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Flauer [41]
3 years ago
10

10 points

Chemistry
1 answer:
Murljashka [212]3 years ago
7 0

Answer:

2. The metal would lose one electrons and the non metal would gain one electrons

Explanation:

An atom of a certain element reacts with the atoms of other elements in order to fullfill its outermost shell (called valence shell).

We notice the following:

- The elements in Group 1 (which are metals) have only 1 electron in their valence shell

- The elements in Group 17 (which are non-metals) have 1 vacancy (lack of electron) in their valence shell

This means that in order for both an atom of group 1 and an atom of group 17 to fullfill the valence shell, they have to:

- The atom in group 1 has to give away its only electron of the valence shell

- The atom in group 17 has to gain one electron in order to fullfill the shell

Therefore, the correct option is

2. The metal would lose one electrons and the non metal would gain one electrons

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How much pressure would 0.8 moles of a gas at 370K exert if it occupied 17.3L of space
dezoksy [38]

Answer:

1.40 atm is the pressure for the gas

Explanation:

An easy problem to solve with the Ideal Gases Law:

P . V = n . R .T

T° = 370K

V = 17.3L

n = 0.8 mol

Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K

P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm

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2 years ago
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What is the volume of a solution that has a specific gravity of 1. 2 and a mass of 185g.
muminat
Volume 2 if I am correct
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2 years ago
If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer
Murljashka [212]

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

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Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

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3 0
3 years ago
Mercury thermometers work because mercury ____ when it is warmed
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Mercury expands when it is heated. This process is called thermal expansion.
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35.0 mL of acid with an unknown concentration is titrated with 24.6 mL of 0.432 M base. What is the concentration of the acid? A
dusya [7]

Given:

35.0 mL of acid with an unknown concentration

24.6 mL of 0.432 M base

Required:

Concentration of the acid

Solution:

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V1 = (0.432 M base) (24.6 mL of base) / (35.0 mL of acid)

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