Therefore, we need to add 173.49 grams of potassium iodide to 2.00 kg of water to prepare a 0.523 m aqueous solution.
How is 1% potassium iodide solution made?Potassium iodide solution is made by dissolving 1 litre of water in 1 gramme of potassium iodide and 1 gramme of hydroxyammonium chloride. Solution of potassium iodide, about 0.2 M: 33 grammes of potassium iodide should be dissolved in 1 litre of water.
We must apply the following formula to get the mass of potassium iodide required to create a 0.523 m aqueous solution:
molarity=moles of solute/liters of solution
First, we must determine the solution's litre volume:
1 kg of water=1000 mL of water
2.00 kg of water = 2000 mL of water
Volume of solution = 2000 mL = 2.00 L
Next, we need to rearrange the formula to solve for the moles of solute:
moles of solute=molarity x liters of solution
moles of solute = 0.523 mol/L x 2.00 L = 1.046 mol
Finally, we can use the molar mass of potassium iodide (166.0028 g/mol) to convert the moles of solute to grams:
mass of potassium iodide = moles of solute x molar mass
mass of potassium iodide = 1.046 mol x 166.0028 g/mol = 173.49 g
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If the salt created from the reaction of a strong acid and a weak base is added to a solution, what will happen to the solution? A. The pH will decrease. B. The pH will increase. C. The pH will stay the same. D. More OH- ions will form. E. The H3O+ ions will decrease.
HELP ASAP
If you started with 20.0 g of a radioisotope and waited for 3 half-lives to pass, then how much would remain? 2.50 g 5.00 g 10.0 g 15.0 g
The amount that would remain, given that 3 half-lives has pass when you started with 20.0 g is 2.50 grams (1st option)
How do i determine the amount that would remain?The following data were obtained from the question:
Original amount of radioisotope (N₀) = 20.0 gramsNumber of half-lives that has passed (n) = 3Amount remaining after 3 half-lives (N) = ?The amount remaining can be obtained as shown below:
N = N₀ / 2ⁿ
N = 20 / 2³
N = 20 / 8
N = 2.50 grams
Thus, we can conclude from the above calculation that the amount that would remain after 3 half-lives to pass is 2.50 grams (1st option)
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Answer:
2.50g
Explanation:
Question 4 of 10
Based on information from the periodic table, what does this image
represent?
= 9 Protons
= 10 Neutrons
= 10 Electrons
A.Neutral fluorine
B. A positively charged fluoride ion
C. A negatively charged fluoride ion
D. A neutral neon atom
A 1.5 L sample of a 0.44 M H Br solution is mixed with 2.2 L of a 0.080 M HCIO4 solution. What is the pH of the mixture?
Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction free energy of the following chemical reaction 2NH3 =N2H4 +H2
The reaction [tex]2NH_3=N_2H_4+H_2[/tex] has a standard reaction free energy of -62.4 kJ/mol.
What is energy?Energy is the capacity to carry out tasks or affect change. It comes in a variety of shapes and sizes, including kinetic energy (energy of motion), potential energy (stored energy of position), thermal energy (heat), electrical, chemical, and nuclear energy. Energy is required for the survival and growth of all living things. Additionally, it is necessary for the operation of industries and equipment.
Equation can be used to get a reaction's standard reaction free energy.
ΔG°rxn = ΣΔG°f (products) - ΣΔG°f (reactants).
For the reaction [tex]2NH_3=N_2H_4+H_2[/tex], the ΔG°f values are as follows:
ΔG°f ([tex]2NH_3[/tex]) = -46.2 kJ/mol
ΔG°f ([tex]N_2H_4[/tex]) = -20.8 kJ/mol
ΔG°f ([tex]H_2[/tex]) = 0 kJ/mo
The equation yields the average reaction free energy (ΔG°rxn) of the reaction is:
ΔG°rxn = (2 x -20.8) - (-46.2)
= -62.4 kJ/mol
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Calculate the standard change in Gibbs free energy for the reaction at 25 °C. Refer to the ΔG°f values.
Fe2O3(s)+2Al(s)⟶Al2O3(s)+2Fe(s)
Δ°rxn=
The standard change in Gibbs free energy for the given reaction at 25°C is -757.9 kJ/mol.
Describe Gibbs Energy.Gibbs energy, also known as Gibbs free energy, is a thermodynamic quantity used to determine the maximum amount of work that can be obtained from a system at a constant temperature and pressure. It is denoted by the symbol G and is named after the American physicist Josiah Willard Gibbs who introduced the concept in the late 19th century.
Gibbs energy is defined as the difference between the enthalpy of a system and the product of the temperature and the entropy of the system:
G = H - TS
where H is the enthalpy, T is the temperature in Kelvin, and S is the entropy of the system.
The Gibbs energy is related to the equilibrium constant of a reaction through the following equation:
ΔG = -RTlnK
To calculate the standard change in Gibbs free energy for the given reaction at 25°C, we need to use the ΔG°f values (standard Gibbs free energy of formation) for the reactants and products involved in the reaction.
The ΔG°f values for Fe₂O₃(s), Al(s), Al₂O₃(s), and Fe(s) can be found in a table of thermodynamic data and are:
ΔG°f [Fe₂O₃(s)] = -824.2 kJ/mol
ΔG°f [Al(s)] = 0 kJ/mol
ΔG°f [Al₂O₃(s)] = -1582.3 kJ/mol
ΔG°f [Fe(s)] = 0 kJ/mol
The standard change in Gibbs free energy for the reaction can be calculated using the following equation:
Δ°rxn = ΣΔG°f(products) - ΣΔG°f(reactants)
Substituting the values, we get:
Δ°rxn = [ΔG°f(Al₂O₃(s)) + 2ΔG°f(Fe(s))] - [ΔG°f(Fe₂O₃(s)) + 2ΔG°f(Al(s))]
Δ°rxn = [(-1582.3 kJ/mol) + 2(0 kJ/mol)] - [(-824.2 kJ/mol) + 2(0 kJ/mol)]
Δ°rxn = -757.9 kJ/mol
Therefore, the standard change in Gibbs free energy for the given reaction at 25°C is -757.9 kJ/mol.
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The combustion of ethane (C2H6) produces CO2 and H2O:
2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g)
How many moles of O2 are needed to react with 15.5 g of ethane?
When 3 g of a petroleum are burned with more oxygen than is required, 8.8 g of Atmospheric carbon dioxide plus 5.4 g liquid [tex]H_{2}O[/tex] are produced.
What happens when ethane gas ([tex]C_{2}H_6}[/tex]) and oxygen gas ([tex]O_{2}[/tex]) burn together?
Oxygen gas and ethane ([tex]C_{2}H_6}[/tex]) react to create water as well as carbon dioxide. Find the total quantity of carbon dioxide created when the reaction yield is 60% when 5 mol of methane is burned and 16 mol of oxygen initially. 2[tex]C_{2}H_4}[/tex]+7[tex]O_{2}[/tex] →4[tex]CO_{2}[/tex]+6[tex]H_{2}O.[/tex]
How many tumours of CO2 are created when 2.2 blackheads of [tex]C_{2}H_4}[/tex]are burned?
As a result, 2 m of carbon dioxide is generated from mole of a substance of [tex]C_{2}H_4} .[/tex] Thus, multiply 2.2 over 2 by 1 pot to get 2.2 moles for c into h or just a cross. It consists of 4.4 moles of carbon.
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In order to estimate the population size of the woodlice in her garden, Jessica used the mark-recapture method: she trapped, marked and released 20 woodlice. Three days later, she trapped 15 woodlice and observed that 10 had marks. Use the following formula to estimate the woodlice population size: population size = 1st sample × 2nd sample ÷ 2nd sample that had been marked.
NEED HELP ASAP!!!!!!
To determine the size of the entire population: In the equation N = M, replace the variables of M solutions, R (number all marked recaptured), with T (total recaptured during second visit). T R = frac M T R N = RMT.
What is the mark-recapture method's formula?In order for the mark-recapture method to function, it must be assumed that the proportion the marked organisms that are recaptured inside the second sample corresponds to that of the original marked in the entire population. This equation R(recaptured)/C(captured in second sample)=M(marked initially)/N) illustrates this (total number in population).
Why do people employ the capture-recapture technique?It has been suggested that capture-recapture techniques be used to gauge a register's degree of completeness. These techniques were initially created to determine how big a confined animal colony was. The process aims to capture, tag, and release as many animals is possible in a given region all at once. This is known as the "capture" stage.
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why does glass containing an iced beverage feel cold
Answer:
Molecules in the skin are moving faster than molecules in the glass.
Explanation:
Put it in your own words.
Answer:
The energy from the ice transferred to the glass.
the rate constant for a specific reaction has been determined at the following temps. 82 C, k= 3.97x10^-3 M^-1 x s^-1 , 125 C, k= 2.07x10^-2 ,M^-1 x s^-1. a) calculate the activation energy for the reaction
b) If data points were plotted in Arrhenius plot, what would the slope of the line be?
c) at what temp will the rate constant have k=1.00x10^-3 M^-1 x s^-1
A) the activation energy is = 66.5 kJ/mol
B) the slope of the line in this case would be -8000 K^-1
C) the temperature at which the rate constant is 1.00x10^-3 M^-1 x s^-1 is 408 K (135°C).
The activation energy, slope and temperaturea) To calculate the activation energy, we can use the Arrhenius equation:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol x K), and T is the temperature in Kelvin.
Taking the natural logarithm of both sides of the equation, we get:
ln(k) = ln(A) - (Ea/RT)
We can use the two sets of data to set up two equations:
ln(k1) = ln(A) - (Ea/RT1)
ln(k2) = ln(A) - (Ea/RT2)
Solving for Ea by taking the difference between the two equations:
ln(k2/k1) = (Ea/R) [(1/T1) - (1/T2)]
Ea = -R ln(k2/k1) / [(1/T1) - (1/T2)]
Plugging in the values:
Ea = -8.314 J/mol x K x ln(2.07x10^-2 / 3.97x10^-3) / [(1/398 K) - (1/398 K)]
Ea = 66.5 kJ/mol
b) In an Arrhenius plot, ln(k) is plotted against 1/T, and the slope of the line is equal to -Ea/R. Therefore, the slope of the line in this case would be:
slope = -Ea/R = -(66.5 x 10^3 J/mol) / (8.314 J/mol x K) = -8000 K^-1
c) To solve for the temperature at which the rate constant is 1.00x10^-3 M^-1 x s^-1, we can rearrange the Arrhenius equation:
k = Ae^(-Ea/RT)
ln(k) = ln(A) - (Ea/RT)
1/T = (ln(k) - ln(A)) / (-Ea/R)
T = -R / (Ea ln(k) - ln(A))
Plugging in the values:
T = -8.314 J/mol x K / [(66.5 x 10^3 J/mol) ln(1.00x10^-3) - ln(3.97x10^-3)]
T = 408 K
Therefore, the temperature at which the rate constant is 1.00x10^-3 M^-1 x s^-1 is 408 K (135°C).
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15.0 moles of gas are in a 6.00 L tank at 20.3 ∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.
Under the given circumstances, the pressure difference between methane and an ideal gas is 58.5 atm.
What is an ideal gas?An ideal gas is a theoretical gas composed of a large number of small particles that have zero volume, do not interact with each other, and are in constant random motion. The behavior of an ideal gas is described by the ideal gas law, which relates the pressure, volume, temperature, and number of moles of the gas.
The van der Waals equation can be used to figure out the pressure difference between methane and an ideal gas under these circumstances:
(P + a n² / V²)(V - n b) = n R T
where P is the pressure, n is the number of moles, V is the volume, T is the temperature in Kelvin, R is the ideal gas constant (0.08206 L·atm/K·mol), a and b are the van der Waals constants for methane.
First, we can calculate the pressure of an ideal gas under these conditions using the ideal gas law:
P = n R T / V
P = (15.0 mol) (0.08206 L·atm/K·mol) (293.45 K) / (6.00 L)
P = 299.8 atm
Next, we can use the van der Waals equation to calculate the pressure of methane under these conditions:
(P + a n² / V²)(V - n b) = n R T
(P + (2.300 L²·atm/mol²) (15.0 mol)² / (6.00 L)²) ((6.00 L) - (15.0 mol) (0.0430 L/mol)) = (15.0 mol) (0.08206 L·atm/K·mol) (293.45 K)
Simplifying the equation gives:
P + 1.319 atm = 359.6 atm
P = 358.3 atm
As a result, under these circumstances, the pressure difference between methane and an ideal gas is:
ΔP = P (methane) - P (ideal gas) = 358.3 atm - 299.8 atm = 58.5 atm.
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what is the compound formed between two or more non metal element that share electron
A compound formed between two or more non-metal elements that share electrons is called a covalent compound or a molecular compound.
In covalent compounds, the atoms share electrons in order to achieve a full outer shell of electrons and become more stable. Covalent compounds typically have low melting and boiling points, are often gases or liquids at room temperature, and are poor conductors of electricity. Examples of covalent compounds include water (H2O), carbon dioxide (CO2), and methane [CH4].
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Lattice energy is an estimate of the bond
Answer:
strength in ionic compounds.
Explanation:
:)
Lattice energy is an estimate of the bond of strength. It denotes the amount of energy required to break down one mole of a solid ionic compound into its constituent gaseous ions.
What does the lattice energy represent?The strength of an ionic compound's ionic bonds is measured by lattice energy. It explains several properties of ionic solids, including their volatility, solubility, and hardness. An ionic solid's lattice energy cannot be measured directly.
How does lattice energy influence ionic compound solubility in water?The higher an ionic compound's lattice energy, the more difficult it is to disassemble the crystal lattice structure and dissolve it in water. As a result, compounds with high lattice energies are less soluble in water than those with low lattice energies.
What factors influence lattice energy?The value of lattice energy is determined by the charges on the two ions as well as the distance between them. The distance between the ions is directly proportional to their size.
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Complete question:
Lattice energy is an estimated bond of the bond:
A. conductivity
B. group
C. length
D. strength
Choose the correct option.
Given:
2KI → 2K + I2
Calculate the number of moles of iodine produced in the decomposition of 7.0 moles of potassium iodide (KI). Express your answer to the correct number of significant figures.
Since 2 moles of KI produce 1 mole of I2, then 7.0 moles of KI will produce 7.0/2 moles of I2. In other words, 3.5 moles of I2 will be produced
48 grams of MgCl2 is dissolved in 500g of water, DENSITY OF
WATER = 1 kg/1L. What is the MOLALITY of this solution? Please
show your steps.
mol
kg
m =
or
m = mol = kg
Step1: given information:
solute=
(UNIT=
Solvent =
(UNIT =
)
)
The molality of the solution is 1.080 mol/kg of a 48 grams of MgCl2 is dissolved in 500g of water, density of water = 1 kg/1L.
How to calculate molality?Taking the moles of solute and dividing it by the kilograms of solvent yields the molality of a solution.
Molality is calculated as follows: kg of solvent/kg of solute
500 g, or 0.500 kg, is the mass of the solvent.
MgCl2 molecular weight divided by its mass gives the amount of moles.
48 g / 95.2 g m o l 1 equals the quantity of moles of magnesium chloride.
0.504 moles of MgCl2 are present in one mole.
Molality is calculated as 0.504moles per kilogram.
Molality is equal to 1.080 mol k g 1.
As stated in the definition, molality is the "total moles of a solute contained in a kilogram of a solvent." The terms "molality" and "molal concentration" are synonymous. It is a measurement of a solvent.
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what is a type of atom that can be bonded with water and hydrogen
Answer: Oxygen (??)
Explanation: Im not sure but smth like H2O (my brain is not braining T-T)
I need to find the average percent recovery
Consider that after recrystallization, you obtained 7.0 g of dry pure substance from 10.0 g of impure material. Your recovery rate is then 70% (7/10 × 100).
What is the typical recrystallization percentage?Recrystallization recovery percentages are often lower than 100%, though occasionally they might be higher (see the following issue). This is brought on by impurity loss, material that was left in the liquid solution after dissolving, and "mechanical losses."
What drives our % recovery calculations?The amount if pure compound that is present in the finished chemical synthesis product is calculated using the percent recovery. The ratio of actual yield to theoretical yield is used to compute percent yield. The ratio of the pure compound to the starting compound is used to calculate percent recovery.
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The calcium and magnesium in a urine sample were precipitated as oxalates. A mixed precipitate of calcium oxalate (CaC2O4) and magnesium oxalate (MgC2O4) resulted and was analysed by gravimetry. The formed precipitate mixture was heated to form calcium carbonate (CaCO3) and magnesium oxide (MgO) with a total mass of 0.0433 g. The solid precipitate mixture was ignited to form CaO and MgO, the resulting solid after ignition weighed 0.0285 g. What was the mass of calcium in the original sample? All answers should be reported with the correct significant figures
The mass of calcium in the original urine sample would be 0.0140 g.
Stoichiometric problemFirst, we need to find the masses of calcium and magnesium oxalates in the original sample. Let x be the mass of calcium oxalate and y be the mass of magnesium oxalate. Then we have:
x + y = mass of the mixed oxalate precipitate
Next, we need to use the information given to find the mass of calcium in the original sample. The mass of calcium oxide formed after ignition is equal to the mass of calcium oxalate in the original sample. We can calculate the mass of calcium oxide using the mass of calcium carbonate formed and the molar mass ratio of calcium carbonate to calcium oxide.
The balanced chemical equations for the reactions are:
CaC2O4 -> CaCO3 + CO2
CaCO3 -> CaO + CO2
The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CaO is 56.08 g/mol.
From the given information, we have:
0.0433 g = (x + y)(100.09 g/mol + 80.15 g/mol) / (128.10 g/mol + 80.15 g/mol)
0.0285 g = x(56.08 g/mol) + y(40.31 g/mol)
Solving these equations simultaneously, we get:
x = 0.0140 g
y = 0.0053 g
Therefore, the mass of calcium in the original sample (which is equal to the mass of calcium oxide formed after ignition) is:
0.0140 g
So the mass of calcium in the original sample is 0.0140 g.
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Which of the following atoms has the largest atomic radius?
Sr
At
I
Ba
I think it is Ba since the radius increases along the group.but decreases along the period
Answer:
Ba
Explanation:
Ba
CaCO3 + 2HCI =CaCl2 + H₂O + CO2
5. Calcium carbonate (CaCO3) combines with HCl to produce calcium chloride (CaCl₂),
water, and carbon dioxide gas (CO₂). How many grams of HCI are required to react with
6.35 mol CaCO3?
463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.
What is meant by molar mass?Mass of one mole of substance is referred to as the molar mass. The molar mass of a substance can be calculated by adding up the atomic masses of all the atoms in a molecule.
Balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
6.35 mol CaCO₃ * 2 mol HCl / 1 mol CaCO₃ = 12.7 mol HCl
Now, we use the molar mass of HCl (36.46 g/mol) to convert from moles to grams: 12.7 mol HCl * 36.46 g/mol = 463.5 g HCl
Therefore, 463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.
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Which reactions have a positive Δrxn?
o A(g)+B(g)⟶C(g)
o A(g)+B(g)⟶3C(g)
o A(s)+B(s)⟶C(g)
o 2A(g)+B(g)⟶C(g)
The reactions with a positive Δrxn are:
A(g) + B(g) ⟶ 3C(g)
A(s) + B(s) ⟶ C(g)
What is (Δrxn) of a reaction?The entropy change (Δrxn) of a reaction indicates the change in the degree of randomness or disorder of the system during the reaction. If the number of product molecules is greater than the number of reactant molecules, the disorder of the system usually increases, resulting in a positive Δrxn. Therefore, we can determine the answer by analyzing the stoichiometry of each reaction:
A(g) + B(g) ⟶ C(g)
In this reaction, the number of product molecules is less than the number of reactant molecules, so the disorder of the system decreases. Therefore, this reaction has a negative Δrxn.
A(g) + B(g) ⟶ 3C(g)
In this reaction, the number of product molecules is greater than the number of reactant molecules, so the disorder of the system increases. Therefore, this reaction has a positive Δrxn.
A(s) + B(s) ⟶ C(g)
In this reaction, the solid reactants are combining to form a gaseous product. The disorder of the system is expected to increase, resulting in a positive Δrxn.
2A(g) + B(g) ⟶ C(g)
In this reaction, the number of product molecules is less than the number of reactant molecules, so the disorder of the system decreases. Therefore, this reaction has a negative Δ rxn.
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A flask filled to the 25.0 ml mark contain 29.97 g of a concentrated salt water solution. What is the density of the solution?
A concentrated saltwater solution weighing 29.97 g and fitting into a flask to the mark of 25.0 ml has a density of about 1199.2 g/L.
How is the density of the solution determined?By dividing the solution's mass by its volume, we may get its density: density = mass/volume
We need to know the density of water at the solution's temperature as well as the capacity of the flask up to the 25.0 ml level in order to calculate the volume of the solution.
Since 1 mL = 0.001 L, volume is equal to 25.0 mL, or 0.0250 L.
Now, we may determine the solution's density as follows:
1199.2 g/L or 29.97 g/0.0250 L is what is referred to as density.
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What are two basic aspects of science?
The Two Aspects of Science: Control over nature and understanding of nature must both be held in equal honor.
The equation below represents the dissociation of vinegar which is a weak acid. How can you tell that it is an acid and it is weak? Just from looking at the equation.
Because it is not very effective at transferring [tex]H^{+}[/tex] ions to water, vinegar is a weak acid. Less than 0.4% of the [tex]CH_{3}CO_{2}H[/tex]molecules in a 1 M solution interact with water to create [tex]H_{3}O^{+}[/tex]and [tex]CH_{3}CO_{2}^{-}[/tex] ions. More than 99.6% of the acetic acid molecules are still whole.
Weak acidsAcids that partially dissociate in solution are referred to as weak acids. To put it another way, a weak acid is any acid that is not a strong acid. A weak acid's strength is influenced by how much it dissociates; the more it dissociates, the stronger the acid.In comparison to weak acids, strong acids have a lower pH. 2) Strong acids dissociate more, resulting in a lower pH (greater concentration of [tex]H^{+}[/tex] ions in solution). 3) This can be verified by usingFor more information on weak acid kindly visit to
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what volume of SO3 would be produced by complete reaction of 100cm3 of HO2 with O2?
The volume of [tex]SO_{3}[/tex] would be produced by complete reaction of 100cm3 of [tex]H_{2}O[/tex]with [tex]O_{2}[/tex] is at STP is 60.03L.
Volume22.4 L divided by 2.68 moles per mole yields 60.03 L [tex]SO_{3}[/tex].
describing the elements contributing to the response,
[tex]2SO_{2(g)}+ O_{2} = 2SO_{3}[/tex]
In this instance, [tex]SO_{3}[/tex] and [tex]O_{2}[/tex] have a mole ratio of 2:1. Assume the reaction takes place at STP, where 1 mole of any gas has a volume of 22.4 L. Consequently, 30 [tex]dm^{3}[/tex]of [tex]O_{2}[/tex] (1 dm3 = 1 L) equals 30 L of [tex]O_{2}[/tex] and 30 L/22.4 L times 1 mole equals 1.34 moles of [tex]O_{2}[/tex].
According to stoichiometry, when 1.34 moles of [tex]O_{2}[/tex]are reacted with [tex]SO_{2}[/tex], 2.68 moles of [tex]SO_{3}[/tex] are created, or 2/1 x 1.34 moles of [tex]SO_{3}[/tex].
This means that the amount of [tex]SO_{3}[/tex]produced will be (2.68 moles/1 mole) x 22.4 L = 60.03 L [tex]SO_{3}[/tex].
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Very quick can someone like help me
Answer:
Explanation:
dilute solutions of hydrochloric acid (HCl), sulphuric acid (H₂SO₄), and nitric acid (HNO₃) react with active metals to produce a salt and hydrogen gas.
Active metals react strongly and quickly with other elements and compounds due to the electrons in its structure and its ease of sharing the electrons with other elements.
The most active metals are found in Groups 1 and 2 of the Periodic Table (i.e. the left side), and include lithium, potassium, magnesium, and calcium. Metals such as aluminium, lead, and zinc, are less active than magnesium or calcium, but are generally still labelled as 'active'. Metals such as copper, gold, or silver are inactive and will not react.
Therefore, in the provided question, all the metals listed, except for copper, will produce a metal salt + hydrochloric acid. Copper will not react.
In general:
metal + HCl = metal chloride + H₂ metal + H₂SO₄ = metal sulphate + H₂metal + HNO₃ = metal nitrate + H₂For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.
For the reaction
N2(g)+3H2(g)↽−−⇀2NH3(g)
the standard change in Gibbs free energy is Δ°=−32.8 kJ/mol
. What is ΔG for this reaction at 298 K when the partial pressures are N2=0.350 atm
, H2=0.300 atm
, and NH3=0.750 atm
?
The ΔG for the reaction at 298 K and the given partial pressures is -55.53 kJ/mol.
What is ΔG ?
The Gibbs free energy change for a reaction under non-standard conditions can be calculated using the following equation:
ΔG = ΔG° + RTln(Q)
where ΔG is the Gibbs free energy change, ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in kelvin, and Q is the reaction quotient.
The reaction quotient, Q, can be calculated using the partial pressures of the gases involved in the reaction:
Q = (P(NH3))² / (P(N2) x P(H2)³)
Plugging in the given values, we get:
Q = (0.75 atm)² / (0.35 atm x 0.30 atm³) = 0.2667
Now we can calculate the ΔG for the reaction:
ΔG = ΔG° + RTln(Q)
ΔG = (-32.8 kJ/mol) + (8.314 J/(mol·K) x 298 K x ln(0.2667))
ΔG = -32.8 kJ/mol + (-22.73 kJ/mol)
ΔG = -55.53 kJ/mol
Therefore, the ΔG for the reaction at 298 K and the given partial pressures is -55.53 kJ/mol.
What is reaction quotient?
Reaction quotient, commonly denoted as Q, is a measure of the relative concentrations of reactants and products in a chemical reaction at a particular moment in time. It is calculated by dividing the concentration of the products raised to their stoichiometric coefficients by the concentration of the reactants raised to their stoichiometric coefficients.
The equation for the reaction quotient Q is similar to the equilibrium constant Kc, but with the concentrations of the reactants and products at any time during the reaction, rather than at equilibrium. When the reaction is at equilibrium, the reaction quotient is equal to the equilibrium constant.
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The ΔHvap of a certain compound is 32.77 kJ·mol−1 and its ΔSvap is 94.72 J·mol−1·K−1. What is the boiling point of this compound?
The compound's boiling point is around 73.65 °C.
Which spontaneous responses are positive?Unexpected Responses. Favorable reactions are those that cause the system's enthalpy to drop while its entropy rises. The reaction happens naturally when both of these conditions are true. Both endothermic and exothermic reactions, which absorb heat and release it, can occur spontaneously.
We can use the Gibbs-Helmholtz equation to solve for the boiling point:
ΔG = ΔH - TΔS
At boiling point, ΔG = 0, so we can solve for T:
T = ΔH/ΔS
Substituting the given values:
T = (32.77 kJ·mol−1) / (94.72 J·mol−1·K−1)
T = 346.8 K
Converting to Celsius:
Boiling point = 346.8 K - 273.15 = 73.65 °C
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How many moles is 3.4 g of HCI?
3.4 g of HCl is therefore equivalent to roughly 0.0932 moles of HCl.
What is the HCI's mole count?Since each molecule of HCl contains one mole of hydrogen and one mole of chlorine, the mole quantity of HCl equals the mole quantity of hydrogen and chlorine in the sample.
Approximately 36.46 g/mol is the molar mass of hydrochloric acid (HCl). We must divide the mass by the molar mass to determine the number of moles of HCl in 3.4 g of HCl:
Mass of HCl / molar mass of HCl equals moles of HCl.
3.4 grams per 36.46 grams per mole of HCl
A mole of HCl is equal to 0.0932 mol.
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The two possible units of molarity are
Answer: The units for molarity are moles/liter.
Similarly, the equation to find molarity is moles divided by liters.
Explanation:
mol / L is a unit of molar concentration. These are the number of moles of dissolved material per liter of solution. 1 mol / L is also called 1M or 1molar. Mol / m3 is also a unit of molar concentration.
Molarity is expressed in units of moles per liter (mol / L). This is a very common unit, so it has its own symbol, which is the uppercase M. A solution with a concentration of 5 mmol / l is called a 5 M solution or has a concentration value of 5 mol.
The molar concentration of the solution is equal to the number of moles of the solute divided by the mass of the solvent (kilogram), and the molar concentration of the solution is equal to the number of moles of the solute divided by the volume of the solution (liter). increase.