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Answer:
The three isomers having the molecular formula are drawn in the figure below.
Explanation:
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Answer:
12 mol CO₂
General Formulas and Concepts:
Atomic Structure
- Compounds
- Moles
- Mole Ratio
Stoichiometry
- Analyzing reactions rxn
- Using Dimensional Analysis
Explanation:
Step 1: Define
Identify
[rxn] 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
[Given] 2 mol C₆H₁₂O₆
[Solve] mol CO₂
Step 2: Identify Conversions
[rxn] 6CO₂ → C₆H₁₂O₆
Step 3: Convert
- [DA] Set up:
- [DA] Multiply [Cancel out units]:
Answers
It states that pressure of gas is directly proportional to the temperature at constant volume.
P/T = k
Where k - constant
P1/T1 = P2/T2
Where parameters for the first instance are on the left side and parameters for the second instance on the right side of the equation.
3.74 atm /394 K = P/789 K
P = 7.49 atm
Answer:
7.49 atm
Explanation:
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In this case, the problem is asking for the balance of a redox reaction in acidic media, in which nickel is reduced to a metallic way and nitrogen oxidized to an ionic way.
Thus, according to the given information, it turns out possible for us to balance this equation in acidic solution by firstly setting up the half reactions:
Next, we cross multiply each half-reaction by the other's carried electrons:
Finally, we add them together to obtain:
Which can be all simplified by a factor of 2 to obtain:
Hence, the coefficients in front of Ni and H⁺ are 4 and 10 respectively.
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Answer:
The activation energy is
Explanation:
The gas phase reaction is as follows.
The rate law of the reaction is as follows.
The reaction is carried out first in the plug flow reactor with feed as pure reactant.
From the given,
Volume "V" =
Temperature "T" = 300 K
Volumetric flow rate of the reaction
Conversion of the reaction "X" = 0.8
The rate constant of the reaction can be calculate by the following formua.
Rearrange the formula is as follows.
The feed has Pure A, mole fraction of A in feed is 1.
= change in total number of moles per mole of A reacte.
Substitute the all given values in equation (1)
Therefore, the rate constant in case of the plug flow reacor at 300K is
The rate constant in case of the CSTR can be calculated by using the formula.
The feed has 50% A and 50% inerts.
Hence, the mole fraction of A in feed is 0.5
= change in total number of moles per mole of A reacted.
Substitute the all values in formula (2)
Therefore, the rate constant in case of CSTR comes out to be
The activation energy of the reaction can be calculated by using formula
In the above reaction rate constant at the two different temperatures.
Rearrange the above formula is as follows.
Substitute the all values.
Therefore, the activation energy is
Answers
Answer:
pH at the equivalence point is 8.6
Explanation:
A titulation between a weak acid and a strong base, gives a basic pH at the equivalence point. In the equivalence point, we need to know the volume of base we added, so:
mmoles acid = mmoles of base
60 mL . 0.1935M = 0.2088 M . volume
(60 mL . 0.1935M) /0.2088 M = 55.6 mL of KOH
The neutralization is:
HBz + KOH ⇄ KBz + H₂O
In the equilibrum:
HBz + OH⁻ ⇄ Bz⁻ + H₂O
mmoles of acid are: 11.61 and mmoles of base are: 11.61
So in the equilibrium we have, 11.61 mmoles of benzoate.
[Bz⁻] = 11.61 mmoles / (volume acid + volume base)
[Bz⁻] = 11.61 mmoles / 60 mL + 55.6 mL = 0.100 M
The conjugate strong base reacts:
Bz⁻ + H₂O ⇄ HBz + OH⁻ Kb
0.1 - x x x
(We don't have pKb, but we can calculate it from pKa)
14 - 4.2 = 9.80 → pKb → 10⁻⁹'⁸ = 1.58×10⁻¹⁰ → Kb
Kb = [HBz] . [OH⁻] / [Bz⁻]
Kb = x² / (0.1 - x)
As Kb is so small, we can avoid the quadratic equation
Kb = x² / 0.1 → Kb . 0.1 = x²
√ 1.58×10⁻¹¹ = [OH⁻] = 3.98 ×10⁻⁶ M
From this value, we calculate pOH and afterwards, pH (14 - pOH)
- log [OH⁻] = pOH → - log 3.98 ×10⁻⁶ = 5.4
pH = 8.6
Final answer:
To calculate the pH at equivalence in a titration, we need to consider the concentration of the excess strong base in the solution. First, we calculate the moles of the acid and the base, then we find the moles of the excess base. Using this information, we can find the concentration of the excess base and subsequently calculate pOH. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.
Explanation:
pH at the equivalence point in a titration can be determined by considering the concentration of the excess strong base present in the reaction mixture. In this case, the excess strong base is KOH. We can calculate [OH-] using the stoichiometry of the reaction and the given concentrations. Then, we can find the pOH using the formula -log[OH-]. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.
Given:
- Volume of benzoic acid solution (HC (H5CO2)): 60.0 mL
- Concentration of benzoic acid solution: 0.1935 M
- Concentration of KOH solution: 0.2088 M
Step 1: Determine the amount of benzoic acid (HC (H5CO2)) in moles:
moles of HC (H5CO2) = volume (L) × concentration (M) = 0.0600 L × 0.1935 M = 0.01161 mol
Step 2: Determine the amount of KOH in moles:
moles of KOH = volume (L) × concentration (M) = 0.0600 L × 0.2088 M = 0.01253 mol
Step 3: Determine the amount of excess KOH in moles:
moles of excess KOH = moles of KOH - moles required for neutralizing HC (H5CO2) = 0.01253 mol - 0.01161 mol = 9.2 × 10-4 mol
Step 4: Determine the concentration of excess KOH:
concentration of excess KOH = moles of excess KOH / volume (L) = 9.2 × 10-4 mol / 0.0600 L = 0.0153 M
Step 5: Determine the pOH of the solution:
pOH = -log[OH-] = -log(0.0153) ≈ 1.82
Step 6: Determine the pH of the solution:
pH = 14 - pOH = 14 - 1.82 ≈ 12.18
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Answers
Answer:
See explanation below
Explanation:
To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:
m = m₀e^-kt (1)
In this case, k will be the constant rate of this element. This is calculated using the following expression:
k = ln2/t₁/₂ (2)
Let's calculate the value of k first:
k = ln2/2.7 = 0.2567 d⁻¹
Now, we can use the expression (1) to calculate the remaining mass:
m = 8.1 * e^(-0.2567 * 2.6)
m = 8.1 * e^(-0.6674)
m = 8.1 * 0.51303
m = 4.16 mg remaining
Final answer:
The half-life of gold-198 is the time it takes for half of it to decay. Given that the half-life is 2.7 days, and the period in consideration is 2.6 days, approximately half of the original amount of 8.1 mg, which is 4.05 mg, will remain.
Explanation:
This problem is related to the concept of half-life in radioactive decay. The half-life of a substance is the time it takes for half of it to decay. As the half-life of gold-198 is 2.7 days and we are considering a period of 2.6 days, which is almost one half-life, therefore, approximately half the substance should have decayed.
So, if you start with 8.1 mg of gold-198, at the end of one half-life (or close to it at 2.6 days), you should have approximately half of this amount remaining. Half of 8.1 mg is 4.05 mg, thus, approximately 4.05 mg remains after 2.6 days.
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