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3 years ago

What is called oxidizing agent and reducing agent?

1 answer:

4 0

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reducedExamples of oxidizing agents include halogens, potassium nitrate, and nitric acid. A reducing agent, or reductant, loses electrons and is oxidized in a chemical reaction. A reducing agent is typically in one of its lower possible oxidation states, and is known as the electron donor.

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Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lun

Lapatulllka [165]

Answer:

P = 0.0166 mm Hg

Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁) (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

Hope this helps

8 0

2 years ago

: An unknown metal crystallizes in the cubic crystal structure. The metal has a radius 140pm, atomic mass of 135 g/mol, and dens

ki77a [65]

The cubic unit cell this metal crystallize as is BCC structure .

The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell.

The unit cell consists of lattice points that represent the locations of atoms or ions.

The entire structure then consists of this unit cell repeating in three dimensions

$\rm \rho =\dfrac{nM}{N_{0} a^{3}} \\\\\\\\\rm n =\dfrac{\rho N_{0} a^{3}}{M} \\\\\\\\Assuming\; it \; to \; be \; a\; BCC\; structure \\\\\\ r = \dfrac{\sqrt{3} \times a}{4} \\\\\\Therefore\; a = 3.23 \times 10^{-8}\\\\\\\\\rm n =\dfrac{13.3 \times 6.022 \times 10^{23}\times 3.23^{3}\times (10^{-8})^{3}}{135} \\\\$

n= 2

Hence our assumption was correct

It is a BCC structure .

Therefore the cubic unit cell this metal crystallize as is BCC structure .

To know more about unit cell

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5 0

2 years ago

Please help! Which Zeros are significant in the measurement .0506010 kg?

Vitek1552 [10]

It would be a,b,c as the answer. hope this helps!

7 0

2 years ago

Methanol (CH3OH) has a heat of fusion of 3.16 kJ/mol. Which of the following is the heat of solidification that occurs when 64 g

Alexandra [31]

To determine the heat dissipated when a substance freezes, we multiply the heat of fusion of the substance to the mass of the substance that freezes. We calculate as follows:

Heat = -3.16 (64/32.06) = - 6.32 kJ

Hope this answers the question.

5 0

3 years ago

Read 2 more answers

Find the ph of a buffer that consists of 0.18 m ch3nh2 and 0.73 m ch3nh3cl (pkb of ch3nh2 = 3.35)?

ozzi

Hello!

First, we need to determine the pKa of the base. It can be found applying the following equation:

$pKa=14-pKb=14-3,35=10,65$

Now, we can apply the Henderson-Hasselbach's equation in the following way:

$pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04$

So, the pH of this buffer solution is 10,04

Have a nice day!

8 0

3 years ago