Response:
The pKa value is 13.0.
Clarification:
pKa + pKb = 14
For trimethylamine, Kb = 6.3 × 
Calculating pKb: pKb = - log (6.3 × )
= 1.0
Thus, pKa = 14 - pKb = 14 - 1.0
pKa = 13.0
Verification: The typical range for pKa in weak acids is from 2 to 13.
Solution:
The gas's new temperature is 604K
Justification:
Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;
Step 1: Formulate the general gas law equation
P1V1/T1 = P2V2/T2
Step 2: Insert the values, converting as needed to standard units.
P1 = 0.800 atm
V1 = 0.180 L
T1 = 29°C = 273 + 29 = 302K
P2 = 3.20 atm
V2 = 90 mL = 90 * 10^-3 L = 0.09 L
Step 3: Solve for T2
The new gas temperature T2 is calculated as:
T2 = P2V2T1/(P1V1)
T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)
T2 = 86.976 / 0.144
T2 = 604K
The gas's new temperature is 604K.
Answer:
Explanation:
Diethyl malonate possesses greater acidity compared to monocarbonyl substances (pKa=13) because its alpha hydrogens are linked to two carbonyl groups. Consequently, the malonic ester can be readily changed into its enolate ion by reacting it with sodium ethoxide in ethanol. When the malonic ester undergoes alkylation, a hydrogen atom in the alpha position becomes acidic, permitting another round of alkylation to yield a dialkylated malonic ester.
In this scenario, when diethyl malonate interacts with urea in the presence of sodium ethoxide base, the second alkylation step occurs within the molecule, producing a cyclic compound known as barbituric acid.
In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.
To find the answer, start by calculating the total mass of the copper utilized:
Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu
Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:
1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.
Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu
= 4.72 moles CuFeS2
The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole
CuFeS2 = 866.49 g CuFeS2
