77.78 kJ of energy is released when 1 mol of glucose is burned, 2802.5 kJ of energy is released.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Calculate the moles of 5.00g of glucose.
Given mass = 5.00g
The molar mass of glucose = 180.156 g/mol
Moles =0.02775372455
The quantity of energy released to a person by eating 5.00g of glucose in a candy.
0.02775372455 x 2802.5 kJ
77.77981305 kJ =77.78 kJ
Hence, 777.78 kJ of energy is released.
Learn more about moles here:
brainly.com/question/8455949
#SPJ1
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
Hot liquid rock comes in contact with Earth´s cold crust
Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?
- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4
Answer:
<h3>The answer is 2 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 48 g
volume = 24 cm³
We have
We have the final answer as
<h3>2 g/cm³</h3>
Hope this helps you
