Answer:
The bottle is labeled pure so it probably is pure.
Explanation:
The second volume : V₂= 0.922 L
<h3>
Further explanation
</h3><h3>Given
</h3>
7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
Answer:
A)
,
, 
A = 1.5×
, A = 1.9×
, A=1.5×
B) 4.469
Explanation:
From Arrhenius equation

where; K = Rate of constant
A = Pre exponetial factor
= Activation Energy
R = Universal constant
T = Temperature in Kelvin
Given parameters:




taking logarithm on both sides of the equation we have;

since we have the rate of two different temperature the equation can be derived as:


= 19846.04×7.544×
= 1.497
=
= 4.469
Products are copper+ aluminium chloride
reactants are aluminium+copper chloride