Answer:
Aluminium was named after alum, which is called 'alumen' in Latin. This name was given by Humphry Davy, an English chemist, who, in 1808, discovered that aluminium could be produced by electrolytic reduction from alumina (aluminium oxide), but did not manage to prove the theory in practice.
Explanation:
Answer:
eukaryotic
Explanation:
all human cells—including those found in the brain, the heart, the muscles, and so on—are also eukaryotic.
Answer:
Rate = -1/2 Δ[SO<sub>2</sub>]/Δt
so its gonna be (in more simple terms) rate= -1/2Δ(SO2)/Δt
Explanation:
Part A
75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF
This combination will form a buffer.
Explanation
Here, weak acid HF and its conjugate base F- is available in the solution
Part B
150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
This combination cannot form a buffer.
Explanation
Here, moles of HF = 0.15 x 0.1 = 0.015 moles
Moles of HCl = 0.135 x 0.175 = 0.023
Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution
Part C
165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
This combination will form a buffer.
Explanation
Moles of HF = 0.165 x 0.1 = 0.0165 moles
Moles of KOH = 0.135 x 0.05 = 0.00675 moles
Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer
Part D
125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
This combination will form a buffer
Explanation
Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer
Part E
105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
This combination will form a buffer
Explanation
Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles
Moles of HCl = 0.095 x 0.1 = 0.0095 moles
Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
Answer:
0.04 mol
Explanation:
Given data:
Mass of barium = 5.96 g
Moles of barium = ?
Solution:
Formula:
Number of moles = mass/molar mass
Molar mass of barium = 5.96 g/ 137.33 g/mol
Number of moles = 0.04 mol
Thus the number of moles of barium in 5.96 g are 0.04 moles. The chemist weight out the 0.04 moles .
