The elements in the blocks of period 4-7 and group 3-12 in the periodic table are in the d-block and d-block elements are known as transition metals or transition elements.
Arsenic is a metalloid and it has four electron shells and five valence electrons.
the s and p orbital of Arsenic have 2 and 3 electrons respectively and which makes five valence electrons.
Answer:
The awnser is D
Explanation:
Since the black pieces absorb light better than white pieces, the white pieces will reflect light better. But since color doesn't affect sound, the sound waves are reflected the same.
Sulfuric acid is prepared industrially by means of the response of water with sulfur trioxide which in turn is made by way of a chemical combination of sulfur dioxide and oxygen both by way of the touch system or the chamber system.
H2SO4 (l) H2O (g) + SO3 (g).
The reaction is highly exothermic as an enormous amount of heat is liberated.
The usual approach is to dilute the sulfur trioxide in sulphuric acid. This produces oleum. SO3 (g) + H2SO4 → H2S2O7 (1) Oleum can be in addition diluted in water to acquire concentrated sulphuric acid.
An acid catalyst is added to protonate the carbonyl carbon. How does this catalyze the response, robust acid catalysts catalyze the hydrolysis and transesterification of esters which enables the mechanism with a view to boom the electrophilicity of the carbonyl carbon to assist protonate the carbonyl oxygen.
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The answer is c hope i helpd
This question is quite vague, as the initial concentration of ethanol is not provided. However, from experience I can tell you that most laboratory work is done with 98% ethanol, and not absolute ethanol (100%). So in order to calculate the final concentration, we need to take the given values, which includes the initial concentration (98%), the initial volume (50.0mL) and the final volume (100.0mL). We apply the following equation to calculate the final concentration:
C1V1 = C2V2
C1 = Initial concentration
C2 = Final concentration
V1 = Initial volume
V2 = Final volume
(98%)(50.0mL) = (C2)(100.0mL)
Therefore, the final concentration (C2) = 49%
