Answer:
C. gold atoms
Explanation:
Metallic bonds are found between atoms of metals. Since gold is the only metal given here, it is the solution to the problem.
Metallic bonds join atoms of metals and alloys together.
The formation of this type of bond is favored by large atomic radius, low ionization energy and a large number of electrons in the valence shell.
The metallic bond is actually an attraction between the positive nuclei of all the closely packed atoms in the lattice and the electron cloud jointly formed by all atoms losing their outermost shell electrons.
Answer:
The heat absorbed is hence 10.751.21 J
Explanation:
The heat absorbed when 88g of water is heated from 5.8°C to 35 °C is;
Heat = m c ΔT
m = 88.0 g
c = specific heat of water = 4.184 J/g°C
ΔT = ( change in temperature) = ( 35 - 5.8)°C = 29.2°C
Equating these values into the formula, we obtain;
Heat = 88* 4.184 * 29.2
Heat = 10 751.2064 J
Heat = 10 751.21 J (2 d.p)
The heat absorbed is hence 10.751.21 J
solid carbon dioxide, iodine, arsenic, and naphthalene
Explanation:
Examples of substances that undergo sublimation
Examples of solids that sublime are dry ice (solid carbon dioxide), iodine, arsenic, and naphthalene (the stuff mothballs are made of).
The net amount of energy produced can be obtained from a table of enthalpy change of formation, available online.
The enthalpy change of formation indicate how much energy the 1 mole of the product (H2O) has relative to the elemental reactants (H2 and O2). In other words, the "lost" energy equals the heat/energy released.
For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released. This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.
Since water is in liquid form under standard conditions, the first value (285.8 kJ/mol) is generally appropriate.
