The concentration of S in the 10.0 mL solution is calculated by considering the initial 8.48 mM concentration and the dilution factor due to the increase in volume to 10.0 mL. The final concentration of S is found to be 0.848 mM.
Explanation:To determine the concentration of S in the 10.0 mL solution, you must take into account the initial concentration of S and how it changes with dilution. Initially, 1.00 mL of 8.48 mM (millimolar) S was added to the solution. The dilution can be calculated using the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume after dilution.
The initial volume of S (V1) is 1.00 mL and the final volume of the diluted solution (V2) is 10.0 mL. Therefore, using the initial concentration of S (C1) as 8.48 mM, we can solve for the final concentration (C2) as follows:
C1V1 = C2V2
(8.48 mM)(1.00 mL) = (C2)(10.0 mL)
Solving for C2 gives us:
C2 = (8.48 mM)(1.00 mL) / (10.0 mL)
C2 = 0.848 mM
The final concentration of S in the 10.0 mL solution is 0.848 mM (millimolar).
n a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. 2A⟶3B How many molecules of substance B are produced when 27.6 g of substance A reacts? The molar mass of substance A is 27.8 g/mol. Step 1: Convert the mass of A to moles. moles of A: mol Step 2: Convert the number of moles of A to the number of moles of B. moles of B: mol Step 3: Convert the number of moles of B to molecules of B. molecules of B:
Answers:
Step 1: Convert the mass of A to moles. moles of A: 0.993 mol Step 2: Convert the number of moles of A to the number of moles of B. moles of B: 1.49 mol Step 3: Convert the number of moles of B to molecules of B. molecules of B: 8.97 × 10²³ moleculesExplanation:
1) Chemical equation (given):
2A⟶3B2) Follow the given steps:
Step 1: Convert the mass of A to moles.
moles of A:
n = mass in grams / molar massn = 27.6 g / 27.8 g/mol = 0.993 mol of AStep 2: Convert the number of moles of A to the number of moles of B.
Stoichiometric ratio:2 mol A : 3 mol B
Proportion:2 mol A / 3 mol B = 0.993 mol A / x
⇒ x = 0.993 mol A × 3 mol B / 2 mol A = 1.49 mol of B
Step 3: Convert the number of moles of B to molecules of B.
Avogadro's number: 6.022 × 10²³ molecules / mol1.49 mol × 6.022 × 10²³ molecules / mol = 8.97 × 10²³ molecules of B.To find the number of molecules of substance B produced from 27.6 g of substance A, calculate the moles of A, convert those moles to moles of B using the stoichiometry of the reaction, and then convert the moles of B to molecules using Avogadro's number. This results in 8.95 × 10²³ molecules of B.
To determine the number of molecules of substance B produced when 27.6 g of substance A reacts, we need to follow three steps:
Convert the mass of A to moles by dividing it by the molar mass of A.Use the balanced chemical equation to convert moles of A to moles of B.Convert the moles of B to molecules using Avogadro's number.Step 1: Moles of A = mass of A / molar mass of A = 27.6 g / 27.8 g/mol = 0.992 mol A
Step 2: According to the balanced equation 2A ⟶ 3B, the mole ratio is 2 moles of A : 3 moles of B. So, moles of B = moles of A × (3 moles B / 2 moles A) = 0.992 mol A × (3/2) = 1.488 mol B
Step 3: Molecules of B = moles of B × Avogadro's number = 1.488 mol B × 6.022 × 10²³ molecules/mol = 8.95 × 10²³ molecules of B.
Write an equation for the reaction of solid aluminum metal with hydrochloric acid (hydrogen monochloride) dissolved in water to form aluminum chloride dissolved in water and hydrogen gas. When balanced, what is the coefficient for the hydrochloric acid (hydrogen monochloride)?
Answer:
The coefficient for the hydrochloric acid is 6.
Explanation:
Stoichiometric coefficient are the number written in front of the reactants and products in balanced chemical reaction.
When solid aluminum reacts with hydrochloric acid to give aluminum chloride and hydrogen gas.
[tex]2Al(s)+6HCl(s)\rightarrow 2AlCl_3(aq)+3H_2(g)[/tex]
The coefficient for the aluminum is 2.
The coefficient for the hydrochloric acid is 6.
The coefficient for the aluminum chloride acid is 2.
The coefficient for the hydrogen gas is 3.
Before performing the titration of your unknown acid, your partner accidentally dilutes the sample by adding water. Consider how this error would effect the following aspects of the titration curve. Part A How would this error effect the volume of base necessary to reach the equivalence point?
Answer:
It would not affect the volume of the base needed to reach the equivalence point
Explanation:
Once the unknown acid has been prepared and placed in the conical flask ready for titration, adding water to it cannot alter how much base is needed to neutralise it. Adding water to the acid merely dilutes the acid i.e if 20 parts of acid were in say 100 parts of water now 20 parts of acid are in 300 parts of water because you added more water. You will still need the exact amount of base to neutralise the 20 particles that are still in the conical flask awaiting titration.
Adding water does not make acid particles increase or decrease, it just means the base particles will have to collide with a lot more water particles to interact with the acid particles which will slow down the reaction. So the rate of the reaction can be affected, not volume of base needed to reach equivalence point.
The reaction between potassium chlorate (KCIO,) and red phosphorus (P.) takes place when one strikes a match. The products of the reaction are tetraphosphorus decoxide and potassium chloride. If 56.0 grams of KCIO, are reacted with an excess amount of red phosphorus, how many grams of P0o and KCI can be produced? How much red phosphorus is consumed in the reaction? (15 pts) Write the balanced reaction first!
Answer: Amount of tetraphosphorus decoxide formed is 38.60g, amount of potassium chloride formed is 33.99g and amount of red phosphorus consumed in the reaction is 81.48 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For potassium chlorate:Given mass of potassium chlorate = 56.0 g
Molar mass of potassium chlorate = 122.55 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of potassium chlorate}=\frac{56.0g}{122.55g/mol}=0.456mol[/tex]
For the given chemical reaction:
[tex]10KClO_3+12P\rightarrow 3P_4O_{10}+10KCl[/tex]
Red phosphorus is given in excess . So, it is considered as an excess reagent and potassium chlorate is considered as a limiting reagent.
For tetraphosphorus decoxide:By Stoichiometry of the reaction:
10 moles of potassium chlorate reacts with 3 moles of tetraphosphorus decoxide
So, 0.456 moles of potassium chlorate will react with = [tex]\frac{3}{10}\times 0.456=0.136moles[/tex] of tetraphosphorus decoxide
Calculating the mass of tetraphosphorus decoxide by using equation 1, we get:
Molar mass of tetraphosphorus decoxide = 283.886 g/mol
Moles of tetraphosphorus decoxide = 0.136 moles
Putting values in equation 1, we get:
[tex]0.136mol=\frac{\text{Mass of tetraphosphorus decoxide}}{283.886g/mol}\\\\\text{Mass of tetraphosphorus decoxide}=38.60g[/tex]
For potassium chloride:By Stoichiometry of the reaction:
10 moles of potassium chlorate reacts with 10 moles of potassium chloride
So, 0.456 moles of potassium chlorate will react with = [tex]\frac{10}{10}\times 0.456=0.456moles[/tex] of potassium chloride
Calculating the mass of potassium chloride by using equation 1, we get:
Molar mass of potassium chloride = 74.55 g/mol
Moles of potassium chloride = 0.456 moles
Putting values in equation 1, we get:
[tex]0.456mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=33.99g[/tex]
For Red phosphorus (excess reagent)By Stoichiometry of the reaction:
10 moles of potassium chlorate reacts with 12 moles of red phosphorus.
So, 0.456 moles of potassium chlorate will react with = [tex]\frac{12}{10}\times 0.456=2.631moles[/tex] of red phosphorus
Calculating the mass of red phosphorus by using equation 1, we get:
Molar mass of red phosphorus = 30.97 g/mol
Moles of red phosphorus = 2.631 moles
Putting values in equation 1, we get:
[tex]2.631mol=\frac{\text{Mass of red phosphorus}}{30.97g/mol}\\\\\text{Mass of red phosphorus}=81.48g[/tex]
Hence, amount of tetraphosphorus decoxide formed is 38.60g, amount of potassium chloride formed is 33.99g and amount of red phosphorus consumed in the reaction is 81.48 g.
The combustion of ethane in the presence of excess oxygen yields carbon dioxide and water: 2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (l) The value of ΔS° for this reaction is ________ J/K⋅mol. The combustion of ethane in the presence of excess oxygen yields carbon dioxide and water: 2C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) The value of S° for this reaction is ________ J/Kmol. -151.0 +718.0 -620.1 +151.0 -718.0
Answer: The value of [tex]\Delta S^o[/tex] for the given reaction is -620.1 J/Kmol.
Explanation:
Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.
Mathematically,
[tex]\Delta S_{rxn}=\sum [n\times S^o_{products}]-\sum [n\times S^o_{reactants}][/tex]
For the given chemical equation:
[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)[/tex]
Taking the standard entropy of formation for the following:
[tex]S^o_{C_2H_6}=229.6Jmol^{-1}K^{-1}\\S^o_{CO_2}=213.6Jmol^{-1}K^{-1}\\S^o_{H_2O}=69.9Jmol^{-1}K^{-1}\\S^o_{O_2}=205Jmol^{-1}K^{-1}[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(4\times S^o_{CO_2})+(6\times S^o_{H_2O})]-[(2\times S^o_{C_2H_6})+(7\times S^o_{O_2})][/tex]
[tex]\Delta S^o=[(4\times 213.6)+(6\times 69.9)]-[(2\times 229.6)+(7\times 205)]=-620.1Jmol^{-1}K^{-1}[/tex]
Hence, the [tex]\Delta S^o[/tex] of the reaction is [tex]-620.1Jmol^{-1}K^{-1}[/tex]
The standard entropy change for the complete combustion of ethane is -620.8 J/K.mol.
What is standard entropy change?The standard entropy change is equal to the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants.
Let's consider the combustion of ethane.
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l)
The standard entropy change (ΔS°) is:
ΔS° = n(products) × S°products - n(reactants) × S°reactants
where,
n is the stoichiometric coefficient.S° is the standard entropy.ΔS° = 4 × 213.74 J/mol.K + 6 × 69.91 J/mol.K - 2 × 229.60 J/mol.K - 7 × 205.14 J/mol.K = -620.8 J/K.mol
The standard entropy change for the complete combustion of ethane is -620.8 J/K.mol.
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Rank the following atoms by number of valence electrons. Rank from most to fewest valence electrons. To rank items as equivalent, overlap them. Rb Xe Sb I In
To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.
Rb: [Kr] 5s¹ - 1 valence electron
Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons
Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons
I: [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons
In: [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons
Rank from most to fewest valence electrons:
Xe > I > Sb > In > Rb
Rank from most to fewest valence electrons:
Xe > I > Sb > In > Rb. Each atom in the bond contributes one valence electron.
A valence electron is a kind of electron that is part of an atom's outer shell in chemistry and physics. If the outer shell is open, the valence electron can take part in the creation of a chemical bond, forming a shared pair within a single covalent bond.We examine the electronic configuration then add the electrons from the electronic shells with the largest main quantum number to determine the amount of valence electrons.
Rb: [Kr] 5s¹ - 1 valence electron
Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons
Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons
I: [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons
In: [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons
Rank from most to fewest valence electrons:
Xe > I > Sb > In > Rb
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Express the equilibrium constant for the combustion of ethanol in the balanced chemical equation. C2H5OH(g)+3O2(g)⇌2CO2(g)+3H2O(g) Express the equilibrium constant for the combustion of ethanol in the balanced chemical equation. A) K=[C2H5OH][O2][CO2][H2O] B) K=[CO2]2[C2H5OH][O2]3 C) K=[CO2]2[H2O]3[C2H5OH][O2]3 D) K=[CO2][H2O][C2H5OH][O2]
Answer:
The correct answer is C where K=[CO2]^2[H2O]^3/[C2H5OH][O2]^3
Explanation:
Equilibrium constant (K) is expressed as the product of all the products divided by the product of all reactants. the coefficient in front of each reactants or products are used as their exponents. Solids and liquids are considered to be 1 hence they are not shown, but in this case everything is in gaseous form thus they are all expressed in the equilibrium formula formula.
Final answer:
The equilibrium constant for the combustion of ethanol is expressed using the balanced chemical equation, providing the correct formula for K from the given options.
Explanation:
Equilibrium constant (K) for the combustion of ethanol:
Write the balanced equation: C₂H₅OH(g) + 3O₂(g) ⇌ 2CO₂(g) + 3H₂O(g).
Express the equilibrium constant: K = [CO₂]²[H₂O]³ / [C₂H₅OH][O₂]³.
Hence, the correct choice is Option C.
Given that the drift mobility of electrons in copper is 32 cm2 V–1 s–1 , and using values of the electrical conductivity, atomic weight, and density of copper from C&R, compute the number of conduction electrons per atom in copper. (Hint: compute the number of atoms per unit volume in copper, and then compare this to the number of conduction electrons per unit volume (i.e., the carrier density n) computed from copper’s electrical conductivity.) b) In compounds, copper is known to exhibit two common oxidation states, +1 and +2. Discuss whether this fact is consistent with your answer for part a).
At 25 °C and 775 Torr, carbon dioxide has a solubility of 0.0347 M in water. What is its solubility at 25 °C and 1470 Torr?
Answer : The solubility at [tex]25^oC[/tex] is, 0.0658 M
Explanation :
According to the Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
[tex]S\propto P[/tex]
or,
[tex]\frac{S_1}{S_2}=\frac{P_1}{P_2}[/tex]
where,
[tex]S_1[/tex] = initial solubility of carbon dioxide gas = 0.0347 M
[tex]S_2[/tex] = final solubility of carbon dioxide gas = ?
[tex]P_1[/tex] = initial pressure of carbon dioxide gas = 775 torr
[tex]P_2[/tex] = final pressure of carbon dioxide gas = 1470 torr
Now put all the given values in the above formula, we get the final solubility of the carbon dioxide gas.
[tex]\frac{0.0347M}{S_2}=\frac{775\text{ torr}}{1470\text{ torr}}[/tex]
[tex]S_2=0.0658M[/tex]
Therefore, the solubility at [tex]25^oC[/tex] is, 0.0658 M
Using Henry's Law, the solubility of carbon dioxide in water at 25 °C would double from 0.0347 M at 775 Torr to 0.0694 M at 1470 Torr.
Explanation:To determine the solubility of carbon dioxide at 25 °C and 1470 Torr, we can use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Given that the solubility of carbon dioxide is 0.0347 M at 775 Torr, we can calculate its solubility at 1470 Torr (which is double the initial pressure) by simply doubling the initial solubility. Therefore, at 25 °C and 1470 Torr, the solubility of carbon dioxide in water would be 0.0694 M.
A unit of heat energy that was formerly used frequently was the calorie. Look up the definition of the calorie in your textbook or a handbook and record it here. Calculate the calorimeter constant for your calorimeter in cal/°C .
Answer and Explanation:
Calorie is the unit of heat energy . There are 2 units with the same name 'calorie' which is widely used.
'The amount of heat energy required to increase the temperature of 1 gram of water by mass by [tex]1^{\circ}C[/tex] or 1 K is known as small calorie or gram calorie'.
Another one is large calorie which can be defined as :
'The amount of heat energy required to make arise in temperature of water 1 kg by mass by [tex]1^{\circ}C[/tex] or 1 K is known as large calorie or kilcalorie and is represented as Cal or kcal'.
After the adoption of SI system, thee units of the metric system cal, C or kilocal are considered deprecated or obsolete with the SI unit for heat energy as 'joule or J'
1 cal = 4.184 J
1C or 1 kilocal = 4184 J
Calorimeter constant:
Calorimeter constant, represented as '[tex]C_{cal}[/tex]' is used to quantify the heat capacity or the amount of heat of a calorimeter.
It can be calculated by ther given formula:
[tex]{\displaystyle C_{cal}}={\frac {\Delta {H}}{\Delta {T}}}}}[/tex]
where,
[tex]{\Delta {T}}[/tex] = corresponding temperature change
[tex] {\Delta {H}[/tex] = enthalpy change
Its unit is J/K or J/1^{\circ}C[/tex] which can be convertyed to cal/1^{\circ}C[/tex] by dividing the calorimeter constant by 4.184 or 4184 accordingly.
A calorie is the amount of heat energy required to raise the temperature of one gram of water by one degree Celsius, equating to 4.184 Joules. The calorimeter constant is the heat capacity of the calorimeter, calculated by dividing the amount of heat absorbed by the temperature increase during a calorimetry experiment.
Explanation:A calorie is a conventional unit of heat, defined as the quantity of heat energy required to raise the temperature of one gram of water by one degree Celsius. This definition of a calorie aligns specifically to the temperature change from 14.5°C to 15.5°C.To convert calories into Joules, the SI unit of heat, a constant of 4.184 Joules equals one calorie.
The calorimeter constant (also known as the heat capacity of the calorimeter) is vital in calorimetry, the process of measuring heat energy changes in a chemical system. It is specified in units of cal/°C. To calculate the calorimeter constant for your calorimeter, you would have to know the mass of the calorimeter and the temperature change observed during an experiment. The constant is calculated by dividing the amount of heat absorbed (in calories, calculated using known quantities and specific heat values of substances involved) by the temperature increase.
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Which of the following statements about soaps and detergents is NOT true?
a.Soaps form micelles in solution.
b.The hydrophilic end has the hydrocarbon chain.
c.The hydrophobic end has the hydrocarbon chain.
d.The hydrophilic end is ionic.
Answer:
The answer is B. the hydrophilic end has the hydrocarbon chain.
Explanation:
That answer is wrong because soap and detergent form a bilayer with hydrophilic and hydrophobic ends.
the hydrophilic end interacts with polar compound such as water because that end is ionic and charged, it is composed of carbon and Oxygen and forms dipole interaction and electrostatic interactions.
the hydrophobic end is the non polar end, it is composed of carbon and hydrogen chain (Hydrocarbon) which are non-polar due to equal electronegativity between carbon and hydrogen.
that is why answer B is not true, the hydrophilic end does not have the hydrocarbon chain, the hydrocarbon chain is on the hydrophobic end.
A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 mmHg) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of M.
Based on the data provided, it appears that the metal M with a molar mass of 27.0 g/mol reacts with hydrochloric acid to produce hydrogen gas, indicating the metal is likely aluminum. The reaction equation would be 2Al + 6HCl → 2AlCl3 + 3H2. The oxide of aluminum is Al2O3 and the sulfate is Al2(SO4)3.
Explanation:To find the equation for the reaction of the metal M with hydrochloric acid, we should first understand the typical reaction that occurs, which is a metal reacting with an acid to produce a salt and hydrogen gas:
M + 2HCl → MCl2 + H2
Given the data, we can calculate the moles of hydrogen gas (H2):
n(H2) = PV / RT
Where P is the pressure in atmospheres (741 mmHg is equivalent to 741/760 atm), V is the volume (0.303 L), R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (17°C = 290 K).
Once we have the moles of H2, we can find the ratio of metal M to H2 in the reaction and then use the molar mass of M to determine how many moles of M reacted, which should be half the moles of H2 produced since the metal is likely to be in Group 1 or 2. Lastly, using the molar mass of the metal (27.0 g/mol), we can confirm the stoichiometry.
For the formulas of the oxide and sulfate of M:
Oxide of M: M2O or MO, where the subscripts depend on the valency of the metal M.Sulfate of M: M2(SO4)3 or M(SO4), again depending on valency.In summary, the molar mass given and typical valence of metals may suggest that metal M is likely aluminum (Al), hence the possible chemical reactions would be:
2Al + 6HCl → 2AlCl3 + 3H2
Formulas for aluminum compounds:
Oxide of Al: Al2O3
Sulfate of Al: Al2(SO4)3
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The metal M is aluminum (Al), and the Balanced reaction is: [tex]2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2[/tex] the Oxide is: [tex]\text{Al}_2\text{O}_3[/tex], and Sulfate is [tex]\text{Al}_2(\text{SO}_4)_3[/tex].
To determine the reaction and write the formulas for the oxide and sulfate of the metal M , we first need to identify the metal and its valency based on the data provided.
1. Determine the number of moles of hydrogen gas liberated:
Using the ideal gas law, PV = nRT :
- P is the pressure in atm. To convert from mmHg to atm: [tex]P = \frac{741 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.974 \, \text{atm}[/tex].
- V is the volume in liters: V = 0.303 L.
- R is the ideal gas constant: R = 0.0821 L·atm/(mol·K.
- T is the temperature in Kelvin: [tex]T = 17^\circ \text{C} + 273.15 = 290.15 \, \text{K}[/tex].
Now, solve for n , the number of moles of H₂ :
[tex]n = \frac{PV}{RT} \\\\ n = \frac{(0.974 \, \text{atm})(0.303 \, \text{L})}{(0.0821 \, \text{L\ . atm/(mol\ . K)})(290.15 \, \text{K})} \\\\ n = \frac{0.295 \, \text{L\ . atm}}{23.81 \, \text{L\ . atm/mol}} \\\\ n \approx 0.0124 \, \text{mol}[/tex]
2. Determine the number of moles of metal M reacted:
Given the mass of the metal and its molar mass:
[tex]\text{Moles of metal} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.225 \, \text{g}}{27.0 \, \text{g/mol}} = 0.00833 \, \text{mol}[/tex]
3. Determine the valency of the metal M :
From the reaction with hydrochloric acid, each mole of metal M produces hydrogen gas. The general reaction is:
[tex]M + xHCl \rightarrow MCl_x + \frac{x}{2}H_2[/tex]
Since 0.00833 moles of M produced 0.0124 moles of H₂:
[tex]0.00833 \, \text{mol} \times \frac{x}{2} = 0.0124 \, \text{mol} \\\\ x = \frac{0.0124 \times 2}{0.00833} \\\\ x \approx 2.98 \approx 3[/tex]
So, the valency of metal M is 3.
4. Write the chemical reaction:
The balanced equation for the reaction is:
[tex]2M + 6HCl \rightarrow 2MCl_3 + 3H_2[/tex]
5. Write the formulas for the oxide and sulfate of M :
- The formula for the oxide of M :
[tex]M_2O_3[/tex]
- The formula for the sulfate of M :
[tex]M_2(SO_4)_3[/tex]
Given that the metal has a valency of 3, and the molar mass of 27 g/mol corresponds to aluminum (Al), the formulas for the oxide and sulfate of M (Aluminum) are:
- Oxide: [tex]\text{Al}_2\text{O}_3[/tex]
- Sulfate: [tex]\text{Al}_2(\text{SO}_4)_3[/tex]
Thus, the metal M is aluminum (Al), and the balanced reaction, oxide, and sulfate formulas are:
- Balanced reaction:
[tex]2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2[/tex]
- Oxide:
[tex]\text{Al}_2\text{O}_3[/tex]
- Sulfate:
[tex]\text{Al}_2(\text{SO}_4)_3[/tex]
What mass (in g) of urea (CO(NH2)2) in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution? The temperature is held constant at 40oC.
Answer: 6.7 g
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}[/tex]
where,
[tex]p^o[/tex] = vapor pressure of pure solvent (water) = 55.32 mmHg
[tex]p_s[/tex] = vapor pressure of solution = 54.21 mmHg
[tex]w_2[/tex] = mass of solute (urea) = ? g
[tex]w_1[/tex] = mass of solvent (water) = 100 g
[tex]M_1[/tex] = molar mass of solvent (water) = 18 g/mole
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mole
Now put all the given values in this formula ,we get the vapor pressure of the solution.
[tex]\frac{55.32-54.21}{55.32}=\frac{x\times 18}{100\times 60}[/tex]
[tex]x=6.7g[/tex]
Therefore, 6.7 g of urea is needed in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution.
Final answer:
6.70 g of urea is required to adjust the vapor pressure of water as specified, using calculations based on Raoult's law and the relations of vapor pressures, mole fractions, and molar masses.
Explanation:
To solve this problem, we can use Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. The formula for Raoult's law is P1 = X1P10, where P1 is the vapor pressure of the solvent in the solution, X1 is the mole fraction of the solvent, and P10 is the vapor pressure of the pure solvent.
First, we calculate the change in vapor pressure: ΔP = P10 - P1 = 55.32 mmHg - 54.21 mmHg = 1.11 mmHg. We use the given vapor pressures to find the mole fraction of water in the solution.
Using the equation for Raoult's law rearranged to find the mole fraction of the solvent (water), we get: X1 = P1 / P10 = 54.21 mmHg / 55.32 mmHg = 0.9799.
Since the mole fraction of the solute (urea) and solvent (water) together is 1, the mole fraction of urea is 1 - X1 = 0.0201. To find the moles of urea, we need the moles of water, which can be calculated from its mass (100.0 g) and molar mass (18.015 g/mol).
The moles of water are 100.0 g / 18.015 g/mol = 5.55 mol. Therefore, the moles of urea required are 5.55 mol × 0.0201 = 0.1115 mol. Finally, calculating the mass of urea required: 0.1115 mol × 60.056 g/mol (molar mass of urea) = 6.70 g.
In summary, 6.70 g of urea is needed to decrease the vapor pressure of water from 55.32 mmHg to 54.21 mmHg when dissolved in 100.0 g of water, keeping the temperature constant at 40°C.
A solution is prepared by dissolving 6.00 g of an unknown nonelectrolyte in enough water to make 1.00 L of solution. The osmotic pressure of this solution is 0.750 atm at 25.0°C. What is the molecular weight (g/mol) of the unknown solute? g
Answer: The molar mass of unknown solute is 195.44 g/mol.
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=icRT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 0.750 atm
i = Van't hoff factor = 1 (for non-electrolytes)
c = concentration of solute = ?
R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[273+25]=298K[/tex]
Putting values in above equation, we get:
[tex]0.750atm=1\times c\times 0.0820\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\c=0.0307mol/L[/tex]
The concentration of solute is 0.0307 mol/L
This means that, 0.0307 moles are present in 1 L of solution.
To calculate the molecular mass of solute, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of solute = 0.0307 mol
Given mass of solute = 6g
Putting values in above equation, we get:
[tex]0.0307mol=\frac{6g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=195.44g/mol[/tex]
Hence, the molar mass of unknown solute is 195.44 g/mol.
Final answer:
To find the molecular weight of the unknown solute in the solution, use the osmotic pressure formula by considering the given values of pressure, temperature, and weight of the solute. By applying the formula, the molecular weight of the solute is calculated to be 180 g/mol.
Explanation:
To calculate the molecular weight of the unknown solute, we use the formula for osmotic pressure: π = iMRT. Given that the osmotic pressure is 0.750 atm, temperature is 25.0°C, and the weight of the solute is 6.00 g in 1.00 L solution, we can find the molecular weight to be 180 g/mol.
The reaction between ethyl bromide (C2H5Br) and hydroxide ion in ethyl alcohol at 330 K, C2H5BR(alc) + OH-(alc) --> C2H5OH(l) + Br-(alc), is first order each in ethyl bromide and hydroxide ion. When [C2H5Br] is 0.0477 M and [OH-] is 0.100 M, the rate of disappearance of ethyl bromide is 1.7 x 10^-7 M/s.
What is the value of the rate constant?
k=?
Answer: [tex]3.5\times 10^{-5}mol^{-1}Lsec^{-1}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]C_2H_5Br(alc)+OH^-(alc)\rightarrow C_2H_5OH(l)+Br^-(alc)[/tex]
Given: Order with respect to [tex]C_2H_5Br[/tex] = 1
Order with respect to [tex]OH^-[/tex] = 1
Thus rate law is:
[tex]Rate=k[C_2H_5Br]^1[OH^-]^1[/tex]
k= rate constant
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
[tex]Rate=-\frac{1d[C_2H_5Br]}{dt}=k[C_2H_5Br]^1[OH^-]^1[/tex]
Given: [tex]\frac{d[C_2H_5]}{dt}]=1.7\times 10^{-7}[/tex]
Putting in the values we get:
[tex]Rate=1.7\times 10^{-7}=k[0.0477]^1[0.100]^1[/tex]
[tex]k=3.5\times 10^{-5}mol^{-1}Lsec^{-1}[/tex]
Thus the value of the rate constant is [tex]3.5\times 10^{-5}mol^{-1}Lsec^{-1}[/tex]
The rate constant of the reaction is 3.6 * 10^-5.
What is rate of reaction?We define the rate of reaction as how quickly or slowly that a reaction is occuring. That is, the rate of disappearance of C2H5Br and OH- or rate of appearance of C2H5OH(l) and Br-.
Now;
Rate of disappearance of ethyl bromide = 1.7 x 10^-7 M/s.
Concentration of ethyl bromide = 0.0477 M
Concentration of OH- = 0.100 M
Thus;
-d[C2H5Br]/dt = k [C2H5Br] [OH-]
1.7 x 10^-7 = k [0.0477] [0.100]
k = 1.7 x 10^-7/ [0.0477] [0.100]
k = 3.6 * 10^-5
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5. Tree sap can be a very concentrated solution of solutes in water. These are mostly sugars, with van’t Hoff factors of 1. The root system provides a semi-permeable membrane across which water moves to “dilute” the sap, providing a significant osmotic pressure.
a. If a tree sap has an effective concentration of 37 Molar, what pressure is generate at 298K across the endodermis root membrane? R = 0.08216 L.atm/mol.K.
b. If an osmotic pressure of 1.0 atm can raise a volume of water 10.33 meters high, how high can the sap of this tree rise?
Answer:
a)905,89 atm of pressure will be generated at 298K.
b)The sap of tree can rise upto 9,357.84 meters.
Explanation:
a)
Effective concentration of sap = c = 37 M
Osmotic pressure generate at 298K = [tex]\pi [/tex]
Temperature ,T = 298 K
[tex]\pi =cRT[/tex]
[tex]\pi =37 mol/L\times 0.08216 atm L/mol L\times 298 K[/tex]
[tex]\pi =905.89 atm[/tex]
905,89 atm of pressure will be generated at 298K across the endodermis root membrane.
b)
Given that 1.0 at of pressure raises the volume of water upto height of 10.33 m
Then 905.89 atm of pressure will raise the height of water upto:
[tex]\frac{10.33}{1.0}\times 905.89 m=9,357.84 m[/tex]
The sap of the tree can rise upto 9,357.84 meter.
The osmotic pressure generated across the root membrane of a tree with a 37 Molar sap concentration at 298K is 903.62 atm. This pressure could theoretically raise the sap to a height of about 9334.39 meters, though such a height is unrealistic in actual tree physiology due to factors like structural limits.
To calculate the osmotic pressure generated at 298K across the endodermis root membrane for a tree sap with an effective concentration of 37 Molar, we use the formula
c = iMRT, where:
i is the vant Hoff factor (which is 1 for sugars)
M is the molarity of the solution (37 M)
R is the ideal gas constant (0.08216 L.atm/mol.K)
T is the temperature in Kelvins (298K)
Plugging in the numbers, we get:
c = (1)(37 M)(0.08216 L.atm/mol.K)(298K) = 903.62 atm
To determine how high the sap can rise given an osmotic pressure of 1.0 atm can raise a column of water 10.33 meters, we set up a proportion:
(1 atm / 10.33 m) = (903.62 atm / x m)
Solving for x gives us x = 903.62 atm * 10.33 m/atm ≈ 9334.39 meters. Therefore, theoretically, the sap could rise to a height of approximately 9334.39 meters due to the generated osmotic pressure. However, this scenario is not realistic due to limitations such as hydrostatic pressure and structural integrity of the tree.
Osmotic pressure and root pressure are essential in understanding the transport of water and nutrients in plants, particularly in how they work together with transpiration and cohesive forces in water molecules to move sap through the xylem vessels from the roots to the leaves.
Consider 0.10 M solutions of acetic acid, ammonia, hydrochloric acid, and sodium hydroxide. Rank these solutions in order of increasing pH (starting from the lowest, and ending with the highest pH value). Please also make a statement, whether each solution is an acidic or a basic one, and provide an explanation why.
Answer:
HCl < CH₃COOH < NH₃ < NaOH
Explanation:
Given compounds:
Acetic acid: CH₃COOH
Ammonia; NH₃
Hydrochloric acid: HCl
Sodium hydroxide: NaOH
All the solutions are of the same molarity which is 0.1M. We need to see how these compounds dissociate to form solutions in order to establish their pH value:
For Acetic acid;
CH₃COOH + H₂O ⇄ H₃O⁺ + CH₃COO⁻
Acetic acid is a weak acid and it ionizes slightly in solutions. It would have a pH close to 7
For Ammonia;
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Ammonia is a weak base and it ionizes slightly in solutions. It sets up an equilibrium in the process. It's would be slightly above 7
For HCl:
HCl + H₂O → H₃O⁺ + Cl⁻
HCl is a strong acid and ionizes completely in solutions. It has a very low pH
For NaOH:
NaOH → Na⁺ + OH⁻
NaOH ionizes also completely in solutions and it breaks down into sodium and hydroxide ions. It is a strong base and it would have a high PH value.
HCl < CH₃COOH < NH₃ < NaOH
This is the trend of increasing pH
Answer:
Explanation:hi
Zoe used two examples to represent two different states of matter. Example 1: Fish swimming around one another in an aquarium Example 2: Fish swimming away from one another in the ocean Which statement is correct? Example 1 represents a gas and Example 2 represents a solid. Example 1 represents a liquid and Example 2 represents a gas. Example 1 represents a gas and Example 2 represents a liquid. Example 1 represents a solid and Example 2 represents a gas.
Answer:
the 4th choice.
Explanation:
the fish can't swim through the aquarium
the fish can't swim in the air
Answer:
B.Example 1 represents a liquid and example 2 represents a gas
Explanation:
Solid : It is that states of matter in which atom or molecules very close to each other. The distance between atoms or molecules is negligible. The atom or molecules can't escape out of solids.
Liquid: It is that states of matter in which atoms or molecules are close to each other but not very close as in solid.The distance between atoms or molecules is small.
Gas: It is that states of matter in which atoms or molecules are very far to each other .The distances between atoms or molecules are very large. The two two atoms can not come closer to each other at normal condition.
We are given that Zoe used two examples to represent two different states of matter
1.Example :Fish swimming around one another in an aquarium
It means fish can not escape out of aquarium. Hence, it represent the liquid states of matter.
2.Example: Fish swimming away one another in the ocean .
When fish swimming away one another ,Then they can'not come close to each other .Hence, it represents gas states of matter.
Therefore, option D is true.
Answer:Example 1 represents a solid and example 2 represents a gas
Calculate the osmotic pressure of a solution containing 1.502 g of (Nh4)2SO4 in 1 L at 36.54 Degrees Celcius. (The gas constant is .08206 Latm/molK. The molar mass of (Nh4)2SO4 is 132.16)
Answer:
0.2886 atm is the osmotic pressure of a solution.
Explanation:
Osmotic pressure of solution =[tex]\pi[/tex]
Concentration of the solution = c
Mass of the ammonium sulfate = 1.502 g
Moles of ammonium sulfate = [tex]\frac{1.502 g}{132.16 g/mol}=0.01136 mol[/tex]
Volume of the solution = 1 L
Concentration of the solution:
[tex]=\frac{\text{Moles of ammonium sulfate}}{\text{Volume of the solution}}[/tex]
[tex]c=\frac{0.01136 mol}{1 L}=0.01136 mol/L[/tex]
Temperature of the solution ,T= 36.54°C = 309.69 K
R = universal gas constant = 0.08206 L atm/mol K
[tex]\pi=cRT[/tex]
[tex]\pi=0.01136 mol/L\times 0.08206 L atm/mol K\times 309.69 K[/tex]
[tex]\pi=0.2886 atm[/tex]
0.2886 atm is the osmotic pressure of a solution.
To find the osmotic pressure, the temperature was converted to Kelvin, the moles of (NH₄)₂SO₄ were calculated using its molar mass, and the van't Hoff factor for the dissociation of (NH₄)₂SO₄ was determined. The gas constant and these values were then used in the osmotic pressure formula to calculate a pressure of approximately 0.08452 atm.
To calculate the osmotic pressure of a solution containing 1.502 g of (NH₄)₂SO₄ in 1 L at 36.54 Degrees Celsius, we use the formula:
π = i × M × R × T
where:
π is the osmotic pressure in atm,
i is the van't Hoff factor (number of particles the solute dissociates into),
M is the molarity of the solution,
R is the gas constant (0.08206 L·atm/mol·K), and
T is the temperature in Kelvin.
First, convert the temperature to Kelvin:
T(K) = 36.54 + 273.15 = 309.69 K
Then, calculate the moles of (NH4)2SO4 using its molar mass:
Moles = {1.502 g}÷{132.16 g/mol} ≈ 0.01136 mol
Since the solution's volume is 1 L, the molarity (M) is 0.01136 M. Ammonium sulfate ((NH₄)₂SO₄) dissociates into 2NH₄⁺ and SO₄²⁻. Therefore, the van't Hoff factor (i) is 3.
Finally, substitute the values into the formula to find the osmotic pressure:
π = 3 × 0.01136 M × 0.08206 L·atm/mol·K × 309.69 K
π ≈ 0.08452 atm
The osmotic pressure of the solution is approximately 0.08452 atm.
Write the molecular equation and the net inonic equation for each of the following aqueous reactions. If no reaction occurs, write NR after the arrow. AgNO3 + Nal --->, Ba(NO3)2 + K2SO4--->, NH4NO3 + K2SO4--->, LiCl + Al(NO3)3--->
Answer: The equations are given below.
Explanation:
For the given options:
Option 1: [tex]AgNO_3+NaI[/tex]The molecular equation for this follows:
[tex]AgNO_3(aq.)+NaI(aq.)\rightarrow AgI(s)+NaNO_3(aq.)[/tex]
Ionic form of the above equation follows:
[tex]Ag^+(aq.)+NO_3^-(aq.)+Na^+(aq.)+I^-(aq.)\rightarrow AgI(s)+Na^+(aq.)+NO_3^-(aq.)[/tex]
As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:
[tex]Ag^+(aq.)+I^-(aq.)\rightarrow AgI(s)[/tex]
Option 2: [tex]Ba(NO_3)_2+K_2SO_4[/tex]The molecular equation for this follows:
[tex]Ba(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+BaSO_4(s)[/tex]
Ionic form of the above equation follows:
[tex]Ba^{2+}(aq.)+2NO_3^-(aq.)+K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2K^+(aq.)+2NO_3^-(aq.)+BaSO_4(s)[/tex]
As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:
[tex]Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)[/tex]
Option 3: [tex]NH_4NO_3+K_2SO_4[/tex]The molecular equation for this follows:
[tex]2NH_4NO_3(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+H_2SO_4(aq.)+NH_3(g)[/tex]
Ionic form of the above equation follows:
[tex]2NH_4^+(aq.)+2NO_3^-(aq.)+2K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)+SO_4^{2-}(aq.)+2K^+(aq.)+2NO_3^-(aq.)[/tex]
As, nitrate, potassium and sulfate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:
[tex]2NH_4^+(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)[/tex]
Option 4: [tex]LiCl+Al(NO_3)_3[/tex]The molecular equation for this follows:
[tex]3LiCl(aq.)+Al(NO_3)_3(aq.)\rightarrow 3LiNO_3(aq.)+AlCl_3(s)[/tex]
Ionic form of the above equation follows:
[tex]3Li^+(aq.)+3Cl^-(aq.)+Al^{3+}(aq.)+NO_3^-(aq.)\rightarrow 3Li^+(aq.)+3NO_3^-(aq.)+AlCl_3(s)[/tex]
As, lithium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:
[tex]Al^{3+}(aq.)+3Cl^-(aq.)\rightarrow AlCl_3(s)[/tex]
Hence, the molecular and ionic equations are given above.
2. Consider the reaction 2NO(g) + O2(g) → 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.066 M/s. a. At what rate is NO2 being formed? (3 points) b. At what rate is molecular oxygen reacting? (3 points)
Answer :
(a) The rate of [tex]NO_2[/tex] formed is, 0.066 M/s
(b) The rate of [tex]O_2[/tex] formed is, 0.033 M/s
Explanation : Given,
[tex]\frac{d[NO]}{dt}[/tex] = 0.066 M/s
The balanced chemical reaction is,
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
The rate of disappearance of [tex]NO[/tex] = [tex]-\frac{1}{2}\frac{d[NO]}{dt}[/tex]
The rate of disappearance of [tex]O_2[/tex] = [tex]-\frac{d[O_2]}{dt}[/tex]
The rate of formation of [tex]NO_2[/tex] = [tex]\frac{1}{2}\frac{d[NO_2]}{dt}[/tex]
As we know that,
[tex]\frac{d[NO]}{dt}[/tex] = 0.066 M/s
(a) Now we have to determine the rate of [tex]NO_2[/tex] formed.
[tex]\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}[/tex]
[tex]\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s[/tex]
The rate of [tex]NO_2[/tex] formed is, 0.066 M/s
(b) Now we have to determine the rate of molecular oxygen reacting.
[tex]-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}[/tex]
[tex]\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s[/tex]
The rate of [tex]O_2[/tex] formed is, 0.033 M/s
The rate of the formation is the time taken and required by the reaction to yield the product. The rate of formation of nitrogen dioxide is, 0.066 M/s and oxygen is 0.033 M/s.
What is the rate of formation?The rate of formation is the time derivative of the chemical reaction. The balanced chemical reaction can be given as:
[tex]\rm 2 NO + O_{2} \rightarrow 2NO_{2}[/tex]
From the reaction:
The rate of disappearance of [tex]\rm NO[/tex] = [tex]-\dfrac{1}{2} \dfrac{d[\rm NO]}{dt}[/tex]The rate of disappearance of [tex]\rm O_{2}[/tex] = [tex]-\dfrac{d[\rm O_{2}]}{dt}[/tex]The rate of formation of [tex]\rm NO_{2}[/tex] = [tex]\dfrac{1}{2} \dfrac{d[\rm NO_{2}]}{dt}[/tex]Given,
Rate of reaction [tex]\dfrac{d[\rm NO]}{dt}[/tex] = 0.066 M/s
Calculate the rate of formation of [tex]\rm NO_{2}[/tex] :
[tex]\begin{aligned}\dfrac{1}{2} \dfrac{d[\rm NO_{2}]}{dt} &= \dfrac{1}{2} \dfrac{d[\rm NO]}{dt} \\\\\dfrac{d[\rm NO_{2}]}{dt} &=\dfrac{d[\rm NO]}{dt}\\\\&= 0.066 \;\rm M/s\end{aligned}[/tex]
Calculate the rate for molecular oxygen:
[tex]\begin{aligned}-\dfrac{d[\rm O_{2}]}{dt} &= -\dfrac{1}{2} \dfrac{d[\rm NO]}{dt} \\\\\dfrac{d[\rm O_{2}]}{dt} &=\dfrac{1}{2} \times 0.066 \\\\&= 0.033\;\rm M/s\end{aligned}[/tex]
Therefore, the rate of formation of nitrogen dioxide is, 0.066 M/s and oxygen is 0.033 M/s.
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(hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles in liquid water according to the equation 2HgO(s)+H2O(l)+2Cl2(g)⇌2HOCl(aq)+HgO⋅HgCl2(s) What is the equilibrium-constant expression for this reaction?
The equilibrium constant expression for this reaction is [tex]K_{eq} = \frac{[HOCl]^{2}}{[Cl_{2}]^{2}}[/tex]. This is an example of a heterogeneous equilibrium wherein the states of the reactants and products are different. In such a case, only the concentration of the gaseous and aqueous substances are included in the equilibrium constant expression.
Further Explanation:
The equilibrium constant expression is the ratio of the concentration of the products and the concentration of reactants.
The guidelines in writing the equilibrium constant expressions are as follows:
Write molar concentration of each product in the numerator of the Keq expression. Write the coefficient of the substance as the exponent of the molar concentration of the substance in the Keq expression. Write the molar concentration of each reactant raised to its coefficient in the denominator.Note: Pure substances (i.e. solids and liquids) are not included in the equilibrium constant expression as their concentrations are constant.
The numerical equivalent of the Keq expression is called the equilibrium constant. It has a specific value for a given temperature. The equilibrium constant provides information about the spontaneity and progress of an equilibrium reaction.
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As we climb a mountain to a higher altitude, we experience a pressure increase.(T/F)
Answer:
As we climb a mountain to a higher altitude, we experience a pressure increase. -True
Consider the following reaction at 25 °C: CaCrO4s)-Ca2*la)+CrO,2(aq) Ko 7.1 104 What are the equilibrium concentrations of Ca2 and CrO42?
Answer: The equilibrium concentration of [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] are 0.0266 M.
Explanation:
The chemical equation for the ionization of calcium chromate follows:
[tex]CaCrO_4\rightleftharpoons Ca^{2+}+CrO_4^{2-}[/tex]
The expression for equilibrium constant is given as:
[tex]K_c=\frac{[Ca^{2+}][CrO_4^{2-}]}{[CaCrO_4]}[/tex]
We are given:
[tex]K_c=7.1\times 10^{-4}[/tex]
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Let the equilibrium concentration for [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] be 'x'
Putting values in above equation, we get:
[tex]7.1\times 10^{-4}=x^2\\\\x=0.0266M[/tex]
Hence, the equilibrium concentration of [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] are 0.0266 M.
Which of the following definitions of pesticides in NOT correct? a. Herbicides are chemicals that kill plants. b. Insecticides are chemicals that kill insects. c. Fertilizers are chemicals that kill plants. d. Fumigants are chemicals that fungi.
Answer:
Fertilizers are chemicals that kill plants.
Explanation:
Fertilizers are not pesticides. Fertilizers helps to improve plant yield by supplying nutrients needed for plant growth and food formation. When fertilizers are applied to plants, the nutrient pool increases and plants can readily grow without any deficiency resulting from lack of one mineral or the other.
Herbicides are used to kill plants.
6. A sample of a gas has a mass of 0.527 g. Its volume is 0.35 L at a temperature of 88 degree Celsius and a pressure of 945 mm Hg. Find it's molar mass (show your work)
Answer: The molar mass of the gas is 35.87 g/mol.
Explanation:
To calculate the mass of gas, we use the equation given by ideal gas:
PV = nRT
or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = Pressure of gas = 945 mmHg
V = Volume of the gas = 0.35 L
m = Mass of gas = 0.527 g
M = Molar mass of gas = ? g/mo
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
T = Temperature of gas = [tex]88^oC=[88+273]=361K[/tex]
Putting values in above equation, we get:
[tex]945mmHg\times 0.35L=\frac{0.527g}{M}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 361K\\\\M=35.87g/mol[/tex]
Hence, the molar mass of the gas is 35.87 g/mol.
In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.
This motion is associated by mechanical means such as pumps, gravity or fans, to produce the movement of the fluid: A. Free convection B. Forced conduction C. Forced convection D. Radiation INDOORS
The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.8 kJ/mol
Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
[tex]\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ][/tex]
Where,
[tex]\Delta H_v[/tex] is the Heat of vaoprization (J/mol)
[tex]T_b[/tex] is the normal boiling point of the gas (K)
[tex]T_c[/tex] is the Critical temperature of the gas (K)
[tex]P_c[/tex] is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:
[tex]T_b=307.4\ K[/tex]
[tex]T_c=466.7\ K[/tex]
[tex]P_c=36.4\ bar[/tex]
Applying the above equation to find heat of vaporization as:
[tex]\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ][/tex]
[tex]\Delta H_v=26400 J/mol[/tex]
The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,
[tex]\Delta H_v=26.4 kJ/mol[/tex]
Option C is correct
You are given a food containing 6 g protein per serving (30g), how much of the food you needto add to 100 ml solution to make a solution with 500 μg protein/ml? What is the dilution factor?
Hey there!
500 mg of protein is present in 100 mL of solvent as per the concentration 0.5 mg/mL or 500 g/mL ,
So, 250 mg (0.25 g) of serving food need to be added to 100 mL solvent in order to prepare 50 mg of protein/100 mL solution.
Dilution factor = initial amount of protein / final amount of protein
= 6 g / 0.05 g = 120
Hope this helps!