Answer:
The molar mass in g/mol is 121.4 g/m
Explanation:
Let's apply the Ideal Gases Law to solve this:
P . V = n . R. T
V = 125 mL → 0.125L
P = 754 Torr
760 Torr ___ 1 atm
754 Torr ____ (754 / 760) = 0.992 atm
Moles = Mass / Molar mass
0.992 atm . 0.125L = (0.495 g / MM) . 0.082 . 371K
(0.992 atm . 0.125L) / (0.082 . 371K) = (0.495 g / MM)
4.07x10⁻³ mol = 0.495 g / MM
MM = 0.495 g / 4.07x10⁻³ mol → 121.4 g/m
The element iodine (I) is important for the fast and hastened metamorphosis of frog-tadpoles. As amphibians, the tadpoles can live in water and land but when they are born they are iodine-deficient. Tadpoles that do not receive ample amount of iodine become tadpoles until the end of their days.
Answer:
5.8μg
Explanation:
According to the rate or decay law:
N/N₀ = exp(-λt)------------------------------- (1)
Where N = Current quantity, μg
N₀ = Original quantity, μg
λ= Decay constant day⁻¹
t = time in days
Since the half life is 4.5 days, we can calculate the λ from (1) by substituting N/N₀ = 0.5
0.5 = exp (-4.5λ)
ln 0.5 = -4.5λ
-0.6931 = -4.5λ
λ = -0.6931 /-4.5
=0.1540 day⁻¹
Substituting into (1) we have :
N/N₀ = exp(-0.154t)----------------------------- (2)
To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:
N = 5.0 μg
N₀ = Unknown
t = 1 day
Substituting into (2) we have
[5/N₀] = exp (-0.154 x 1)
5/N₀ = 0.8572
N₀ = 5/0.8572
= 5.8329μg
≈ 5.8μg
The Chemist must order 5.8μg of 47-CaCO3
<span>A pulse with an amplitude of 3+ would be considered as increased.
Peripheral Pulse Assessment Grading System is measured in 0 - 3 Scale.
0 = absent
1+ = Weak/thready pulse
2+ Normal Pulse
3+ = Full, firm pulse.
from the above scale we can find that the 3+ reading shows that the pulse is increased.</span>
Answer:
The correct option is c
Explanation:
The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image
Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers
