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Korvikt [17]
2 years ago
5

Ethene is a useful substance that can form polymers. It has a melting point of 169°C and a boiling point of 104°C. At which temp

erature would ethene be a liquid?
Chemistry
1 answer:
blondinia [14]2 years ago
4 0

Answer:

-169°C to -104°C

Explanation:

Ethene, also known as ethylene exists in solid, liquid and gaseous states. Ethene is an aliens with condensed structural formula C2H4. Athens is a colourless gas. It is flammable and is also a sweet smelling gas in its pure form. It is the monomer in the production of polyethylene which is of great importance in the plastic industry. In agriculture, it is used to induce the ripening of fruits. It can be hydrated in order to produce ethanol.

The liquid range of ethene refers to the temperatures at which ethene is found in the liquid state of matter. It is actually the difference between the melting point and the boiling points of ethene. Hence the liquid range of ethene is -169°C to -104°C

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Consider the following oxides: SO2, Y2O3, MgO, Cl2O, and N2O5. How many are expected to form acidic solutions in water? Consider
Snowcat [4.5K]

Answer:

Three of the five oxides are expected to form acidic solutions in water

Explanation:

We have different types of oxides : Acidic oxides, Basic oxides, Amphoteric oxides, Peroxides and Higher oxides.

Only acidic oxides will dissolve in water to give an acidic solution.

Considering the given oxides carefully,

  • SO2 will dissolve in water to produce H2SO3 which is acidic.

  • Y2O3 will dissolve in water to produce Yttrium(III) hydroxide which is basic.

  • MgO will dissolve in water only to produce Mg(OH)2 which is also basic.

  • Cl2O dichlorine mono oxide will dissolve in water to produce HClO which is acidic.

  • N2O5 will dissolve in water to produce HNO3 which is also acidic.

5 0
2 years ago
A solution of HCl has StartBracket upper H superscript plus EndBracket. = 0.01 M. What is the pH of this solution?
hoa [83]

Answer:

2

Explanation:

Data:

[H⁺] = 0.01 mol·L⁻¹

Calculation:

pH = -log[H₃O⁺] = -log(0.01) = -log(1) - log(10⁻²) = -0 - (-2) = 0 + 2 = 2

7 0
2 years ago
When 64.0 g of methanol (CHOH) is burned, 1454 kJ of energy is produced. What is the heat of combustion for methanol?
andreev551 [17]

 The heat  of combustion  for  methanol   is 727  kj/mol


    <em><u>calculation</u></em>

 calculate the moles  of methanol (CH3OH)

moles = mass/molar  mass

molar mass of methanol =  12 +( 1 x3)  +16 + 1= 32 g /mol

moles is therefore= 64.0 g / 32 g/mol =  2 moles


Heat of combustion  is therefore = 1454 Kj / 2 moles =  727  Kj/mol

6 0
2 years ago
127) Thirty-six colonies grew in nutrient agar from 1.0 ml of undiluted sample in a standard plate count. How many cells were in
SIZIF [17.4K]

Answer:

36

Explanation:

Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew  colonies then  0.1 ml  yielded  6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.

6 0
2 years ago
a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin
Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0
2 years ago
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