Explanation:
Relation between heat energy and specific heat is as follows.
[tex]q_{bomb} = c \times \Delta t[/tex]
or, c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]
where, c = specific heat
[tex]q_{bomb}[/tex] = heat energy
[tex]\Delta t[/tex] = change in temperature
Putting the given values into the above formula we will calculate the specific heat as follows.
c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]
= [tex]\frac{5.23 kJ}{7.33^{o}C}[/tex]
= 0.713 [tex]kJ^{o}C[/tex]
Thus, we can conclude that heat capacity C of the calorimeter is 0.713 [tex]kJ^{o}C[/tex].
Final answer:
The heat capacity C of the calorimeter is calculated using the formula C = q / ΔT, with the given values resulting in a C of 0.7138 kJ/°C.
Explanation:
The heat capacity C of the calorimeter can be calculated using the amount of heat energy released and the resultant temperature change. The formula to calculate heat capacity is C = q / ΔT, where q is the heat energy released and ΔT is the change in temperature.
In this case, 5.23 kJ of heat energy was released, causing a temperature rise of 7.33 °C. Therefore, the heat capacity of the calorimeter is calculated as follows:
C = 5.23 kJ / 7.33 °C = 0.7138 kJ/°C
It's crucial to note that the heat capacity is commonly expressed in kJ/°C for bomb calorimeters, reflecting the amount of energy required to raise the temperature of the entire calorimeter setup by one degree Celsius.
In fractional distillation, liquid can be seen running from the bottom of the distillation column back into the distilling flask. What effect does this returning condensate have on the fractional distillation?
Answer:
substances with a higher boiling point are returning back to the flask which allows another substances with the specific context temperature (lower boiling point) to boil over and be purified.
Explanation:
The reason it happens because the lower boiling point substance vaporizes and crosses over while the other substance is waiting for its boiling point to reach
In fractional distillation, the returning condensate from the bottom of the distillation column to the flask enhances the efficiency of the process. The returned condensate serves as a mixing agent, increasing temperature gradients and refining the separation of components. More 'reflux' or returned condensate means more stages of distillation and better separation.
Explanation:In the process of fractional distillation, when condensate returns to the distilling flask from the bottom of the distillation column, it serves as a mixing agent. It enhances the efficiency of the fractional distillation process. This returning condensate mixes with the rising vapor which leads to a thorough exchange of heat. This increases the temperature gradient in the column, making the distillation more effective in separating the chemical components according to their boiling points. Each 'reflux' or return of condensate causes more 'theoretical plates' or stages of distillation, refining the separation.
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Which one of the following choices describes most accurately the actual, internal reaction temperature (in other words, the temperature of the reaction mixture inside the reaction vial) for the Fischer esterification experiment of 1-butanol with acetic acid to form n-butyl acetate? Select one, and explain your answer.
a) Sand bath temperature (160-180 °C)
b) Boiling point of 1-butanol (116-118 °C)
c) Boiling point of the reaction mixture (reflux temperature)
d) Boiling point of acetic acid (117 °C)
e) Boiling point of n-butyl acetate (124-126 °C)
Answer:
c) Boiling point of the reaction mixture (reflux temperature)
Explanation:
Hello,
At first, it is important to remember that esterification is an organic chemical reaction related with the neutralization of organic acids and alcohols to form esters, in this case from 1-butanol and acetic acid to n-butyl acetate as shown below:
[tex]CH_3COOH+CH3(CH_2)2CH_2OH \rightleftharpoons CH_3COOCH_2(CH_2)2CH3+H_2O[/tex]
It is shown that is a reaction which equilibrium condition is present since the n-butyl acetate is likely to come back to the acetic acid and the 1-butanol. Moreover, it is necessary to catalyze esterification with sulfuric acid and including constant heating and stirring, nonetheless, such heating induces boiling of the reacting mixture containing the acetic acid and the 1-butanol which are likely to boil. Therefore, reflux must be implemented as shown on the attached picture to prevent reactant lost which shift the reaction leftwards, diminishing n-butyl acetate yield, thus, the most accurately way to describe the actual temperature is c) boiling point of the reaction mixture (reflux temperature) since acetic acid and 1-butanol have a composition which modifies their boiling point into an only one that is the mixture's boiling point which is also related with the temperature at which the reflux is performed.
Best regards.
The option that best describes most accurately the actual, internal reaction temperature is Boiling point of the reaction mixture (reflux temperature).
What is Reflux Reactions about?It is said to be a very hard task when one is trying to monitor and control the temperature of a reaction chamber when a person do not have expensive and well equipped laboratory tools.
People often uses phase when the above is not in place as it is Phase said to be a form of melting or boiling occur at particular temperatures, and all heat exchanged are said to be done in a phase transition goes into the phase transition.
The main reason of refluxing a solution is done so as to heat a solution in a manner where one can control the outcome at a constant temperature and this is the option that is best for the Fischer esterification experiment of 1-butanol with acetic acid to create n-butyl acetate.
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At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re‑established?
Answer: The equilibrium concentration of NO after it is re-established is 0.55 M
Explanation:
For the given chemical equation:
[tex]N_2(g)+O_(g)\rightleftharpoons 2NO(g)[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[NO]^2}{[N_2][O_2]}[/tex] .....(1)
We are given:
[tex][NO]_{eq}=0.400M[/tex]
[tex][N_2]_{eq}=0.200M[/tex]
[tex][O_2]_{eq}=0.200M[/tex]
Putting values in expression 1, we get:
[tex]K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4[/tex]
Now, the concentration of NO is added and is made to 0.700 M
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
The equilibrium will shift in backward direction.
[tex]N_2(g)+O_(g)\rightleftharpoons 2NO(g)[/tex]
Initial: 0.200 0.200 0.700
At eqllm: 0.200+x 0.200+x 0.700-2x
Putting values in expression 1, we get:
[tex]4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075[/tex]
So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M
Hence, the equilibrium concentration of NO after it is re-established is 0.55 M
Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S (s) 3F 2(g) SF 6 (g) The maximum amount of SF 6 that can be produced from the reaction of 3.5 g of sulfur with of fluorine is ________ g.
Answer:
15.95 g
Explanation:
Calculation of the moles of sulfur as:-
Mass = 3.5 g
Molar mass of sulfur = 32.065 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{3.5\ g}{32.065\ g/mol}[/tex]
[tex]Moles= 0.1092\ mol[/tex]
From the reaction,
[tex]S+3F_2\rightarrow SF_6[/tex]
1 mole of sulfur on reaction forms 1 mole of sulfur hexafluoride
0.1092 mole of sulfur on reaction forms 0.1092 mole of sulfur hexafluoride
Molar mass of sulfur hexafluoride = 146.06 g/mol
Mass= Moles*Molar mass = 0.1092*146.06 g = 15.95 g
15.95 g is the maximum amount of [tex]SF_6[/tex] that can be produced from the reaction of 3.5 g of sulfur with of fluorine.
The maximum amount of SF6 that can be produced from the reaction of 3.5 g of sulfur with fluorine is 47.7 g.
The question is asking about the maximum amount of sulfur hexafluoride (SF6) produced from the reaction of 3.5 g of sulfur with an unknown amount of fluorine. To determine the maximum amount, we need to calculate the limiting reactant and use its stoichiometry to find the amount of SF6 produced. The balanced equation for the reaction is:
S (s) + 3F2 (g) → SF6 (g)
First, we need to find the molar mass of sulfur (S) and calculate the moles of sulfur:
Molar mass of sulfur (S) = 32.06 g/mol
Moles of sulfur = 3.5 g / 32.06 g/mol = 0.109 mol
Next, we need to find the molar mass of fluorine (F2) and calculate the moles of fluorine:
Molar mass of fluorine (F2) = 38.00 g/mol
Using the stoichiometry of the balanced equation, we can see that 1 mole of sulfur reacts with 3 moles of fluorine to produce 1 mole of SF6. Therefore, the moles of fluorine needed to react with 0.109 mol of sulfur is:
Moles of fluorine = 3 moles of fluorine/mol of sulfur × 0.109 mol of sulfur = 0.327 mol
Finally, we can use the moles of fluorine and the molar mass of SF6 to calculate the mass of SF6 produced:
Molar mass of SF6 = 146.06 g/mol
Mass of SF6 produced = moles of SF6 × molar mass of SF6 = 0.327 mol × 146.06 g/mol = 47.7 g (rounded to two decimal places)
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Which of the pairs below would be the best choice for a pH 5 buffer? HF/NaF, K a (HF) = 3.5 × 10-4 HC2H3O2/KC2H3O2, K a (HC2H3O2) = 1.8 × 10-5 NH3/NH4Cl, K b (NH3) = 1.8 × 10-5
Answer:
HC₂H₃O₂/KC₂H₃O₂
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the basic buffer solution as:
[tex] pH=pK_b+log\frac{[salt]}{[acid]} [/tex]
For a best pair, the pKa value must be equal to pH.
NH₃/NH₄Cl forms a basic buffer and cannot account for pH = 5
out of the acidic buffer given,
So, HF , Ka = 3.5 × 10⁻⁴ , So pKa = 3.46
HC₂H₃O₂ , Ka = 1.8 × 10⁻⁵ , So pKa = 4.77
The best pair to show pH = 5 is HC₂H₃O₂/KC₂H₃O₂
The pair for the best choice for a pH 5 buffer is:
HC₂H₃O₂/KC₂H₃O₂
Henderson- Hasselbalch equation:The equation that is used for calculation of the pH of the basic buffer solution as:
pH= pkb + log [salt]/ [acid]
For a best pair, the pKa value must be equal to pH.
NH₃/NH₄Cl forms a basic buffer and cannot account for pH = 5
Out of the acidic buffer given,
So, HF , Ka = 3.5 × 10⁻⁴ , So pKa = 3.46
HC₂H₃O₂ , Ka = 1.8 × 10⁻⁵ , So pKa = 4.77
The best pair to show pH = 5 is HC₂H₃O₂/KC₂H₃O₂
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The reaction of a carboxylic acid and thionyl chloride produces an acid chloride plus the gases SO2 and HCl. In the boxes, draw the mechanism arrows for the reaction. Be sure to add lone pars of electrons and nonzero formal charges on all species.
Answer:
[tex]S_{N}i[/tex] is the major step in forming acid chloride from carboxylic acid and thionyl chloride
Explanation:
In the first step, -OH group in carboxylic acid gives nucleophilic substitution reaction at S center in thionyl chloride and substitutes -Cl atomIn the second step, deprotonation takes place by chloride ion.In the third step, an intramolecular nucleophilic substitution reaction ([tex]S_{N}i[/tex]) takes place where bond electrons rearranges to produce [tex]SO_{2}[/tex], HCl and thionyl chloride.This rearrangement is highly favorable due to formation of gaseous species [tex]SO_{2}[/tex]Reaction mechanism has been shown below.In the organic combustion reaction of 41.9 g of octane (C8H18) with excess oxygen, what volume (in L) of carbon dioxide is produced if the reaction is performed at STP?
Answer:
The volume CO2 produced is 65.8 L
Explanation:
Step 1: Data given
Mass of octane = 41.9 grams
Molar mass octane = 114.23 g/mol
Step 2: The balanced equation
2C8H18 + 25O2 → 16CO2 + 18H2O
Step 3: Calculate moles octane
Moles octane = mass octane / molar mass octane
Moles octane = 41.9 grams / 114.23 g/mol
Moles octane = 0.367 moles
Step 4: Calculate moles CO2
For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O
For 0.367 moles octane we need 8*0.367 = 2.936 moles
Step 5: Calculate volume of CO2
1 mol = 22.4 L
2.936 moles = 22.4 * 2.936 = 65.8 L
The volume CO2 produced is 65.8 L
Carbon tetrachloride can be produced by this reaction: Suppose 1.1 mol and 3.3 mol are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains 0.82 mol . Calculate .
The question is incomplete, complete question is:
Carbon tetrachloride can be produced by this reaction:
[tex]CS_2(g) + 3Cl_2(g)[/tex] ⇌ [tex]S_2Cl_2(g) + CCl_4 (g)[/tex]
Suppose 1.1 mol and 3.3 mol are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains 0.82 mol .
Calculate [tex]K_c[/tex].
Answer:
The value of the [tex]K_c[/tex] of the reaction is 4.05.
Explanation:
[tex]Concentration=\frac{\text{Moles of solute}}{\text{Volume od solution}(L)}[/tex]
Initial concentration of [tex]CS_2[/tex]:
[tex][CS_2]=\frac{1.1 mol}{1 L}=1.1 M[/tex]
Initial concentration of [tex]Cl_2[/tex]:
[tex][Cl_2]=\frac{3.3mol}{1 L}=3.3M[/tex]
Equilibrium concentration of [tex]CCl_4[/tex]:
[tex][CCl_4]=\frac{0.82 mol}{1 L}=0.82 M[/tex]
[tex]CS_2(g) + 3Cl_2(g)[/tex] ⇌ [tex]S_2Cl_2(g) + CCl_4 (g)[/tex]
initially :
1.1 M 3.3 M 0 0
At equilibrium
(1.1-0.82) M (3.3-3 × 0.82) M 0.82 M 0.82 M
0.28 M 0.84 0.82 0.82
The expression of equilibrium constant [tex]K_c[/tex] is given by :
[tex]K_c=\frac{[S_2Cl_2][CCl_4]}{[CS_2][Cl_2]^3}[/tex]
[tex]=\frac{0.82 M\times 0.82 M}{0.28M\times (0.84 M)^3}=4.05[/tex]
The value of the [tex]K_c[/tex] of the reaction is 4.05.
Consider the reaction 2CO(g) + 2NO(g)2CO2(g) + N2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.00 moles of CO(g) react at standard conditions. S°surroundings = J/K Are You Sure? Please check your answer for mistakes. Submit Answer
Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of CO gas is reacted is 197.77 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]2CO(g)+2NO(g)\rightarrow 2CO_2(g)+N_2(g)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})+(1\times \Delta S^o_{(N_2(g))})]-[(2\times \Delta S^o_{(CO(g))})+(2\times \Delta S^o_{(NO(g))})][/tex]
We are given:
[tex]\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(N_2(g))}=191.61J/K.mol\\\Delta S^o_{(CO(g))}=197.67J/K.mol\\\Delta S^o_{(NO(g))}=210.76J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(2\times (213.74))+(1\times (191.61))]-[(2\times (197.67))+(2\times (210.76))]\\\\\Delta S^o_{rxn}=-197.77/K[/tex]
Entropy change of the surrounding = - (Entropy change of the system) = -(-197.77) J/K = 197.77 J/K
We are given:
Moles of CO gas reacted = 2.00 moles
By Stoichiometry of the reaction:
When 2 mole of CO gas is reacted, the entropy change of the surrounding will be 197.77 J/K
So, when 2.00 moles of CO gas is reacted, the entropy change of the surrounding will be = [tex]\frac{197.77}{2}\times 2.00=197.77J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of CO gas is reacted is 197.77 J/K
The entropy change for the surroundings when 2.00 moles of CO(g) react at standard conditions is -96.94 J/K.
To calculate the entropy change for the surroundings, we can use the following equation:
ΔSsurroundings = -ΔSsystem - ΔSuniverse
where ΔSsystem is the entropy change of the system and ΔSuniverse is the entropy change of the universe.
The entropy change of the system can be calculated from the standard molar entropies of the reactants and products:
ΔSsystem = ΣS°products - ΣS°reactants
The standard molar entropies of the reactants and products can be found in a standard thermodynamics data table. For the reaction given in the question, the standard molar entropies are as follows:
| Species | S° (J/mol·K) |
|---|---|---|
| CO(g) | 197.69 |
| NO(g) | 210.76 |
| CO2(g) | 213.64 |
| N2(g) | 191.50 |
Substituting these values into the equation for ΔSsystem, we get:
ΔSsystem = (2 mol × 213.64 J/mol·K) + (1 mol × 191.50 J/mol·K) - (2 mol × 197.69 J/mol·K) - (2 mol × 210.76 J/mol·K)
ΔSsystem = -96.94 J/K
The entropy change of the universe is always positive for a spontaneous process. Since the reaction given in the question is spontaneous, the entropy change of the universe is positive. Therefore, the entropy change for the surroundings is negative:
ΔSsurroundings = -ΔSsystem - ΔSuniverse
ΔSsurroundings = -(-96.94 J/K) - (+)
ΔSsurroundings = -96.94 J/K
Therefore, the entropy change for the surroundings when 2.00 moles of CO(g) react at standard conditions is -96.94 J/K.
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Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H 2C 2O 4), by mass, then a 0.6543-g sample of that solid will require ________ mL of 0.3483 M NaOH for neutralization. 11.19 97.78 28.59 1.119 22.39
Answer: The volume of NaOH required is 22.39 mL
Explanation:
We are given:
Mass of sample = 0.6543 g
Mass percent of oxalic acid = 53.66 %
This means that 53.66 grams of oxalic acid is present in 100 grams of sample
Mass of oxalic acid in the given amount of sample = [tex]\frac{53.66}{100}\times 0.6543=0.351g[/tex]
To calculate the number of moles, we use the equation:[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of oxalic acid = 0.351 g
Molar mass of oxalic acid = 90 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxalic acid}=\frac{0.351g}{90g/mol}=0.0039mol[/tex]
The chemical equation for the reaction of oxalic acid and NaOH follows:
[tex]C_2H_2O_4+2NaOH\rightarrow Na_2C_2O_4+2H_2O[/tex]
By Stoichiometry of the reaction:
1 mole of oxalic acid reacts with 2 moles of NaOH
So, 0.0039 moles of oxalic acid will react with = [tex]\frac{2}{1}\times 0.0039=0.0078mol[/tex] of NaOH
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Moles of NaOH = 0.0078 moles
Molarity of solution = 0.3483 M
Putting values in above equation, we get:
[tex]0.3483M=\frac{0.0078\times 1000}{V}\\\\V=\frac{0.0078\times 1000}{0.3483}=22.39mL[/tex]
Hence, the volume of NaOH required is 22.39 mL
To determine the volume of NaOH needed to neutralize a 0.6543-g sample of a solid containing 53.66% oxalic acid, we calculate the moles of oxalic acid in the sample and use stoichiometry to determine the moles of NaOH required for neutralization. Finally, we use the formula for molarity to find the volume of NaOH needed.
Explanation:Oxalic acid, H2C2O4, is a diprotic acid, meaning it can donate two protons (H+) in an acid-base reaction. To determine the volume of 0.3483 M NaOH needed for neutralization of a 0.6543-g sample of the solid containing 53.66% oxalic acid by mass, we need to first calculate the moles of oxalic acid in the sample.
First, we calculate the moles of H2C2O4 in the sample:
Moles of H2C2O4 = mass of H2C2O4 / molar mass of H2C2O4
Next, we use stoichiometry to calculate the moles of NaOH required for neutralization:
Moles of NaOH = 2 * moles of H2C2O4
Finally, we use the formula for molarity to determine the volume of NaOH needed:
The volume of NaOH (L) = moles of NaOH / molarity of NaOH
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A sample of 0.281 gg of an unknown monoprotic acid was dissolved in 25.0 mLmL of water and titrated with 0.0950 M NaOH NaOH. The acid required 30.0 mLmL of base to reach the equivalence point.What is the molar mass of the acid?
Answer:
98.6 g/mol.
Explanation:
Equation of the reaction
HX + NaOH--> NaX + H2O
Number of moles = molar concentration × volume
= 0.095 × 0.03
= 0.00285 moles
By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.
Molar mass = mass ÷ number of moles
= 0.281 ÷ 0.00285
= 98.6 g/mol.
A(g) + 2B(g) → C(g) + D(g)If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?
Answer: The value of [tex]K_p[/tex] for the reaction is 0.169
Explanation:
We are given:
Initial partial pressure of A = 1.00 atm
Initial partial pressure of B = 1.00 atm
The given chemical equation follows:
[tex]A(g)+2B(g)\rightleftharpoons C(g)+D(g)[/tex]
Initial: 1.00 1.00
At eqllm: 1-x 1-2x x x
We are given:
Equilibrium partial pressure of C = 0.211 atm = x
So, equilibrium partial pressure of A = (1.00 - x) = (1.00 - 0.211) = 0.789 atm
Equilibrium partial pressure of B = (1.00 - 2x) = (1.00 - 2(0.211)) = 0.578 atm
Equilibrium partial pressure of D = x = 0.211 atm
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{0.211\times 0.211}{0.789\times (0.578)^2}\\\\K_p=0.169[/tex]
Hence, the value of [tex]K_p[/tex] for the reaction is 0.169
What volume will o.128 g of propane, c3h8 occupy at a pressure golf 485 mm Hg and a temperature of 30.0 c
Answer:
113 mL
Explanation:
Let's apply the Ideal Gases Law for this propose:
P . V = n . R .T → (n. R . T) / P = V
We need to convert the T° to Absolute T° and pressure from mmHg to atm
T° K = 30°C + 273 = 303K
485 mmHg . 1atm / 760 mmHg = 0.638 atm
Let's replace the information obtained:
V = (n . 0.082 . 303K) / 0.638 atm
n = number of moles → 0.128 g . 1mol / 44g = 0.00291 moles
V = (0.00291 moles . 0.082 . 303K) / 0.638 atm → 0.113 L
The value can be written as 113 mL
Some CH2Cl2 is placed in a sealed flask and heated to 517 K. When equilibrium is reached, the flask is found to contain CH2Cl2 (3.42×10-2 M), CH4 (3.69×10-2 M), and CCl4 (4.12×10-2 M). What is the value of the equilibrium constant for this reaction at 517 K?
Answer: The value of equilibrium constant for the given reaction at 517 K is 1.30
Explanation:
The chemical equation for the dissociation of [tex]CH_2Cl_2[/tex] follows:
[tex]2CH_2Cl_2(g)\rightleftharpoons CH_4(g)+CCl_4(g)[/tex]
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=\frac{[CH_4][CCl_4]}{[CH_2Cl_2]^2}[/tex]
We are given:
[tex][CH_4]_{eq}=3.69\times 10^{-2}M[/tex]
[tex][CCl_4]_{eq}=4.12\times 10^{-2}M[/tex]
[tex][CH_2Cl_2]_{eq}=3.42\times 10^{-2}M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(3.69\times 10^{-2})\times (4.12\times 10^{-2})}{(3.42\times 10^{-2})^2}\\\\K_{eq}=1.30[/tex]
Hence, the value of equilibrium constant for the given reaction at 517 K is 1.30
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons are transferred in the reaction
Answer:
In the above reaction, the oxidation state of tin changes from 2+ to 4+.
10 moles of electrons are transferred in the reaction
Explanation:
Redox reaction is:
2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻
SnO₂²⁻ → SnO₃²⁻
Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.
SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation
BrO₃⁻ → Br₂
First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.
6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ Reduction
In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton
(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5
(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1
5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O
We can cancel the e⁻ and we substract:
12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)
6H₂O - 5H₂O = H₂O (on the left side)
2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻
Enter the complete ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing
Answer: The complete ionic equation is written below.
Explanation:
Complete ionic equation is defined as the equation in which all the substances that are strong electrolyte are present in an aqueous are represented in the form of ions.
The balanced molecular equation for the reaction of lead (II) nitrate and potassium sulfate follows:
[tex]Pb(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+PbSO_4(s)[/tex]
The complete ionic equation for the above equation is:
[tex]2Pb^{2+}(aq.)+2NO_3^{-}(aq.)+2K^{+}(aq.)+SO_4^{2-}(aq.)\rightarrow 2K^+(aq.)+2NO^{3-}(aq.)+PbSO_4(s)[/tex]
Hence, the complete ionic equation is written above.
Draw a resonance structure, complete with all formal charges and lone (unshared) electron pairs, that shows the resonance interaction of the carboxy with the para position in benzoic acid.
Answer:see the picture attached
Explanation:
Final answer:
Resonance structures for benzoic acid highlight electron delocalization, particularly the interaction between the carboxyl group and the para position on the benzene ring. Valence electrons are shown to resonate, creating multiple valid forms which collectively define the resonance hybrid of the molecule.
Explanation:
Resonance Structures of Benzoic Acid
When drawing resonance structures for benzoic acid, especially illustrating the resonance interaction with the para position (position opposite to the carboxyl group), we focus on the delocalization of π-electrons within the benzene ring and the adjacent carboxyl group. For the carboxyl group, one of the oxygen atoms will have a double bond with the carbon to fulfill the octet rule. However, due to the equivalent nature of the oxygen atoms, the double bond can resonate between the two oxygens, creating two resonance forms. At the para position, another resonance form is created by the movement of π-electrons from the benzene ring towards the carboxyl group, thus extending the conjugation and delocalization of electrons throughout the molecule.
Each resonance structure will be carefully drawn ensuring that all atoms obey the octet rule and formal charges are correctly assigned. It's important to note that the actual molecule is better represented by a resonance hybrid, which is an average of all valid resonance forms. This concept signifies that electrons are not strictly localized in one structure, but are distributed across the molecule in a delocalized π-electron cloud.
The air in a bicycle tire is bubbled through water and collected at 25 ∘C. If the total volume of gas collected is 5.65 L at a temperature of 25 ∘C and a pressure of 765 torr , how many moles of gas was in the bicycle tire?
Final answer:
To find the number of moles of gas in the bicycle tire, we can use the ideal gas law equation: PV = nRT. Given the pressure and volume of the gas, we can solve for n using the equation n = PV / RT. The number of moles of gas in the bicycle tire is approximately 0.185 moles.
Explanation:
To find the number of moles of gas in the bicycle tire, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas in atmV is the volume of the gas in Ln is the number of moles of gasR is the ideal gas constant (0.0821 L·atm/mol·K)T is the temperature of the gas in KelvinGiven that the pressure is 765 torr (which is equivalent to 1.01 atm) and the volume is 5.65 L, we can rearrange the equation to solve for n:
n = PV / RT
Substituting the given values, we get:
n = (1.01 atm) x (5.65 L) / (0.0821 L·atm/mol·K x (25 + 273) K)
Simplifying, we find that the number of moles of gas in the bicycle tire is approximately 0.185 moles.
Liquid methanol is fed to a space heater at a rate of 12.0 L/h and burned with excess air. The product gas is analyzed and the following dry-basis mole percentages are determined: CH3OH = 0.45%, CO2 = 9.03%, and CO = 1.81%. (a) After drawing and labeling a flowchart, verify that the system has zero degrees of freedom. (b) Calculate the fractional conversion of methanol, the percentage excess air fed, and the mole fraction of water in the product gas. (c) Suppose the combustion products are released directly into a room. What potential problems do you see and what remedies can you suggest?
Answer:
(a) The bellow flow chart shows that the system has 0° of freedom.
(b) i - Fractional conversion of menthol: 0.960 mol CH₃OH reacted/mol fed
ii - The percentage excess air fed: 28.5%
iii - molecular fraction of water in the product gas: 0.178 mol H₂O/mol
(c) Potential Problems: Remedies
Conflagration: The gas should be vented outside to prevent fire outbreak.
Toxicity: Put a CO detection alarm in the room.
Explanation:
See picture for the flow chat and calculation of other answers.
A space heater fed with liquid methanol and burned with excess air is analyzed to determine its fractional conversion, percentage excess air fed, and mole fraction of water. The system has zero degrees of freedom. Potential problems when the combustion products are released directly into a room include the release of harmful gases and increased humidity.
Explanation:To determine if a system has zero degrees of freedom, we need to analyze the material balance and the component balance. In this case, the flow chart shows that the only input is the liquid methanol and the only outputs are the product gases. Since there are no unknown variables or degrees of freedom in the system, we can conclude that the system has zero degrees of freedom.
To calculate the fractional conversion of methanol, we can use the mole percentages of CO2 and CO in the product gas. The fractional conversion is the difference between the initial mole percentage of methanol and the mole percentage of CO and CO2. The percentage excess air fed can be calculated by comparing the mole percentage of O2 in the product gas to the stoichiometric requirement. Finally, the mole fraction of water in the product gas can be found by subtracting the sum of the mole percentages of other components from 100%.
When the combustion products are released directly into a room, potential problems include the release of harmful gases such as CO and the increase in humidity due to the presence of water vapor. To remedy these problems, it is important to ensure proper ventilation and monitoring of indoor air quality. It is also advisable to use a flue or chimney to exhaust the combustion products safely.
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A water sample has a pH of 8.2 and a bicarbonate concentration of 97 mg/L. What is the alkalinity of the sample in moles/L and in mg CaCO3/L?
Answer:6.94
Explanation:
Molar mass of CaCO3=40+12+16×3
=40+12+48=100g/mol
Moles=mass of substance/molar mass
=97mg/100g=0.097/100=0.00097moles/L.
PH=-log[CaCo3]=-log(0.00097)=6.94
P.s it's log to base e
Final answer:
The alkalinity of the water sample with a pH of 8.2 and bicarbonate concentration of 97 mg/L is 0.00159 moles/L when calculated as bicarbonate, and 159.14 mg CaCO3/L when expressed as equivalent calcium carbonate.
Explanation:
A student has asked about calculating the alkalinity of a water sample with a pH of 8.2 and a bicarbonate concentration of 97 mg/L, both in moles/L and in mg CaCO3/L. To find the alkalinity in moles/L, we need to consider that the molecular weight of HCO3- (bicarbonate) is approximately 61.01 g/mol. Therefore, 97 mg/L can be converted to moles/L by dividing by the molecular weight:
Alkalinity (as HCO3-) = 97 mg/L / (61.01 g/mol * 1000 mg/g) = 0.00159 mol/L.
To convert this alkalinity to mg CaCO3/L, we use the equivalent weight of CaCO3 (100.087 g/mol) since 1 mol of HCO3- is chemically equivalent to 1 mol of CaCO3 for neutralization purposes:
Alkalinity (as CaCO3) = 0.00159 mol/L * 100.087 g/mol * 1000 mg/g = 159.14 mg CaCO3/L.
Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.
Answer: The equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
Explanation:
We are given:
Molarity of oxalic acid solution = 0.20 M
Oxalic acid [tex](H_2C_2O_4)[/tex] is a weak acid and will dissociate 2 hydrogen ions.
The chemical equation for the first dissociation of oxalic acid follows:[tex]H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)[/tex]
Initial: 0.20
At eqllm: 0.20-x x x
The expression of first equilibrium constant equation follows:
[tex]Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}[/tex]
We know that:
[tex]Ka_1\text{ for }H_2C_2O_4=0.059[/tex]
Putting values in above equation, we get:
[tex]0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = x = 0.083 M
The chemical equation for the second dissociation of oxalic acid:[tex]HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)[/tex]
Initial: 0.083
At eqllm: 0.083-y 0.083+y y
The expression of second equilibrium constant equation follows:
[tex]Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}[/tex]
We know that:
[tex]Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}[/tex]
Putting values in above equation, we get:
[tex]6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = y = 0.0000639 M
Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M
Hence, the equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
The equilibrium concentration of H3O+ in a solution of oxalic acid is linked to the acid dissociation constant Ka. Calculation of this requires specific values related to the nature of oxalic acid, a weak acid, and its behaviour in solution.
Explanation:In order to find the equilibrium concentration of H3O+ in a solution of oxalic acid, we must first understand that oxalic acid is a weak acid, and thus does not fully ionize in solution. In the ionization process of oxalic acid H2C2O4, it would donate a proton (H+) to water (H2O), forming a hydronium ion (H3O+) and a mono hydrogen oxalate ion. The equilibrium concentration of H3O+ (hydronium ions) thus corresponds to the acid dissociation constant Ka, with the equation Ka=[H3O+][HC2O4-]/[H2C2O4]. Considering the initial concentration of the oxalic acid, specific calculations will be required to find an accurate equilibrium concentration which is not possible without the value of Ka. So, the specific answer depends on the Ka value of oxalic acid.
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Thermal energy is Question 2 options: A) solar energy, i.e. energy that comes from the sun. B) the energy stored within the structural units of chemical substances. C) energy available by virtue of an object's position. D) the energy associated with the random motion of atoms and molecules.
Answer:
D) the energy associated with the random motion of atoms and molecules.
Explanation:
Thermal energy is the manifestation of energy in the form of heat. In all materials, the atoms that make up their molecules are in continuous movement, either moving or vibrating, which implies that the atoms have a certain kinetic energy that we call heat or thermal energy. In a way, thermal energy is the internal energy of a body.
Final answer:
Thermal energy is the energy associated with the random motion of atoms and molecules, a type of kinetic energy that increases with object's temperature, making Option D the correct choice.
Explanation:
The question poses multiple options for defining thermal energy. Option A suggests solar energy, which isn't correct as solar energy is a form of radiant energy. Option B refers to the energy within chemical substances, known as chemical energy. Option C discusses energy by virtue of an object's position, commonly known as potential energy. The correct answer is Option D, which states that thermal energy is the energy associated with the random motion of atoms and molecules.
More specifically, thermal energy is related to the kinetic energy resulting from the random translational motions of particles and also includes vibrational and rotational energies within molecules. An increase in this motion, as measured by temperature, corresponds to an increase in thermal energy. Hence, thermal energy is a form of internal energy of an object, manifesting as the microscopic motion of its constituent particles.
Progesterone is a hormone that contains two ketone groups. The oxygen in the ketone group can function as a hydrogen bond acceptor. Select the amino acids that have side chains that can form a hydrogen bond with progesterone at pH 7.
a. threonine.
b. cysteine.
c. alanine.
d. aspartate.
e. arginine.
f. tryptophan.
Amino acids that can donate hydrogen bonds include -
a. threoninee. argininef. tryptophanWith its -NH group, Tryptophan can serve as a hydrogen-bond donor, but its aromatic ring can also serve as an acceptor.
The side chains of three amino acids—arginine, lysine, and tryptophan—contain hydrogen donor atoms.
The side chains of 2 amino acids (aspartic acid and glutamic acid) include hydrogen acceptor atoms.
The side chains of six amino acids—asparagine, glutamine, histidine, serine, threonine, and tyrosine—contain both hydrogen donor and acceptor atoms.
The side chain of threonine has the capacity to serve as a hydrogen bond donor and acceptor. An oxygen atom that is part of the side chain of the amino acid threonine has two possible hydrogen bonding roles: acceptor and donor.
Therefore, from the given list of amino acids, three amino acids can act as hydrogen donor, these include – a. threonine, e. arginine, f. tryptophan.
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At pH 7, the amino acids that can form a hydrogen bond with progesterone are threonine, aspartate, and arginine. Threonine has a polar uncharged group while aspartate and arginine carry a charge, allowing them to participate in hydrogen bonding.
Explanation:
The amino acids that have side chains capable of forming a hydrogen bond with progesterone at pH 7 include threonine, aspartate, and arginine. These amino acids have side chains with polar, uncharged groups or charged groups. To be specific, threonine has a polar, uncharged hydroxyl group which can act as a hydrogen donor or acceptor. Aspartate carries a negative charge at pH 7 and can act as a hydrogen bond acceptor.
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An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 700 kPa and 59°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.
Answer:
[tex]P_2=350\ kPa[/tex]
[tex]T_2=59^{\circ}\ C[/tex]
Explanation:
Given that
mass , m = 4 kg
Initial pressure ,[tex]P_1=700\ kPa[/tex]
Initial temperature ,[tex]T_1=59^{\circ}\ C[/tex]
The volume of rigid tanks are same
[tex]V_1=V[/tex]
[tex]V_2=2 V[/tex]
Let's take final temperature[tex] =T_2[/tex]
Given that tank is insulated that is why heat transfer in the tank will be zero.
By using energy balance
[tex]E_{in}-E_{out}=\Delta U[/tex]
[tex]\Delta U[/tex]= Change in the internal energy of the gas
[tex]0 = m C_V(T_2-T_1[/tex]) ( Cv=Specific heat capacity at constant volume)
[tex]0 = T_2-T_1[/tex]
Therefore [tex]T_1=T_2[/tex]
[tex]T_2=59^{\circ}\ C[/tex]
We know that ideal gas equation for gas
P V = m R T
P=pressure ,V=Volume ,m=mass ,R= gas constant ,T=temperature
By using mass conservation
[tex]m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}[/tex]
Now by putting the values in the above equation
[tex]\dfrac{700\times V}{RT_1}=\dfrac{P_2\times 2V}{RT_1}[/tex]
[tex]P_2=\dfrac{700}{2}\ kPa[/tex]
[tex]P_2=350\ kPa[/tex]
Therefore the final volume will be 350 kPa and temperature will be 59°C.
The final temperature in the tank is approximately 332.15K. When the volume doubles, the pressure is halved, yielding a final pressure of approximately 350 kPa.
Explanation:In this problem, your task is to determine the final temperature and pressure in a tank after an ideal gas is allowed to expand. Given the system in question is both insulated (adiabatic) and rigid, we can infer that neither heat (Q) nor work (W) is done, as indicated by the equation AEint=Q-W = 0.
Furthermore, as the internal energy does not change, this implies that the temperature will remain constant. So, the final temperature in the tank is the same as the initial, 59°C. We must convert this into Kelvin, as the equation of state of the ideal gas requires temperature to be in Kelvin. The conversion is as follows: T(K) = T(°C) + 273.15. Therefore, the final temperature is approximately 332.15K.
On the other hand, according to the ideal gas law parameters in this problem, when the volume doubles (since the partition is removed), the pressure is halved. This is reflected by the formula P = nRT/V, where 'n' is the amount of the gas, 'R' is the ideal gas constant, 'T' is the temperature and 'V' is the volume. Therefore, the final pressure is the initial pressure divided by 2, yielding a final pressure of approximately 350 kPa.
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A cylindrical specimen of some metal alloy having an elastic modulus of 129 GPa and an original cross-sectional diameter of 4.4 mm will experience only elastic deformation when a tensile load of 1570 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm.
Explanation:
The cross-sectional area of the specimen is calculated as follows.
[tex]A_{o} = \frac{pi}{4} d^{2}[/tex]
= [tex]\frac{3.14}{4} \times (\frac{4.4}{1000})^{2}[/tex]
= [tex]1.5197 \times 10^{-5} m^{2}[/tex]
Equation of stress is as follows.
[tex]\sigma = \frac{F}{A_{o}}[/tex]
And, the equation of strain is as follows.
[tex]\epsilon = \frac{\Delta l}{l_{o}}[/tex]
Hence, the Hook's law is as follows.
E = [tex]\frac{\sigma}{\epsilon}[/tex]
E = [tex]\frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}[/tex]
= [tex]\frac{F \times l_{o}}{A_{o} \times \Delta l}[/tex]
or, [tex]l_{o} = \frac{E \times \Delta l \times A_{o}}{F}[/tex]
= [tex]\frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}[/tex]
= 0.6554 m
or, [tex]l_{o}[/tex] = 655.4 mm
Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.
Please, Please help
A flask containing 550 mL of 0.75 M H2SO4 was accidentally knocked to the floor.
How many grams of NaHCO3 do you need to put on the spill to neutralize the acid according to the following equation?
H2SO4(aq)+2NaHCO3(aq)→Na2SO4(aq)+2H2O(l)+2CO2(g)
Express your answer using two significant figures.
m= g
Answer:
We need 69 grams of NaHCO3
Explanation:
Step 1: Data given
Volume = 550 mL = 0.550 L
Molarity H2SO4 = 0.75 M
Step 2: The balanced equation
H2SO4(aq) + 2NaHCO3(aq) → Na2SO4(aq) + 2H2O(l) + 2CO2(g)
Step 3: Calculate moles of H2SO4
Moles H2SO4 = molarity * volume
Moles H2SO4 = 0.75 M * 0.550 L
Moles H2SO4 = 0.4125 moles H2SO4
Step 4: Calculate moles NaHCO3
For 1 mol H2SO4 we need 2 moles NaHCO3 to produce 1 mol Na2SO4 and 2 moles H2O and 2 Moles CO2
For 0.4125 moles H2SO4 we need 2*0.4125 = 0.825 moles NaHCO3
Step 5: Calculate mass NaHCO3
Mass NaHCO3 = moles * molar mass
Mass NaHCO3 = 0.825 moles * 84.0 g/mol
Mass NaHCO3 = 69.3 grams ≈ 69 grams
We need 69 grams of NaHCO3
At constant temperature and volume, a sample of oxygen gas is added to a sample of nitrogen gas. The pressure of the mixture is found by adding the pressures of the two individual gases. This is an example of:
(A) Boyle's Law
(B) Charles's Law
(C) Avogadro's Law
(D) Dalton's Law
D. Dalton's Law
Explanation:
As the pressure of the gas is related to the sum of the partial pressures of the individual gases present in the mixture was explained by Dalton's law, the given system is an example of Dalton's law.
Boyle's law relates the inverse proportionality of volume and pressure of an ideal gas.
Charles's Law reveals the direct relationship of temperature and volume of an ideal gas.
Avogadro's Law states the relationship between the volume of gas and number of molecules at same pressure as well as temperature.
(D) Dalton's Law states that the pressure of the mixture is found by adding the pressures of the two individual gases.
Dalton’s Law also known as the Law of Partial Pressures states that in a mixture of non-reacting gases the total pressure exerted is equal to the sum of the partial pressures of the gases in the mixture.
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If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, how many moles of Pb2+ were originally in the solution?
Answer:
There were 0.00735 moles Pb^2+ in the solution
Explanation:
Step 1: Data given
Volume of the KI solution = 73.5 mL = 0.0735 L
Molarity of the KI solution = 0.200 M
Step 2: The balanced equation
2KI + Pb2+ → PbI2 + 2K+
Step 3: Calculate moles KI
moles = Molarity * volume
moles KI = 0.200M * 0.0735L = 0.0147 moles KI
Ste p 4: Calculate moles Pb^2+
For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+
For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+
There were 0.00735 moles Pb^2+ in the solution
Calculate the volume in liters of a 0.0015/molL calcium sulfate solution that contains 25.0g of calcium sulfate CaSO4 . Be sure your answer has the correct number of significant digits.
Answer:
The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.
Explanation:
Molarity (M) is the number of moles of solute that are dissolved in a given volume. Molarity is expressed by:
[tex]Molarity=\frac{number of moles of solute}{Dissolution volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]
In this case you have 25.0g of calcium sulfate CaSO₄. First of all you need to know the amount of moles that mass represents. For that you must first know the atomic masses of each element:
Ca: 40 g/molS: 32 g/molO: 16 g/molThen the molar mass of the compound calcium sulfate is:
molar mass= 40 g/mol + 32 g/mol + 4*16 g/mol= 136 g/mol
It is then possible to apply a rule of three as follows: if 136 g represents 1 mol of the compound calcium sulfate, 25 g how many moles are they?
[tex]moles=\frac{25 g*1 mole}{136 g}[/tex]
moles≅0.1838
Now you can apply a rule of three knowing the molarity of 0.0015 [tex]\frac{moles}{liters}[/tex]: if 0.0015 moles represents 1 liter of solution, 0.1838 moles how many liters are they?
[tex]volume=\frac{0.1838moles*1 liter}{0.0015moles}[/tex]
volume=122.53 liters
The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.
To find the volume of a 0.0015 mol/L calcium sulfate solution that contains 25.0g of calcium sulfate, first convert the mass to moles, then use the molarity to find the volume. The approximate volume is 122L.
Explanation:To calculate the volume of the solution, we first need to convert the mass of calcium sulfate in grams to moles. We do this by dividing by the molar mass of calcium sulfate, which is about 136.14 g/mol.
25.0g CaSO4 * (1 mol / 136.14 g) = approx 0.183 mol CaSO4
Then, we use the molarity of the solution to find the volume. Remember that the definition of molarity is moles of solute per liter of solution.
Volume = moles / molarity = 0.183 mol / 0.0015 mol/L = approx 122 L
So, the volume of the calcium sulfate solution is approximately 122 liters.
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Predict the type of bond that would be formed between each of the following pairs of atoms(ionic, polar covalent or nonpolar covalent)
a. H and Cl
b. Mg and F
c. Li and N
d. N and S
Answer:
In between H and Cl the bond will be covalent.
In between Mg and F the bond will be ionic.
In between Li and N the bond will be ionic .
In between N and S the bond will be polar covalent.
Explanation:
Ionic bonds are defined as the bonds which are formed by the complete transfer of electrons from cation (positively charged ions) to anion (negatively charged ions). For Example: NaCl, [tex]MgF_2[/tex] etc.
Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. For Example: HCl, [tex]CH_4[/tex] etc.
Its of two type:
Polar covalent compound: This compound is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. For Example: [tex]CO_2,NO_2[/tex] etc..Non-polar covalent compound: This compound is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example: [tex]N_2,O_2[/tex] etc.In between H and Cl the bond will be covalent.
In between Mg and F the bond will be ionic.
In between Li and N the bond will be ionic .
In between N and S the bond will be polar covalent.
The type of bond that would be formed between each of the following pairs of atoms will be:
H and Cl - Covalent bondMg and F - Ionic bondLi and N - Ionic bondN and S - Polar covalentIonic bonds are formed by the transfer of electrons from cation to anion. An example is NaCl.
Covalent bond is a bond that's formed by the sharing of electrons between atoms. e.g. HCl.
A polar covalent bond is formed when there is the presence of difference in electronegativity between the atoms.
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