Question:
Each liter of air has a mass of 1.80 grams. How many liters of air are contained in 2.5 x103) kg of air?
Answer:
11.48106 L
Answer:
The answer to your question is <u>111 g of CaCl₂</u>
Explanation:
Reaction
2HCl + CaCO₃ ⇒ CaCl₂ + CO₂ + H₂O
Process
1.- Calculate the molecular mass of Calcium carbonate and calcium chloride
CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g
CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g
2.- Calculate the amount of calcium chloride produced using proportions.
The proportion CaCO₃ to CaCl₂ is 1 : 1.
100 g of CaCO₃ ------------- 111 g of CaCl₂
Then 111g of CaCl₂ will be produced.
The answer is a , as aluminium has 13 protons and electrons .
Answer:
120 mol Mg
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
120 moles H₂
<u>Step 2: Identify Conversions</u>
RxN: 3 mol H₂ = 3 mol Mg
<u>Step 3: Stoichiometry</u>
<u />
= 120 mol Mg
Answer:
Metallic bonding
Explanation:
Metals have low ionization energies. Therefore, their valence electrons are easily delocalized (attracted to the neighbouring metal atoms). These delocalized electrons are then not associated with a specific metal atom. Since the electrons are “free”, the metal atoms have become cations, and the electrons are free to move throughout the whole crystalline structure.
We say that a metal consists of an array of cations immersed in a sea of electrons
.
The electrons act as a “glue” holding the cations together.
Metallic bonds are the attractive forces between the metal cations and the sea of electrons.
In an NaK alloy, for example, the Na and K atoms contribute their valence electrons to the "sea". The atoms aren’t bonded to each other, but they are held in place by the metallic bonding.