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2 years ago

Each element in the periodic table can be represented by a chemical symbol. Match each element to its chemical symbol.

Chemistry

1 answer:

Andrej [43]2 years ago

5 0

$\qquad\qquad\huge\underline{{\sf Answer}}$

The Correct pair of Element and it's symbol ~

Mercury - Hg

Sodium - Na

Nitrogen - N

Chlorine - Cl

Hydrogen - H

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How many grarns would 3 moles of carbon dioxide (CO2) weigh? (round to the

BigorU [14]

Mass of CO₂ = 132 g

<h3>Further explanation </h3>

A mole is a number of particles(atoms, molecules, ions) in a substance

This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mole also can be formulated :

$\tt n=\dfrac{mass}{MW}$

moles of CO₂ = 3

mass of CO₂(MW=44.01 g/mol) :

$\tt mass=mol\times MW\\\\mass=3\times 44.01\\\\mass=132.03~g\approx 132~g$

6 0

3 years ago

Non metals are acidic in nature explain with an equation

natka813 [3]

When non metallic oxide for example SO2 dissolved in water it produces sulphurous acid. SO2 + H2O → H2SO3 When metallic oxide reacts with water it produces Metal hydroxides. Na2O + H2O → 2NaOH

6 0

3 years ago

If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP m

Nadusha1986 [10]

Answer:

$128~ATP$

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:

$Number~of~Rounds=\frac{n}{2}-1$

Where "n" is the number of carbons, in this case "18", so:

$Number~of~Rounds=\frac{18}{2}-1~=~8$

We also have to calculate the amount of Acetyl-Coa produced:

$Number~of~Acetyl-Coa=\frac{18}{2}~=~9$

Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 $FADH_2$ and 1 $NADH$ . So, if we have 8 rounds we will have 8 $FADH_2$ and 8 $NADH$ .