Answer:
D
Explanation:
We know that the
reaction catalyzing power of a catalyst ∝ surface area exposed by it
Given
volume V1= 10 cm^3
⇒
hence r= 1.545 cm
also, surface area S1= 
now when the sphere is broken down into 8 smaller spheres
S2= 8×4πr'^2
now, equating V1 and V2 ( as the volume must remain same )

and solving we get
r'= r/2
therefore, S2=
S2=
S2= 2S1
hence the correct answer is
. The second run has twice the surface area.
Answer:
D. 15.8atm
Explanation:
Given parameters:
Initial pressure = 13atm
Initial temperature = 34°C = 34 + 273 = 307K
Final temperature = 100°C = 100 + 273 = 373K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.
The expression is shown mathematically below;
=
P and T pressure and temperature values
1 and 2 are initial and final states
Insert the parameters and solve for T₂;
=
P₂ = 15.8atm
The vertical columns are groups
The horizontal rows are periods
Given that the volume and amount of water are kept constant,
P/T = constant
P₁/T₁ = P₂/T₂
Normal atmospheric pressure is 746 mmHg and normal boiling point of water is 100 °C.
746/100 = 589/T₂
T₂ = 79.0 °C
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.