Answer:
54%
Explanation:
The balanced equation is:
The first step is to determine the limiting reactant. For this, we calculate the moles of each given component and divide the result for the stoichiometric coefficient.
3.4 g octane / 114.23 g/mol = 0.030 mol octane
0.030 mol octane/2=0.015
5.9 g O2 / 32 g/mol = 0.18 mol O2
0.18 mol O2/25= 0.0074 mol
The lower number, in this case oxygen, is the limiting reactant. The value corresponds to the theoretical yield of the reaction.
Similarly, the real yield is calculated from the product.
2.80 g CO2/ 44.01 g/mol = 0.0636 mol CO2
0.0636 mol CO2/16 = 0.00398 mol
The percent yield is the ratio of the 2 multiplied by a hundred, then
Percent yield= 0.0398/0.0074 *100 = 54%
The percent yield of carbon dioxide from the combustion reaction of octane and oxygen, given the provided masses of the reactants and the yield of CO2, is calculated to be 26.7%
The combustion of octane in oxygen yields carbon dioxide and water in a 1:1 ratio, as the balanced chemical equation for the reaction is 2C8H18 + 25O2 -> 16CO2 + 18H2O. To calculate the percent yield of CO2, we first need to determine the theoretical yield of CO2. We can use the provided masses of octane and O2 and their respective molar masses to calculate the number of moles of each reactant:
- Moles of octane = 3.4 g / 114.22 g/mol = 0.0298 mol
- Moles of O2 = 5.9 g / 32.00 g/mol = 0.184 mol
Now, using the stoichiometric relationship from the balanced chemical equation, we can calculate the theoretical yield of CO2:
- Theoretical yield = 0.0298 mol octane x 16 mol CO2/2 mol octane = 0.238 mol CO2
Next, we convert this to grams using the molar mass of carbon dioxide:
- Theoretical yield = 0.238 mol CO2 x 44.01 g/mol = 10.5g CO2
Now that we have both the actual yield (2.80 g) and the theoretical yield (10.5 g), we can calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) x 100% = (2.80 g / 10.5 g) x 100% = 26.7%
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b) is has decreased
c)it has remained uncharged
d) it has at minimum been halved
Answer is: a) is has increased.
There are two types of reaction:
1) endothermic reaction (chemical reaction that absorbs more energy than it releases).
For example, the breakdown of ozone is an endothermic process. Ozone has lower energy than molecular oxygen (O₂) and oxygen atom, so ozone need energy to break bond between oxygen atoms.
2) exothermic reaction (chemical reaction that releases more energy than it absorbs).
For example, ΔH(reaction) = -225 kJ/mol; this is exothermic reaction.
Answer : The concentration of ion, pH and pOH of solution is, , 4.98 and 9.02 respectively.
Explanation : Given,
Concentration of ion =
pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.
The expression used for pH is:
First we have to calculate the pH.
The pOH of the solution is, 9.02
Now we have to calculate the pH.
The pH of the solution is, 4.98
Now we have to calculate the concentration.
The concentration is,
Answer:
pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978
Explanation:
Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C
As, pOH = - log [OH⁻]
⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022
The self-ionisation constant of water is given by
Kw = [H⁺] [OH⁻] and pKw = pH + pOH
Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.
Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴
⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M
also,
pH + pOH = pKw = 14
⇒ pH = 14 - pOH = 14 - 9.022 = 4.978
Answer:
The mixture is not in equilibrium, the reaction will shift to the left.
Explanation:
Based on the equilibrium:
Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺
kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]
Where [] are concentrations at equilibrium. The reaction is in equilibrium when the ratio of concentrations = kc
Q is the same expression than kc but with [] that are not in equilibrium
Replacing:
Q = [10.0M] [1.0M] / [0.1M] [0.1M]
Q = 1000
As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc
The mixture's equilibrium status can be determined by comparing the reaction quotient (Q) with the equilibrium constant (Kc). If Q < Kc, the reaction proceeds to the right (products) to achieve equilibrium. If Q > Kc, the reaction proceeds to the left (reactants) to achieve equilibrium.
To determine if the mixture is initially at equilibrium, we need to calculate and compare the reaction quotient (Q) and the equilibrium constant (Kc) of the reaction. The reaction quotient is a measure of the relative concentrations of products and reactants at any point in time, whereas Kc, is the measure of these concentrations only at equilibrium.
Assuming that the reaction in question is: Fe3+ + HSCN ↔ FeSCN2+ + H + . In this case,
Q = [FeSCN2+]/[Fe3+][HSCN] = 10 / (0.1 * 0.1) = 1000. If Kc is less than 1000, the reaction is not at equilibrium and will need to proceed to the left (reactants) to reach equilibrium. Conversely, if Kc is greater than 1000, the reaction is not at equilibrium and will need to proceed to the right (products).
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Answer:
The final volume
Explanation:
between work ( w), pressure ( P ) and volume ( V ) is the following:
w=−PΔV
where,
ΔV=V2−V1
It was stated that the gas is expanding, then the work is done by the system and it is of a negative value .
Note that work, should be expressed in 1L⋅atm=101.3J
CHECK THE ATTACHMENT FOR DETAILED EXPLATION
Answer:
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
Explanation:
Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.
In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
In a solution it is fulfilled:
Ci* Vi = Cf* Vf
where:
In this case:
Replacing:
1.40 M* 20 mL= 0.088 M* Vf
Solving:
Vf= 318.18 mL
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.
To dilute a solution, you can use the formula:
M1V1 = M2V2
where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. Rearranging the formula, we can solve for V2:
V2 = (M1 · V1) / M2
Plugging in the values given:
V2 = (1.40 M · 20.0 mL) / 0.0880 M = 318.18 mL
To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.
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