primarily ionic include = sodium iodide( NaI) , calcium chloride ( CaCl2)
primarily covalent include - Ammonia (NH3) , Methane ( CH4) and
Glucose (C6H12O6)
Explanation
ionic bond is formed when there is complete transfer of electron between atoms. It occur between metal which donate electrons and a non metal which accept electrons.
for example in formation of CaCl2, ca donate 2 electron to 2 Cl atom, while 2 Cl atom accept the 2 electrons to form ionic bond.
Covalent bond is formed when two or more non metal form bond by sharing electrons pairs.
For example in NH3 3 pairs of electron are shared. to form covalent bond.
Answer:
Explanation:
An empirical formula shows the smallest whole-number ratio of the atoms in a compound.
So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.
<h3>1. Divide </h3>
The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.
There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300
- Sulfur: 0.300/0.300= 1
- Oxygen: 0.900/0.300= 3
<h3>2. Write Empirical Formula</h3>
The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.
This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.
The endotherdic energy raises that its breaks down the bond of the solvent which makes the solute-slovent interaction weaker
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
Answer:
The answer to your question is below
Explanation:
To solve this problem use the general formulas for Alkanes, Alkenes, and Alkynes.
Alkanes Cn H(2n + 2)
Alkenes Cn H (2n) "n" is the number of Carbons
Alkynes Cn H (2n - 2)
a) CH₄ C₁ H 2(1) +2 = C₁H₂ + ₂ = CH₄ Alkane
b) C₂H₄ C₂ H₂(₂) Alkene
c) C₃H₆ C(₃) H₂ (₃) Alkene
d) C₂H₂ C₂ H₂(₂) - 2 = C₂H₄ - ₂ = C₂H₂ Alkyne
e) C₁₀H₂₂ C₁₀ H₂(₁₁) Alkene
