All categories

2 years ago

The standard hydrogen electrode:

Chemistry

1 answer:

tiny-mole [99]2 years ago

3 0

Answer:

is a simple, safe, and easy-to-create electrode

Explanation:

the hydrogen electrode is based on the redox half-cell:

2H+(aq) + 2e- → H2(g)

building:

1) Platinium electrode

2) hydrogen pumping

3) acid solution [H+] = 1M

4) siphon to prevent oxygen presence

5) galvanic cell connector

this electrode is used as the basis for standard potential tables

You might be interested in

Your lab partner named this compound 3-methyl-4-n-propylhexane, but that is not correct.

loris [4]

Answer: The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

Explanation:

The IUPAC nomenclature of alkanes are given as follows:

Select the longest possible carbon chain.
For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
A suffix '-ane' is added at the end of the name.
If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

We are given:

An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms

The image of the given alkane is shown in the image below.

Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

4 0

2 years ago

We are all familiar with the general principles of operation of an internal combustion engine: the combustion of fuel drives out

Alexus [3.1K]

Answer:

The work done by the system is 100 J

Explanation:

Given details

The cross sectional area of the of the container is A = 100.0 cm^2 = 0.01m²

The total distance pushed by the piston is d = 10 cm = 0.10m

The total external pressure by which piston pushed is P = 100 kPa

From above data, the following relation can be used to determine the change in volume of the container

∆V = A * d

∆V = 0.01 * 0.10 = 0.001 m³

By using the following relation, the work done by the system is calculated as;

Work done W = P * ∆V

W = 100 * 0.001 = 0.1 kJ = 100 J

The work done by the system is 100 J

7 0

2 years ago

Question 4 (1 point)

Rainbow [258]

а о

Explanation:

The given cation:

(Rf₂Al₂F₃)³⁺

The oxidation number gives the extent to which a specie is oxidized in a reaction.

This number is assigned based on some rules:

Elements in combined state whose atoms combines with themselves have an oxidation number of zero.
The charge carried on simple ions gives their oxidation number.
Algebraic sum of all the oxidation numbers of atoms in neutral compound is zero. In an ion with more than one kind of atom, the charge on it is the oxidation number.

for the specie given;

Known:

oxidation number of Al = +3

F = -1

charge = +3

let the oxidation number of Rf = k

2k + 2(3) + 3(-1) = +3

2k + 6 - 3 = 3

2k = 0

k = 0

The oxidation state of rutherfodium is 0

learn more:

Oxidation state brainly.com/question/10017129

#learnwithBrainly

8 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Kisachek [45]

Answer: The empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.01 g of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 10.65 g of water, $\frac{2}{18}\times 10.65=1.18g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is $C_9H_{12}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Hence, the empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

6 0

2 years ago

The refractive index, n of a polymer was determined as a function of temperature. The results are given in the table. Determine

mylen [45]

Answer:

Explanation:

The given data shows linear trend between refractive index and temperature .

From 20 degree to 80 degree , every increase of 10 degree decreases refractive index by .0015 . This trend breaks at 90 degree beyond which , refractive index falls at greater rate or sharply.

So transition temperature is 90 degree .

4 0

2 years ago