After three half-lives, the concentration of the reactant will be 1/8 of its initial concentration. This means that the remaining concentration of the reactant after three half-lives will be 8 m.
A second order reaction is one that has a rate proportional to the product of the concentration of two reactants or the square of the concentration of one reactant. In this case, the rate of the reaction is given by the equation:
r = k[A]²
The half-life of a reaction is the amount of time it takes for the concentration of the reactant to decrease by half. The half-life of a second-order reaction is given by the equation:
t½ = 1 / (k[A]₀)
Where k is the rate constant, [A]₀ is the initial concentration of the reactant, and t½ is the half-life of the reaction. After one half-life, the concentration of the reactant will be [A] = [A]₀ / 2
After two half-lives, the concentration of the reactant will be [A] = [A]₀ / 4
After three half-lives, the concentration of the reactant will be [A] = [A]₀ / 8
Given that the initial concentration of the reactant is 64 M, the concentration of the reactant after three half-lives is:
[A] = [A]₀ / 8[A] = 64 / 8[A] = 8 M
Therefore, the concentration of the reactant that is left after three half-lives is 8 M.
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how many electrons does cl want to gain? hint: how many are gained to form a stable noble gas electron configuration, ns2 np6 (octet rule)?
Chlorine (Cl) is a nonmetal, meaning it has the tendency to gain electrons to achieve the electron configuration of a noble gas. The noble gas electron configuration of the nearest noble gas, argon (Ar), is 1s2 2s2 2p6 3s2 3p6, with a total of 18 electrons.
Chlorine has 7 valence electrons, meaning it needs 1 more electron to achieve a stable noble gas electron configuration. Therefore, chlorine wants to gain 1 electron to achieve a stable noble gas configuration.
In terms of bonding, chlorine can either gain 1 electron to form an anion with a 1- charge or it can share electrons with another atom to form a covalent bond. Chlorine most commonly forms a single covalent bond with another atom, such as hydrogen, to form hydrogen chloride (HCl). In this case, both atoms share electrons to form a stable molecule.
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the amount of kinetic energy required to strain the chemical bonds in substrates so they can achieve the transition state is the definition of ?
The amount of kinetic energy required to strain the chemical bonds in substrates so they can achieve the transition state is the definition of activation energy.
What is Activation Energy?
Activation energy is the amount of energy required for a chemical reaction to occur. The energy that must be provided to molecules in order for them to react with one another is known as activation energy.
This can be accomplished in a variety of ways, such as by increasing the temperature or pressure, adding a catalyst, or irradiating the reactants with light.
Activation energy is defined as the energy required for the reaction to begin. It's the energy that molecules require to overcome the initial barrier so that a reaction may proceed.
When a chemical reaction occurs, the reactants must collide with one another with sufficient force and in the appropriate orientation to form products.
It's critical to note that activation energy is a form of potential energy that isn't included in the overall energy change of a reaction.
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2.37-l container is filled with 186 g argon. (a) if the pressure is 10.0 atm, what is the temperature? webassign will check your answer for the correct number of significant figures. k (b) if the temperature is 225 k, what is the pressure?
(a) If the pressure is 10.0 atm, the temperature is 62.0 K.
(b) if the temperature is 225 k, the pressure is 36.3 atm.
a) In order to calculate the temperature, we need to use the ideal gas law, PV = nRT, where P is the pressure, V is the volume of the container, n is the number of moles of argon, R is the ideal gas constant, and T is the temperature.
We can calculate the number of moles, n, by using the molar mass of argon, which is 39.948 g/mol.
We have n = 186 g / 39.948 g/mol = 4.656 mol.
So we can plug in our values and solve for T:
T = (10.0 atm)(2.37 L) / (4.666 mol)(0.08206 L·atm/mol·K) = 62.0 K.
b) To calculate the pressure, we can again use the ideal gas law, PV = nRT. We know the values of n, R, and T from the previous question.
Since the volume of the container is given, we can plug in these values to solve for P:
P = (4.666 mol)(0.08206 L·atm/mol·K)(225 K) / 2.37 L = 36.3 atm.
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does any solid cu(oh)2 form when 0.075 g koh is dissolved in 1.0 l of 1.0 x 10 -3 m cu(no3)2? ksp of cu(oh)2
Yes, a solid Cu(OH)2 will form when 0.075 g KOH is dissolved in 1.0 L of 1.0 x 10^-3 M Cu(NO3)2. 0.107 g of solid Cu(OH)2 will form.
First, we need to determine the amount of Cu2+ ions present in the solution:
1.0 x 10^-3 M Cu(NO3)2 means that there are 1.0 x 10^-3 moles of Cu2+ ions per liter of solution.
Next, we can use stoichiometry to determine the amount of OH- ions that will react with the Cu2+ ions to form Cu(OH)2. The balanced chemical equation for this reaction is:
Cu2+ (aq) + 2OH- (aq) → Cu(OH)2 (s)
For every 1 mole of Cu2+ ions, we need 2 moles of OH- ions. Therefore, the total amount of OH- ions needed to react with all of the Cu2+ ions in the solution is:
2 x 1.0 x 10^-3 mol = 2.0 x 10^-3 mol
Now we can use the Ksp of Cu(OH)2 to calculate the concentration of Cu2+ and OH- ions in the solution. The Ksp expression for Cu(OH)2 is:
Ksp = [Cu2+][OH-]^2
Since we know the Ksp value for Cu(OH)2, we can solve for either [Cu2+] or [OH-]. Let's solve for [OH-]:
Ksp = [Cu2+][OH-]^2
4.8 x 10^-20 = (1.0 x 10^-3 M)[OH-]^2
[OH-]^2 = 4.8 x 10^-17
[OH-] = 2.2 x 10^-9 M
Therefore, the concentration of OH- ions in the solution is 2.2 x 10^-9 M. Since we need 2 moles of OH- ions for every mole of Cu2+ ions, we know that the concentration of Cu2+ ions is half of the concentration of OH- ions:
[Cu2+] = 1.1 x 10^-9 M
Finally, we can use the molar mass of Cu(OH)2 to determine the mass of solid that will form:
Molar mass of Cu(OH)2 = 97.56 g/mol
1 mole of Cu(OH)2 is formed for every mole of Cu2+ ions, so the mass of Cu(OH)2 that will form is:
0.0011 mol x 97.56 g/mol = 0.107 g
Therefore, 0.107 g of solid Cu(OH)2 will form when 0.075 g KOH is dissolved in 1.0 L of 1.0 x 10^-3 M Cu(NO3)2.
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it is found that, when equilibrium is reached at a certain temperature, hi is 40. percent dissociated. calculate the equilibrium constant kc for the reaction at this temperature.
The equilibrium constant (Kc) is the ratio of the concentration of the products to the reactants at equilibrium. The value of Kc changes with the temperature but is constant at a given temperature.
The expression for the equilibrium constant Kc can be defined as follows:-
Kc = [C]^c[D]^d/[A]^a[B]^b
where [ ] denotes the molar concentration of the respective species. a, b, c, and d are the coefficients of the balanced chemical equation for the species A, B, C, and D.
If a chemical reaction is at equilibrium at a given temperature, the concentration of reactants and products remains constant over time. In other words, the rate of the forward reaction and the rate of the reverse reaction is equal.
The reaction for which we need to find the equilibrium constant is:-
HI(g) ↔ H(g) + I(g)
Now, assume that initially there were 'x' moles of HI in the reaction mixture. After the dissociation of HI, the concentration of H and I will be equal to 'x - y' moles. The concentration of HI will be equal to 'x - y' moles.
Here, y is the number of moles of HI that dissociated. According to the given statement, HI is 40% dissociated. Therefore, the number of moles of HI that dissociated will be 0.4x. Similarly, the number of moles of H and I that will be formed will also be 0.4x.
The equation for the dissociation of HI can be written as:-
HI(g) ↔ H(g) + I(g)
The initial number of moles = x Moles dissociated = 0.4x
At equilibrium, the number of moles of HI = x - 0.4x = 0.6x
Number of moles of H = 0.4x
Number of moles of I = 0.4x
Finally, substitute these values in the expression for the equilibrium constant:-
Kc = [H][I]/[HI]
Kc = (0.4x)(0.4x)/(0.6x)²
Kc = 0.16/0.36Kc = 0.4444 (approximately)
Therefore, the equilibrium constant Kc for the given reaction is 0.4444 (approximately).
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calculate the density (in grams per milliliter) for a glass marble with a volume of 7.94 ml and a mass of 15.36 g.
To calculate the density (in grams per milliliter) for a glass marble with a volume of 7.94 ml and a mass of 15.36 g, you must divide the mass by the volume. In this case, the density would be 1.93 g/mL.
To solve this problem mathematically:
Step 1: Identify the mass (m) and volume (v) of the marble.
Mass (m) = 15.36 g
Volume (v) = 7.94 mL
Step 2: Divide the mass by the volume to calculate the density.
Density (d) = m/v
Density (d) = 15.36 g / 7.94 mL
Density (d) = 1.93 g/mL
Therefore, the density of the glass marble is 1.93 g/mL.
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