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2 years ago

5

How is mercury obtained and separated from other nearby substances?

2 answers:

valkas [14]2 years ago

8 0

How is mercury obtained and separated from other nearby substances?

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VLD [36.1K]2 years ago

7 0

Answer:

Most mercury forms in a sulfide ore called cinnabar, but mercury is also frequently found in small amounts in other ores. A common method for separating mercury from cinnabar is to crush the ore and then heat it in a furnace in order to vaporize the mercury. This vapor is then condensed into liquid mercury form.

Explanation:

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1. Using the balanced equation, answer the following questions:

nalin [4]

Answer:

a) 2.53 × 10²³ molecules of O₂

b) 31.90 g of KCl

Explanation:

The balance chemical equation for given decomposition reaction is as follow;

2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

Moles = Mass / M.Mass

Moles = 34.35 g / 122.55 g/mol

Moles = 0.280 moles of KClO₃

Step 2: <u>Find out moles of O₂ produced;</u>

According to eq,

2 moles of KClO₃ produces = 3 moles of O₂

So,

0.280 moles of KClO₃ will produce = X moles of O₂

Solving for X,

X = 0.280 mol × 3 mol / 2 mol

X = 0.42 moles of O₂

Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

No. of Molecules = Moles × 6.022 × 10²³

No. of Molecules = 0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules = 2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

Moles = Mass / M.Mass

Moles = 52.53 g / 122.55 g/mol

Moles = 0.428 moles of KClO₃

Step 2: <u>Find out moles of KCl produced;</u>

According to eq,

2 moles of KClO₃ produces = 2 moles of KCl

So,

0.428 moles of KClO₃ will produce = X moles of KCl

Solving for X,

X = 0.428 mol × 2 mol / 2 mol

X = 0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

Mass = Moles × M.Mass

Mass = 0.428 mol × 74.55 g/mol

Mass = 31.90 g of KCl

6 0

3 years ago

The air we breathe is not considered by chemists to be an element. why not

ohaa [14]

Because Air itself is not an element. it is made up by different elements such as Oxygen and nitrogen. since its made up of a mix of different elements, its rather a homogeneous mixture :)

7 0

3 years ago

Read 2 more answers

What particles would you find in a nucleus of an atom

sammy [17]

Answer:

The nucleus is a collection of particles called protons, which are positively charged, and neutrons, which are electrically neutral. Protons and neutrons are in turn made up of particles called quarks. The chemical element of an atom is determined by the number of protons, or the atomic number, Z, of the nucleus.

5 0

2 years ago

F a shell can hold a maximum of 32 electrons, what is the value of n? 2 3 4 5

Arte-miy333 [17]

Answer: 4

Explanation:

Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n.

Azimuthal Quantum Number: This quantum number describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Given : a f subshell, thus l = 3 , Thus the subshells present would be 3, 2, 1, 0 and thus n will have a value of 4.

Also electrons give are 32.

The formula for number of electrons is $2n^2$ .

$2n^2=32$

$n=4$

Thus principal quantum no will be n= 4.

8 0

3 years ago

Read 2 more answers

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

2

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

So

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

2 years ago