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The second-order rate law for the reaction A -> products is given by:
rate = k[A]^2
where k is the rate constant and [A] is the concentration of A.
We can use the integrated rate law for a second-order reaction to solve for the concentration of A at a given time t:
1/[A] - 1/[A]₀ = kt
where [A]₀ is the initial concentration of A.
We are given k = 0.540 M¯¹s¯¹ and we want to find the time at which [A] = 1/3[A]₀.
So, we can rearrange the integrated rate law equation to solve for t:
t = (1/[A] - 1/[A]₀) / k
Plugging in [A] = 1/3[A]₀ and simplifying:
t = (1/(1/3[A]₀) - 1/[A]₀) / k
t = (3/[A]₀ - 1/[A]₀) / k
t = 2/[A]₀k
Substituting the given values:
t = 2 / (0.540 M¯¹s¯¹ x [A]₀)
So, to find the time at which [A] = 1/3[A]₀, we need to know the initial concentration of A. If we assume an initial concentration of [A]₀ = 1.0 M, then:
t = 2 / (0.540 M¯¹s¯¹ x 1.0 M) = 3.7 s
Therefore, the concentration of A will be 1/3 of its original concentration after 3.7 seconds.
We can use the ideal gas law to determine the pressure of the gas:3. A 17.3 L container was filled with 0.500 mol of a gas at 200C.
What is the pressure of this gas in atm?
We can use the ideal gas law to determine the pressure of the gas:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.08206 L atm K¯¹mol¯¹), and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin:
T = 200°C + 273.15 = 473.15 K
Now we can plug in the values we have been given:
P(17.3 L) = (0.500 mol)(0.08206 L atm K¯¹mol¯¹)(473.15 K)
Solving for P:
P = (0.500 mol)(0.08206 L atm K¯¹mol¯¹)(473.15 K) / (17.3 L) = 6.05 atm
Therefore, the pressure of the gas in the container is 6.05 atm.
We can use the ideal gas law to determine the pressure of the gas:3. A 17.3 L container was filled with 0.500 mol of a gas at 20 C.
What is the pressure of this gas in atm?
We can use the ideal gas law to determine the pressure of the gas:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.08206 L atm K¯¹mol¯¹), and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin:
T = 20°C + 273.15 = 293.15 K
Now we can plug in the values we have been given:
P(17.3 L) = (0.500 mol)(0.08206 L atm K¯¹mol¯¹)(293.15 K)
Solving for P:
P = (0.500 mol)(0.08206 L atm K¯¹mol¯¹)(293.15 K) / (17.3 L) = 2.50 atm
Therefore, the pressure of the gas in the container is 2.50 atm.
The product formed by mixing CaO (calcium oxide) with H2O (water) is Ca(OH)2 (calcium hydroxide). Calcium hydroxide is also known as slaked lime or hydrated lime.7. CaO is mixed in H2O to produce Ca(OH)2.
What is the name of the product?
The product formed by mixing CaO (calcium oxide) with H2O (water) is Ca(OH)2 (calcium hydroxide). Calcium hydroxide is also known as slaked lime or hydrated lime.
To determine the limiting reactant and the maximum theoretical moles of the product, we need to compare the amount of each reactant to their stoichiometric ratios in the balanced chemical equation.Aluminum metal weighing 2.82 g is exposed to 2.56 g oxygen gas to produce aluminum oxide.
What is the limiting reactant and what is the maximum theoretical moles of the product?
The product formed by mixing CaO (calcium oxide) with H2O (water) is Ca(OH)2 (calcium hydroxide). Calcium hydroxide is also known as slaked lime or hydrated lime.
Fe2+ is the ion formed when an iron atom loses two electrons. When Fe loses two electrons, the electronic configuration changes to 1s2 2s2 2P6 3s2 3s2 3p6 3d6.Iron has an atomic numberof26and has an electronic configuration 1s2 2s2 2P6 3s2 3s2 4s2 3d6
How many d electrons are present in Fe 2+ ?
Fe2+ is the ion formed when an iron atom loses two electrons. When Fe loses two electrons, the electronic configuration changes to 1s2 2s2 2P6 3s2 3s2 3p6 3d6.
In Fe2+, the ion has a +2 charge, which means it has lost two electrons. Therefore, the electronic configuration of Fe2+ is:
1s2 2s2 2P6 3s2 3s2 3p6 3d6
In this configuration, there are 6 d electrons present, because Fe2+ has lost the two outermost 4s electrons, but the d electrons remain unchanged. Therefore, there are 6 d electrons present in Fe2+.