Ka x Kb = Kw Kw = 1 x 10⁻¹⁴
(1.3 x 10⁻⁷) x Kb = 1 x 10⁻¹⁴
Kb = ( 1 x 10⁻¹⁴ ) / ( 1.3 x 10⁻⁷)
Kb = 7.7 x 10⁻⁸
Answer D
hope this helps!
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.
2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.
Answer:
36°C
Explanation:
Given parameters:
Mass of aluminum = 725g
Quantity of heat = 2.35 x 10⁴J
Unknown:
Temperature change = ?
Solution:
To solve this problem, we simply use the expression below:
The quantity of energy is given as:
Q = m C Δt
Q is the quantity of energy
m is the mass
C is the specific heat capacity of aluminum = 0.9J/g°C
Δt is the change in temperature
The unknown is Δt;
Δt =
=
= 36°C
Answer:
it's a
Explanation:
<h3> the atomic mass is the number of protons and neutrons in an element added together</h3>
<span>There would be two electrons in the 1s orbital one oriented up and one oriented down. The other electron would be in the 1p orbital up or down.
Remember that 1s can only hold 2 electrons at a time, and when it is filled there one electron is spinning up and the other is spinning down. After 1s, comes the 1p orbital which can hold 6 electrons. Until 3 electrons are spinning either up or down, the next one will have the next orientation.</span>