Answer:
Explanation:
El hielo encierra la primavera antes de espolvorear la sal.
Cuando se rocía sal sobre el hielo, disminuye el punto de fusión del hielo 32 ° F a un poco por debajo de 32 °, por lo tanto, se acumula.
A medida que el hielo se vuelve a congelar, encierra la primavera
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
Answer:
The advantages of using an indicator to inform pH measurements:
It gives a mathematically result of the pH, in addition, it gives the precise pH of solvent, and it also gives an idea of the straight of the solution also.
Now, the advantage of using a pH meter:
It is a rapid method to characterize between acids, bases. However, this method does not show how strong acid or base actually are, plus it tends to gives a range of acidity or basicity not quite accurate as a result.
Answer:
a. 4,00L
b. 16,00L
c. 12,31L
Explanation:
Avogadro's law says:

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

<em>V₂ = 4,00L</em>
b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

<em>V₂ = 16,00L</em>
c. 25,0g of Ne are:
25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

<em>V₂ = 12,31L</em>
I hope it helps!
Given:
Be - Beryllium - 9,3227
C - Carbon - 11,2603
O - Oxygen - 13,6181
Ne - Neon - 21,5645
B - Boron - 8,298
Li - Lithium - 5,3917
F - Fluorine - 17,4228
N - Nitrogen - 14,5341
Arranged from highest ionization energy to lowest ionization energy.
Ne ; F ; N ; O ; C ; Be ; B ; Li