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Chemistry

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

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Answer:

$H_{3}C$ -- $H_{3}C$ -- $CH_{3}$ --Br-- $CH_{3}$

Explanation:

The steps involved in predicting the structure of the alkyl bromide compound are outlined below.

1) An examination of the product shows that the product could only be formed by a substitution reaction.

2) The structure of the alkyl bromide compound can be then predicted by replacing the methoxide group in the product after the substitution of bromine atom. This is because the methoxide ion acts as a strong nucleophile.

Therefore, by consideration the reaction mechanisms of reactions 1 and 2, it can be predicted that the structure of the alkyl bromide compound is $H_{3}C$ -- $H_{3}C$ -- $CH_{3}$ --Br-- $CH_{3}$ . A pictorial diagram of the alkyl bromide compound is also attached.