A75.0-ml volume of 0.200 m nh3 (kb=1.8×10−5) is titrated with 0.500 m hno3. calculate the ph after the addition of 19.0 ml of hno3. express your answer numerically.

8.03.

Explanation:

C of base (NH₃) = [(MV)NH₃ - (MV)HNO₃] / Vtotal.

C of base (NH₃) = [(0.20 M)(75.0 mL) - (0.50 M)(19.0 mL)] / (94.0 mL) = 0.0585 M.

∵ [OH⁻] = √(Kb.C)

∴ [OH⁻] = √(1.8 x 10⁻⁵)(0.0585 M) = 1.053 x 10⁻⁶.

∵ pOH = - log[OH⁻] = - log (1.053 x 10⁻⁶) = 5.97.

∴ pH = 14 - pOH = 14 - 5.97 = 8.03.

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).