Answer:
the answer to the qustion is 0.013089701 na
Explanation:
n/a
At the same temperature, steam burns are often more severe that water burns because of water's high HEAT OF VAPORIZATION.
Water possesses high heat of vaporization. The heat of vaporization refers to the amount of heat that is needed to convert a unit mass of water to gas. After getting to the boiling point, a lot of heat is still needed to be absorbed by a boiling water before it can be converted to the gaseous form. Thus, the heat that is inherent in the steam is greater than that which is found in the boiling water, that is why the steam causes more damages.
Answer:
5400 cans
Explanation:
First we convert the total weight, 1 ton, to grams:

Now we need to know the mass of aluminum:

Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:

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
Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The
MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (
BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (
ABMO). Bond Order is calculated as,
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2Or,
Bond Order = 0.5