Answer:
Thermal energy is taken from heat sink in higher temperature. A thermal power machine does mechanical energy using part of the heat.
Part of heat taken are given in cold reservoir in lower temperature
Explanation:
Answer:
A = Metallic Bond
B = Strong bonding, strong conductor, high melting and boiling points
Explanation:
Since the bond is between two metals (located in groups 11 and 12), they would experience metallic bonding. Metallically bonded molecules have high melting and boiling points due to the strength of the metallic bond. They also experience strong electrical current due to the there delocalized electrons.
<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it.
Crude appearance.
18 18 0
F --> O + e
9 8 1
Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts.
The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons.
Followed by a right arrow to indicate the direction the reaction is going.
Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons.
Followed by a plus sign to indicate more.
Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>
Answer:
Hydrogen Bond
Explanation:
Hydrogen bond interactions are formed between the hydrogen atom bonded to most electronegative atoms (i.e. F, O and N) of one molecule and most electronegative atom (i.e. F, O and N) of another molecule.
In this interaction the hydrogen atom has partial positive charge and electronegative atom has partial negative charge.
Students performed a procedure similar to Part II of this
experiment (Analyzing Juices for Vitamin C Content) as described in the
procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml
of DCP to titrate 10 mL of sample.
Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01
L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124
g/mol)(1000mg/1g)= 14.36 mg ascorbic acid
