Answer:
8,5 E10 nm
Explanation:
⇒ 0.85 E-2 Km * ( 1000 m / Km ) * ( 1 E9 nm / m ) = 8.5 E10 nm
⇒ 0.85 E-2 Km = 8.5 E10 nm
Answer:
Half life is 6 years.
Explanation:
T½ = In2 / λ
Where λ = decay constant.
But N = No * e^-λt
Where N = final mass after a certain period of time
No = initial mass
T = time
N = 0.625g
No = 10g
t = 24 years
N = No* e^-λt
N / No = e^-λt
λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)
λ = -(1 / 24) * In (0.625/10)
λ = -0.04167 * In(0.0625)
λ = -0.04167 * (-2.77)
λ = 0.1154
T½ = In2 / λ
T½ = 0.693 / 0.1154
T½ = 6.00 years.
The half life of radioactive cobalt-60 is 6 years
Explanation:
The given data is as follows.
= 98.70 kPa = 98700 Pa,
T = = (30 + 273) K = 303 K
height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)
Density = 13.534 g/mL =
= 13534
The relation between pressure and atmospheric pressure is as follows.
P =
Putting the given values into the above formula as follows.
P =
=
= 102683.05 Pa
= 102.68 kPa
thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.
I’m pretty sure the answer is D :)