The internal energy : 1310.43 kJ/mol
Internal energy (ΔE) can be formulated for Calorimeter :
C= the heat capacity of the calorimeter
Δt=36.2-25.68=10.52°C
The change in internal energy when 2.67 grams of butane is combusted in a bomb calorimeter, given a temperature increase from 25.68 C to 36.2C and a heat capacity of 5.73 kJ/C for the calorimeter, is approximately 1308 kJ/mol.
To solve the problem of calculating the changes in internal energy when 2.67 grams of butane (C4H10) is combusted in a bomb calorimeter, it is necessary to understand calorimeter's heat capacity and how a bomb calorimeter works.
The first step will be to calculate the change in temperature which here is the final temperature subtracted from the initial temperature: 36.2 C - 25.68 C = 10.52 C.
Then, we multiply this temperature change by the heat capacity of the calorimeter to find the total heat produced by the reaction in kJ: 10.52 C * 5.73 kJ/C = 60.18 kJ.
The final step is to convert grams of butane to moles, because we are asked to find the energy change in kJ/mol. The molar mass of butane (C4H10) is approximately 58.12 g/mol. So we have approximately 2.67 g / 58.12 g/mol = 0.046 mol.
Finally, we divide the heat produced by the number of moles to get the energy change per mole of butane: 60.18 kJ / 0.046 mol = approximately 1308 kJ/mol.
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b. high point
c. speed
d. low point
Answer:
B
Explanation:
Answer:
Explanation:
the molar mass for Na2CO3 is 2*23+12+3*16=106 g/mole
106*0.251=26.606 grames
I got 15.4 once and now I got 61.4? Are they correct? if so, which one?
To neutralize the KOH solution, we need 61.4 mL of 1.33 mol L−1 H2SO4(aq).
To find the volume of the H2SO4 solution needed to neutralize the KOH solution, we can use the equation:
Mole of H2SO4 = Molarity of KOH x Volume of KOH
First, calculate the moles of KOH:
Moles of KOH = Molarity of KOH x Volume of KOH = 0.830 mol/L x (49.3 mL / 1000 mL) = 0.04089 mol
Since H2SO4 is a diprotic acid and KOH is a strong base, the reaction will be:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
Therefore, the ratio between the moles of H2SO4 and KOH is 1:2. This means that twice the moles of KOH will be needed to neutralize the H2SO4. Calculate the moles of H2SO4 needed:
Moles of H2SO4 needed = 2 x Moles of KOH
= 2 x 0.04089 mol
= 0.08178 mol
Finally, calculate the volume of the H2SO4 solution needed:
Volume of H2SO4 = Moles of H2SO4 / Molarity of H2SO4 = 0.08178 mol / 1.33 mol/L
= 0.0614 L
= 61.4 mL
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Answer : The initial rate of the reaction at 298 K is,
Explanation :
The Arrhenius equation is written as:
Taking logarithm on both the sides, we get:
............(1)
where,
k = rate constant
Ea = activation energy
T = temperature
R = gas constant = 8.314 J/K.mole
A = pre-exponential factor
The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.
Thus, if we plot a graph of vs then the graph shows a straight line with negative slope. That means,
Slope of the line =
And,
Intercept =
As we are given that:
Slope of the line = -982.7 =
Intercept = -0.0726 =
Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:
The value of rate constant is,
Now we have to calculate the initial rate of the reaction at 298 K.
As we know that the slow step is the rate determining step. So,
The slow step reaction is,
The expression of rate law for this reaction will be,
As we are given that:
[A] = 0.500 M
[B] = 0.0500 M
k =
Now put all the given values in the rate law expression, we get:
Therefore, the initial rate of the reaction at 298 K is,
b. mmol NaOH Initial concentration of NaOH
c. mL H2O Initial concentration of HCl
Answer:
The following parameters:
Explanation:
The options have been incorrectly arranged in the question statement given here. The three conditions mentioned above are what would be observed to remain the same when this experiment is carried out.
b. 8.5 x 10-6 M
c. 6.3 x 10-11 M
d. 1.3 x 10-12 M
e. 5.0 x 10-2 M
f. 1.8 x 10-5 M
Answer:
c. 6,3x10⁻¹¹M
Explanation:
The solubility of a buffer is defined as the concentration of the dissolved solid in a saturated solution. For the Cd(OH)₂, solubility is:
[Cd²⁺] = S
The dissolution of Cd(OH)₂ is:
Cd(OH)₂ ⇄ Cd²⁺ + 2OH⁻
And the ksp is defined as:
ksp = [Cd²⁺][OH⁻]²
As ksp = 2,5x10⁻¹⁴ and [OH⁻] at pH=12,30 = 10^-(14-12,30) = 0,01995M
2,5x10⁻¹⁴ = [Cd²⁺]×(0,01995M)²
[Cd²⁺] = 6,3x10⁻¹¹M
That means solubility is c. 6,3x10⁻¹¹M
I hope it helps!
The molar solubility of Cd(OH)2 when buffered at a pH of 12.30 can be calculated using the concept of hydrolysis. The correct answer is 6.3 x 10^(-11) M.
To calculate the molar solubility of Cd(OH)2 when buffered at a pH of 12.30, we need to use the concept of hydrolysis. Cd(OH)2 is a slightly soluble salt that undergoes hydrolysis in aqueous solution. At a high pH value, OH- ions react with water to form more OH- ions, shifting the equilibrium towards the hydrolysis reaction.
After performing the calculations, the molar solubility of Cd(OH)2 when buffered at a pH of 12.30 is approximately 6.3 x 10^(-11) M. Therefore, the correct answer is option c. 6.3 x 10^(-11) M.
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