Hellium argon neon xenon krypton randon oxygen fluorine chlorine bromine
So, you need to have same ammount of atoms on the left and on the right side of the equation. You need to count the ammount of attoms of every substance on the left, and make sure that on the right side the ammount is same. For example in the 1st one it’s 6Sn+2P4=2Sn3P4, so that you have 6atoms of Sn on the left and 6 atoms of Sn on the right, same with the P
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
Stars born larger than 8 solar masses usually retain enough mass to undergo core collapse, with the resulting shock wave producing a Type Ib supernova (spectra without Hydrogen or Silicon lines, with Helium lines), a Type Ic supernova (without Hydrogen or Helium or Silicon lines) or a Type II supernova
I hope it’s what you need
