(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
Answer:
D. The atoms are arranged with alternating positive and negative charges. When struck, the lattice shifts putting positives against positives and negatives against negatives.
Explanation:
Metallic crystals takes their properties as a result of metallic bonds in between the atoms.
Metallic bond is actually the attraction between the positive nuclei of all the closely packed atoms in the lattice and the electron cloud jointly formed by all the atoms by losing their outermost shell electrons this is by virtue of their low ionization energy.
Physical properties of metals such as malleability, ductility, electrical conductivity, etc can be accounted for by metallic bonds.
Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)
In this process 3.14 moles of H₂ will be consumed.
The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-
2H₂ (g) + O₂ (g) = 2H₂O (l).
Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.
So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.
Knowing the number of valence electrons in one of the alien elements helps in identifying it because the number of valence electrons can help categorize the alien element. Similar elements have the same valence electrons and knowing the category of the element can help further analyze the element.
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