 2800 g
Explanation:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass = 5 kg = 5000 g
[tex]\text{Number of moles}=\frac{5000g}{100g/mol}=50moles[/tex]
1 mole of [tex]CaCO_3[/tex] produces = 1 mole of [tex]CaO[/tex]
50 moles of [tex]CaCO_3[/tex] produces =[tex]\frac{1}{1}\times 50=50moles[/tex] of [tex]CaO[/tex]
Mass of [tex]CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g[/tex]
2800 g of [tex]CaO[/tex] is produced from 5.0 kg of limestone.
A decomposition reaction splits the reactants into two or more products. The mass of the quicklime produced from 5 kg of limestone is 2800 gm.
What is mass?Mass is the amount or the weight of the substance occupied in the system. It can be calculated in grams or kilograms.
The decomposition reaction of the Limestone can be shown as:
[tex]\rm CaCO_{3} \rightarrow CaO + CO_{2}[/tex]
The number of the mole of limestone is given as:
[tex]\rm Moles = \rm \dfrac {Mass}{Molar \;mass}[/tex]
Here, mass is 5000 gm and the molar mass is 100 g/mol
Substituting values in the equation above:
[tex]\begin{aligned}\rm n &= \dfrac{5000}{100}\\\\&= 50\;\rm mol\end{aligned}[/tex]
The stoichiometry coefficient of the reaction gives:
1 mole of limestone = 1 mole quicklime
So, 50 moles of limestone = x moles of quicklime
Solving for x:
[tex]\begin{aligned}\rm x &= \dfrac{50 \times 1}{1}\\\\&= 50 \;\rm mole\end{aligned}[/tex]
Mass of quicklime is calculated as:
[tex]\begin{aligned}\rm mass &= \rm moles \times \rm molar \; mass\\\\&= 50 \times 56\\\\&= 2800\;\rm gm\end{aligned}[/tex]
Therefore, 2800 gm of quicklime is produced.
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Be sure to answer all parts. One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that is, it promotes the warming of Earth's atmosphere. Calculate the annual production of carbon dioxide in kilograms if there are exactly 40.0 million cars in the United States and each car covers a distance of 5790 mi at a consumption rate of 24.1 miles per gallon. Enter your answer in scientific notation. × 10 kg
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].
Consider the reaction below. The initial concentrations of PCl3 and Cl2 are each 0.0571 M, and the initial concentration of PCl5 is 0 M. If the equilibrium constant is Kc=0.021 under certain conditions, what is the equilibrium concentration (in molarity) of Cl2?PCl5(g)↽−−⇀PCl3(g)+Cl2(g)Remember to use correct significant figures in your answer (round your answer to the nearest thousandth). Do not include units in your response.
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.
Explanation: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.
if the rate of a particular reaction is 4 ties faster at 373K that it was at 323K what is the activation energy for the reaction
Answer:
27.77 kJ/mol is the activation energy for the reaction.
Explanation:
Rate of the reaction at 323 K =[tex]k_1=k[/tex]
Rate of the reaction at 373 K =[tex]k_2=4k[/tex]
Activation energy for the reaction is calculated by formula:
[tex]\log \frac{k_2}{k_1}=\frac{E_a}{2.303\times R}[\frac{T_2-T_1}{T_2\times T_1}][/tex]
[tex]E_a[/tex] = Activation energy
[tex]T_1[/tex] = Temperature when rate of the reaction was [tex]k_1[/tex]
[tex]T_2[/tex] = Temperature when rate of the reaction was [tex]k_2[/tex]
Substituting the values:
[tex]\log \frac{4k}{k}=\frac{E_a}{2.303\times 8.314 J /mol K}[\frac{373 K-323K}{373 K\times 323 K}][/tex]
[tex]E_a=27,776.98 J/mol=27.77 kJ/mol[/tex]
27.77 kJ/mol is the activation energy for the reaction.
Consider the following reaction: 2CH3OH(g) 2CH4(g) + O2(g) ΔH = +252.8 kJ a) Calculate the amount of heat transferred when 24.0 g of CH3OH(g) is decomposed by this reaction at constant pressure. b) For a given sample of CH3OH, the enthalpy change during the reaction is 82.1 kJ. How many grams of methane gas are produced?
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]
For a: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.
For b: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.
Explanation:a) Calculating the amount of heat transferred: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.
b) Calculating the grams of methane gas produced: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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40.0 mL of 1.00 M KOH(aq) and 40.0 mL of 0.500 M H2SO4(aq) are mixed in a coffee-cup calorimeter, both at 21.00°C. The final solution has a density of 1.02 g/mL, a volume of 80.0 mL, a specific heat of 4.00 J g-1 °C -1 , and a temperature of 27.85°C. Calculate the enthalpy change of this reaction per mole of H2O formed.
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.
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 75% of the lead to decay? (Round your answer to two decimal places.)
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]
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: There are now 2.07 moles of gas in the flask.
Explanation:
[tex]PV=nRT[/tex]
P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = ?
n = number of moles = 1.9
T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
[tex]V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L[/tex]
When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.
[tex]PV=nRT[/tex]
P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = 49.8 L
n = number of moles = ?
T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
[tex]n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles[/tex]
Thus the now the container contains 2.07 moles.
To find the new amount of moles after adding gas to the flask, the combined gas law was used, leading to the calculation that there are now 2.126 moles of gas in the flask.
The subject of the student's question is Chemistry, and it involves applying the ideal gas law to determine the amount of gas in moles after a change in pressure and temperature. To solve the problem, we can use the combined gas law which is derived from the ideal gas law: PV = nRT. Considering that the volume and the R (ideal gas constant) do not change, we can compare the two states of the gas with the formula P1/T1 = P2/T2. Here, pressure is given in mm Hg and temperature in degrees Celsius, which we need to convert to Kelvin by adding 273.15.
First state (before adding more gas):
P1 = 697 mm Hg
T1 = 21°C + 273.15 = 294.15 K
n1 = 1.9 moles
Second state (after adding more gas):
P2 = 775 mm Hg
T2 = 26°C + 273.15 = 299.15 K
The number of moles in the second state (n2) is what we want to find. Using the combined gas law:
P1 / T1 = P2 / T2 × (n2 / n1)
n2 = P1 × T2 / (P2 × T1) × n1
n2 = (697 mm Hg × 299.15 K) / (775 mm Hg × 294.15 K) × 1.9 mol
n2 = 2.126 mol
There are now 2.126 moles of gas in the flask.
A solution is prepared by dissolving 42.0 g of glycerin, C3H8O3, in 186 g of water with a final volume of 200.0 mL. a. Calculate the molarity and molality of the solution. b. What would be the molarity if 300.0 mL of water was added to the solution?
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:
For a: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]
To calculate the molarity of solution, we use the equation:[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.
For b: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
Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can form from 4 moles of hydrogen and excess oxygen. 2H2(g)+O2(g)→2H2O(l) Which of the following shows calculations for a correct way to solve this problem? View Available Hint(s) Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can form from 4 moles of hydrogen and excess oxygen. Which of the following shows calculations for a correct way to solve this problem? 4 mol H2×2 mol H2O2 mol H2=4 mol H2O 4 mol H2×2 mol H2O1 mol O2=8 mol H2O 2 mol H2×2 mol H22 mol H2O=2 mol H2O
Answer: The correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]
Explanation:
We are given:
Moles of hydrogen gas = 4 moles
As, oxygen is given in excess. Thus, is considered as an excess reagent and hydrogen is considered as a limiting reagent because it limits the formation of products.
For the given chemical equation:
[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex]
By Stoichiometry of the reaction:
2 moles of hydrogen produces 2 moles of water molecule.
So, 4 moles of hydrogen will produce = [tex]\frac{2molH_2O}{2molH_2}\times 4molH_2=4mol[/tex] of water.
Hence, the correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]
Which of the following pairs is mismatched? Select one: a. synthesis reaction - two reactants combine to form a larger product b. decomposition reaction - large reactant broken into smaller products c. oxidation - gain of electrons d. dehydration reaction - water is a product of the reaction e. hydrolysis - water is used in decomposition reaction
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.
Explanation: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:
Synthesis reaction does indeed combine smaller reactants to form a larger product.The decomposition reaction involves the breakdown of a larger reactant into smaller products.Dehydration reaction does yield water as a product.In a hydrolysis reaction, water is indeed consumed in the process of breaking down a compound.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.
Learn more about Oxidation here:https://brainly.com/question/15578795
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