Answer:
-191.7°C
Explanation:
P . V = n . R . T
That's the Ideal Gases Law. It can be useful to solve the question.
We replace data:
2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°
(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°
T° = 81.3 K
We convert T° from K to C°
81.3K - 273 = -191.7°C
Metals are on the left side of the table and nonmetals are on the left with metalloids between them. And the noble gases are all in group 18 of the periodic table.
They are too small to see with the naked eye
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
Answer:
m= 4,599.145 g
Explanation:
Let m = mass, d = density and V = volume of the osmium block.
m = d x V
m = 22.610 g/cm3 x (6.70 x 9.20 x 3.3) cm3
m = 4,599.145 g
