The volume of the piece of aluminum is 1.96 mL
Explanation:
Density is the relationship of the mass of a substance and its volume.
In this case, the mass of aluminum is 5.30 g and the density is 2.70 g/mL
The formula to apply is;
D=M/V where D is density in g/mL, M is mass in g and V is volume in mL
2.70=5.30/V
V=5.30/2.70 =1.96 mL
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Keywords: volume, aluminum,density
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The hybrid orbital of this molecule is 
. Hence, option C is correct.
<h3>What is hybridisation?</h3>
Hybridization is defined as the concept of mixing two atomic orbitals to give rise to a new type of hybridized orbitals.
In this compound, a hybrid orbital makes I-O bonds. Due to hybridization iodate should have tetrahedral geometry but because of the presence of lone pair of electrons the shape of 
the ion is pyramidal.
The hybrid orbital of this molecule is . Hence, option C is correct.
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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
The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
