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3 years ago

Which element is the most reactive element in group one of the periodic table

Chemistry

1 answer:

Bad White [126]3 years ago

4 0

Answer:

Explanation:

According to periodic table, lithium will be the most reactive element because reactivity decreases down the group

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Determine two complex numbers, (a+bi) and (c+di), where a and d are irrational numbers and b and c are rational number

Dafna11 [192]

Answer:

The two complex numbers are $\sqrt2 + 5i \text{ and } 6 + \sqrt5i$

Explanation:

We have to form two complex numbers of the form

$a + ib\\c + id$

such that and d are irrational numbers and b and c are rational numbers.

We know that $\sqrt2, \sqrt3$ are irrational numbers.

5 and 6 are rational numbers.

We put

$a = \sqrt2\\b = 5\\c = 6\\d = \sqrt5\\a+ib = \sqrt2 + 5i\\c + id = 6 + \sqrt5i$

Thus, the two complex numbers are: $\sqrt2 + 5i \text{ and } 6 + \sqrt5i$

7 0

3 years ago

Read 2 more answers

Aluminium is the most common metal in the Earth's crust.<br> Which is NOT a property of aluminium?

natali 33 [55]

Answer: Low Density

Explanation:

5 0

2 years ago

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

KATRIN_1 [288]

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

8 0

3 years ago

List six elements with only 1 electron in an s orbital

Verdich [7]

Answer:

$Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1$

Explanation:

When it comes to electron configuration and orbitals, it's important to first identify what exactly we are trying to identify. Below is a given example:

$He1s^2$

$He=element$

$1=level$

$s=orbital$

$(exponent)^2=electrons$

Looking at the periodic table, identify the alkali metal family on the periodic table, or group one elements:

$Lithium=Li2s^1\\Sodium=Na3s^1\\Potassium=K4s^1\\Rubidium=Rb5s^1\\Cesium=Cs6s^1\\Francium=Fr7s^1$

Notice how each configuration has an exponent of $1$ , representative of a single electron in their s-orbital.

6 0

3 years ago

Write the proper word equation to express the following chemical reaction: 4Fe (s) + 3Sn(NO3)4 (aq) arrow 4Fe(NO3)3 (aq) + 3Sn (

Lapatulllka [165]

The word equation for this reaction is:

Iron + Tin nitrate → Iron nitrate + Tin

3 0

3 years ago