Answer:
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Explanation:
Mass of gold = m
Volume of gold = v
Surface area on which gold is plated = 
Thickness of the gold plating = h = 0.5 mm = 0.05 cm
1 mm = 0.1 cm

Density of the gold = 

Moles of gold = 

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :
of electrons
Number of electrons = N =
Charge on single electron = 
Total charge required = Q

Amount of current passes = I = 8 Ampere
Duration of time = T



It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, <span><span>
1. </span>Atomic number either atomic mass each element has</span>
<span><span>2. </span>By substance of that matter either pure substance or mixed substance</span> <span>
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.</span>
Answer:
48.5 mL , just did it on edge
Answer:
The answer to your question is: letter D.
Explanation:
Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.
Due to this characteristic, they are stable and do not react with other elements.
a. 1s22s22p4 The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.
b. [Ne]2s22p2 The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.
c. [Ar] 3s1 This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.
d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.
Second one i think.......