All categories

2 years ago

In what areas of the periodic table do you find the most highly reactive elements?

Chemistry

1 answer:

Minchanka [31]2 years ago

5 0

Answer:

The elements toward the bottom left corner of the periodic table are the metals that are the most active in the sense of being the most reactive.

You might be interested in

A solution is prepared by adding 0.0231moles of H3O+ ions to 3.33L of water. What is the pH of this solution

MrMuchimi

Answer:

2.15

Explanation:

For this question, we have to remember the pH formula:

$pH~=~-Log[H_3O^+]$

By definition, the pH value is calculated when we do the -Log of the concentration of the hydronium ions ( $H_3O^+$ ). So, the next step is the calculation of the concentration of the hydronium ions. For this, we have to use the molarity formula:

$M=\frac{mol}{L}$

We already know the number of moles (0.0231 moles) and the volume (3.33 L). So, we can plug the values into the molarity formula:

$M=\frac{0.0231~moles}{3.33~L}=0.00693~M$

With this value, now we can calculate the pH value:

$pH~=~-Log[0.00693~M]~=~2.15$

The pH would be 2.15

I hope it helps!

8 0

2 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0

2 years ago

A student decomposed 3.67g of copper (ii) hydroxide into copper (ii) oxide. how many ml of 3m h2so4 is need to react with all th

vredina [299]

In the presence of heat, copper (II) hydroxide decomposes in to copper (II) oxide.
Cu(OH)₂ (s) ----> CuO (s) + H₂O (l)
upon decomposition, water is removed from Cu(OH)₂
the amount of Cu(OH)₂ decomposed - 3.67 g
number of moles of Cu(OH)₂ - 3.67 g / 97.5 g/mol = 0.038 mol
stoichiometry of Cu(OH)₂ to CuO is 1:1
therefore number of CuO moles formed are - 0.038 mol
CuO reacts with sulfuric acid to form CuSO₄
CuO + H₂SO₄ ---> CuSO₄ + H₂O
stoichiometry of CuO to H₂SO₄ is 1:1
therefore number of H₂SO₄ moles that should react is 0.038 mol
the molarity of H₂SO₄ is 3M
this means that in 1000 ml - 3 mol of H₂SO₄ present
so if 3 mol are present in 1000 ml
then volume for 0.038 mol = 1000/3 * 0.038
= 12.67 ml

6 0

3 years ago

(1) The solubility of Salt AB2(S) IS 5mol/dm^3.

Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s] [s] [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

6 0

2 years ago

What is the chemical formula for the compound formed between sodium and iodine?

IrinaK [193]

Each compound will be neutral

6 0

2 years ago