Answer:
6.1×10^4Pa or 61KPa
Explanation:
The Clausius-Clapeyron equation is used to estimate the vapour pressure at different temperature, once the enthalpy of vaporization and the vapor pressure at another temperature is given in the question. The detailed solution is shown in the image attached. The temperatures were converted to kelvin and the energy value was converted from kilojoule to joule since the value of the gas constant was given in unit of joule per mole per kelvin. The fact that lnx=2.303logx was also applied in the solution.
Answer:
luyxwfuowdqulgwqofugvwfwgvoueugbvfebfvebh
Explanation:
<span>Answer: 8.15s
</span><span />
<span>Explanation:
</span><span />
<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />
<span>r = k [N]
</span><span />
<span>r = - d[N]/dt =
</span><span />
<span>=> -d[N]/dt = k [N]
</span><span />
<span>2) When you integrate you get:
</span><span />
<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />
<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />
<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>
Electrons are orbiting the nucleus in the fxed way paths located in solid sphere
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
