<u>Methylcyclohexane:</u>
The chemical formula of the organic compound, methylcyclohexane is 
. It is branched into the group of saturated hydrocarbon.
Note: Refer the attached image for the structure of methylcyclohexane
<u>Structure of methylcyclohexane:</u>
- It is a monosubstituted cyclohexane because it has one branching through the attachment of single methyl group on one carbon of the cyclohexane ring.
- Being very toxic to the aquatic ecosystem, the methyl group of this compond in its axial position experiences a steric strain or steric crowding.
The density of marble is between 2.6 and 2.8 grams per cm³ .
Density doesn't depend on how much mass or volume of it you have.
The density of a chip of it is the same as the density of a truckload of it.
Answer:
116.88 g
Explanation:
Step 1: Write the balanced equation for the synthesis of NaCl
Na + 1/2 Cl₂ ⇒ NaCl
Step 2: Calculate the moles corresponding to 45.978 g of Na
The molar mass of Na is 22.990 g/mol.
45.978 g × 1 mol/22.990 g = 1.9999 mol
Step 3: Calculate the number of moles of NaCl formed from 1.9999 moles of Na
The molar ratio of Na to NaCl is 1:1. The moles of NaCl formed are 1/1 × 1.9999 mol = 1.9999 mol.
Step 4: Calculate the mass corresponding to 1.9999 moles of NaCl
The molar mass of NaCl is 58.443 g/mol.
1.9999 mol × 58.443 g/mol = 116.88 g
Answer:
Explanation:
conjugate acid, based on Brønsted–Lowry acid–base theory, is a chemical compound that is formed by the reception of a proton by a base
a. CH₃COOH + H₂O ⇌ H₃0⁺ + CH₃C00-
Acid <> CH₃COOH
Base <> H₂O
Conjugate acid <> H₃0 +
Conjugate base <>CH₃C00-
b. HCO₃ + H₂O ⇌ H₂CO₃⁻ + OH⁻
Acid <> H₂O
Base <> HCO₃
Conjugate acid <> H₂CO₃⁻
Conjugate base <>OH⁻
C. HNO₃ + SO₄²⁻ ⇌ HSO₄⁻ + NO₃⁻
Acid <>HNO₃
Base <>SO₄²⁻
Conjugate acid <>HSO₄⁻
Conjugate base <>NO₃⁻
A Bronsted acid is reffered to as a proton donor while a Bronsted base is a proton acceptor
Answer: option <span>D. be given a positive charge produced by the movement of electrons to the other end of the ball.
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Explanation:
This phenomenon is called electrostatic induction.
The excess of negative charge on the end of the rod will repel the electrons on the side of the pith ball that have been approached to it.
Then the electrons on the pith ball will move far away from this end with it will be left an excess of positive charge.
In this way the rod has induced that the ball acquires a positive charge on one end and a negative charge on the other end.
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