Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of 
= 
= 0.0458 M
Concentration of 
= 
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72
pH ≅ 4.80
3) CH₃-COOH + NH₃ → CH₃-COO⁻NH₄⁺
4) 2 FeCl₃ + 3 Ag₂SO₃ → Fe₂(SO₃)₃ + 6 AgCl
5) 2 Al + 3 NiCl₂ → 2 AlCl₃ + 3 Ni
6) 4 LiCl + Pb(NO₂)₄ → 4 LiNO₂ + PbCl₄
7) 3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O
8) Cd(NO₃)₂ + Na₂S → CdS + 2 NaNO₃
9) Cr₂(SO₄)₃ + 3 (NH₄)₂CO₃ → Cr₂(CO₃)₃ + 3 (NH₄)₂SO₄
Answer:
5230J
Explanation:
Mass (m) = 250g
Initial temperature (T1) = 25°C
Final temperature (T2) = 30°C
Specific heat capacity (c) = 4.184J/g°C
Heat energy (Q) = ?
Heat energy (Q) = Mc∇T
Q = heat energy
M = mass of the substance
C = specific heat capacity
∇T = change in temperature = T2 - T1
Q = 250 × 4.184 × (30 - 25)
Q = 1046 ×5
Q = 5230J
The heat energy required to raise the temperature of 250g of water from 25°C to 30°C is 5230J
HCl and NaOH react in a 1:1 ratio, meaning that 1 H+ from HCl will react with 1 OH- from NaOH. Knowing this, and that molarity is mol/liter, all we need to do is use what we have available. First we must find the mols of HCl in our solution, so we set up the following equation in the following steps:
1. 24.75mL x (0.359mol NaOH / 1000mL) = 8.885 x 10^-3mol NaOH
This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
Here we just used the mols of NaOH we found to convert to mols of HCl using the 1:1 ratio described earlier.
From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
8.885 x 10^-3mol HCl/0.03475L = 2.557 x 10^-1M HCl
However, this is the molarity of the HCl and NaOH solution, not the original HCl solution. Using the dilution equation M1V1=M2V2, we can solve for the original molarity.
M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
V1 = the total volume that our mixture has (34.75mL = 0.03475L)
M2 = what we're trying to find
V2 = the amount of the original HCl that we had (10mL = 0.010L)
Simply solving for M2 gives us:
M2 = (M1V1) / V2 or:
M2=((2.557 x 10^-1) x 0.03475L) / 0.010L = 8.89 x 10^-1M HCl. That is your answer.
Answer:
Ethyl heptanoate (see the attached image for the structure).
Explanation:
- The reaction of heptanoic acid and ethanol will result ethyl heptanoate ester.
- Ethyl heptanoate is the ester resulting from the condensation of heptanoic acid and ethanol. It is used in the flavor industry because of its odor that is similar to grape.
- Kindly, see the attached image of the structure and the reaction.
