Answer:
D, potassium hydroxide
Explanation:
Nitric acid us given and is obviously an acid
Answer:
2 mol H
Explanation:
For every 2 mol of NaOH, we're reacting 2 mol of H2O. In order to figure out how many mol of H are needed, it needs to be set up stochiometrically. Starting off with the given value, 1 mol of NaOH, we can then make a mol to mol ratio. For 2 mol of NaOH, we have 2 mol of H2O. For every 2 mol of H2O, we have 4 mol of H (this is because we are multiplying the coefficient by the subscript: 2 × 2). Now, we can solve for our answer.
1 mol NaOH × (2 mol H₂O / 2 mol NaOH) × (4 mol H / 2 mol H₂O)
= 2 mol H
Thus, we get 2 mol of H are needed to completely react 1 mol of NaOH.
Answer:
r= 0.9949 (For 15,000)
r=0.995 (For 19,000)
Explanation:
We know that
Molecular weight of hexamethylene diamine = 116.21 g/mol
Molecular weight of adipic acid = 146.14 g/mol
Molecular weight of water = 18.016 g/mol
As we know that when adipic acid and hexamethylene diamine react then nylon 6, 6 comes out as the final product and release 2 molecule of water.
So
So
Mo= 226.32/2 =113.16 g/mol
Given that
Mn= 15,000 g/mol
So
15,000 = Xn x 113.16
Xn = 132.55
Now by using Carothers equation we know that
By calculating we get
r= 0.9949
For 19,000
19,000 = Xn x 113.16
Xn = 167.99
By calculating in same process given above we get
r=0.995
Group 2 contains soft, silver metals that are less metallic in character than group 1 elements. There is a fairly strong conductivity trend within each row, left to right, and a weaker trend top to bottom. The elements in group 2 are moderately good conductors, while the elements on the right are very poor conductors. <span>As you move vertically between rows, conductivity decreases overall, but slowly.
Group 2 have a generally low electronegativity. Electronegativity decreases moving left and down across the table.</span>
Chemical reaction: C₃H₇COOH → C₃H₇COO⁻ + H⁺.
c(<span>butanoic acid) = 0,100 M.
</span>α = 1,23% = 0,0123.
Ka = α² · c / 1 - α.
Ka = 0,0123² · 0,1 M / 1 - 0,0123.
Ka = 0,0000153 M.
Ka = c(C₃H₇COO⁻) · c(H⁺) / c(C₃H₇COOH).
c(H⁺) = α · c(C₃H₇COOH).
c(H⁺) = 0,0123 · 0,1 M = 0,00123 mol/L.
pH = -log c(H⁺).
pH = 2,91.
