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Valentin [98]
3 years ago
7

Answer the questions about the characteristics of the elements in group 1 (the alkali metals). What happens when the elements in

group 1 react with bromine? No reaction a salt is formed with the general formula MBr2 a salt is formed with the general formula MBr What happens when the elements in group 1 react with water? Hydrogen gas is released no reaction What happens when the elements in group 1 react with oxygen? No reaction an oxide is formed with the general formula MO an oxide is formed with the general formula M2O Which group 1 element reacts the most vigorously? Na Rb Li K Cs Which group 1 element exhibits slightly different chemistry from the others? Li Na Cs K Rb
Chemistry
1 answer:
irinina [24]3 years ago
6 0

Answer:

See explanation

Explanation:

The elements in group form univalent positive ions and element in group 17 form univalent negative ions. Hence, when a group 1 element reacts with a group 17 element, a compound of the sort MX is formed. Hence, when a group 1 element reacts with bromine, a salt is formed with the general formula MBr.

Elements of group 1 are highly electro positive metals. They react with water to form the metal hydroxide and release hydrogen gas. Hence, when group 1 elements react with water, hydrogen gas is released.

A group 1 element forms a univalent positive ion while a group 16 element forms a divalent negative ion. Hence, when a groups 1 element reacts with oxygen, the compound formed must have the general formula M2O.

The reactivity of group 1 metal increases down the group hence Cs is the most reactive group 1 element.

Lithium displays a slightly different chemistry from other group 1 elements because of its small size.

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The moles required were:

1,024M×0,02500L = <em>0,02560 moles NaOH. </em>These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:

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As density is 1,01g/mL:

1,537g CH₃COOH×\frac{1mL}{1,01g}= <em>1,522mL of CH₃COOH</em>

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As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:

\frac{1,522mLCH_{3}COOH}{20,78mL}×100= <em>7,324 (%V/V)</em>

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Hello,

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