The amount of water that will be produced is 50.36 grams
<h3>Stoichiometric problems</h3>
The metabolism of glucose is represented by the following equation:
The mole ratio of glucose metabolized to the water produced is 1:6.
Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles
Equivalent mole of water = 0.4662 x 6 = 2.7975 moles
Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams
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when sodium metal is dropped in water, hydrogen gas in liberated due to extreme heat released as the reaction is exothermic, gas catches fire.
Answer:
El número atómico de cada uno de los átomos es 26
Explanation:
El número de masa es la suma de las masas del protón y el neutrón de un átomo.
El número atómico es el número de protones en el átomo.
Los parámetros dados son;
La suma del número másico de ambos átomos = 110
La suma de los neutrones = 58
Por lo tanto, sea el número de protones y neutrones en un isótopo = P₁ y N₁ y el número de protones y neutrones en el otro isótopo = P₂ y N₂
Tenemos;
P₁ + N₁ + P₂ + N₂ = 110
N₁ + N₂ = 58
Por lo tanto;
P₁ + P₂ = 110 - (N₁ + N₂)
P₁ + P₂ = 110 - 58 = 52
Dado que los isótopos son del mismo elemento, sus protones serán iguales, por lo tanto;
P₁ = P₂
P₁ + P₂ = P₁ + P₁ = 2 × P₁
P₁ + P₂ = 52
2 × P₁ = 52
P₁ = 52/2 = 26 = P₂
El número atómico de ambos átomos es el número de protones en el átomo que es 26.
El número atómico del elemento del átomo es 26
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
