Answer:
Explanation:
a )
m = m₀ 
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ 
b )
m = 50 x 
= 40.21 mg .
c )
40 = 50 
.8 =
= 1.25
.0004358 t = .22314
t = 512 years .
Answer:
Volume will goes to increase.
Explanation:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
So when the temperature goes to increase the volume of gas also increase. Higher temperature increase the kinetic energy and molecules move randomly every where in given space so volume increase.
Now we will put the suppose values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 4.5 L × 348 K / 298 k
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
Hence prove that volume increase by increasing the temperature.
<span>To compute 4.659×104−2.14×104, the first step is the factorization. That is as follows:4.659×104−2.14×104= 10^4.(4.659−2.14), the next step is to compute 4.659−2.14=2.51, so 10^4.(4.659−2.14)=2.51x10^4=2.51x 10000 (because10^4=10000), the last calculus is 2.51x 10000=25100, the final answer is 25,000.Hope this helps. Let me know if you need additional help!</span>
Answer:
The OH group
Explanation:
Benzhydrol contains OH hydroxyl group in its molecule while fluorene does not. At first glance, one would think that OH, which contributes to hydrogen bonding would causes melting point of benzhydrol to be higher than fluorene. <em>However, </em>the structure of benzhydrol, which is 2 benzene rings connected to center hydroxyl carbon (PhCOHPh), allows for each benzene rings in benzhydrol to rotate until both rings are perpendicular to minimize repulsive force. This prevents the molecule from stacking on each other due to its non flat shape, and thus, lowering its melting point in contrast to flat fluorene molecule.
