Answer:
Part A
Ag+ is the Lewis acid and NH3 is the Lewis base.
Part B
AlBr3 is the Lewis acid and NH3 is the Lewis base.
Part C
AlCl3 is the Lewis acid and Cl− is the Lewis base.
Explanation:
A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.
A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .
Answer:
40 moles of O₂
30 moles of CO₂
Explanation:
Given parameters:
Number of moles of C₃H₄ = 10moles
Unknown:
Number of moles of CO₂ = ?
Solution:
The number of moles helps to understand and make quantitative measurements involving chemical reactions.
We start by solving this sort of problem by ensuring that the given equation is properly balanced;
C₃H₄ + 4O₂ → 3CO₂ + 2H₂O
We can clearly see that all the atoms are conserved.
Now, we work from the known to unknown. We know the number of moles of C₃H₄ to be 10moles;
1 mole of C₃H₄ reacted with 4 moles of O₂
10 moles of C₃H₄ will react with 10 x 4 = 40moles of O₂
1 mole of C₃H₄ will produce 3 moles of CO₂
10 moles of C₃H₄ will produce 10 x 3 = 30moles of CO₂
Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold Foil Experiment to observe the effect of alpha particles on matter.
Answer:
88,7 mL of solution
Explanation:
Molarity (Represented as M) is an unit of chemical concentration that is defined as the ratio between moles of solute per liters of solution, that is:
Molarity = moles of solute / Liters of solution
If molarity of KCN solution is 0,0820M and moles of KCN are 7,27x10⁻³ moles:
0,0820M = 7,27x10⁻³ moles / Liters of solution
Liters of solution = 0,0887L = <em>88,7 mL of solution</em>
I hope it helps!
Answer:
15.04 mL
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 21 L
V₂ = ?
P₁ = 9 atm
P₂ = 15 atm
T₁ = 253 K
T₂ = 302 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 15.04 mL</u>