Answer:
3.33 M
Explanation:
It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:
" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>
<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "
In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:
- 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃
Then we<u> convert the 150 mL of the volumetric flask into L</u>:
Finally we <u>divide the moles by the volume</u>:
- 0.500 mol AgNO₃ / 0.150 L = 3.33 M
Answer: Option (A) is the correct answer.
Explanation:
Elements present in group 1 are known as alkali metals. Whereas elements present in group 2 are called alkaline earth metals and elements from group 11 to 12 are transition metals.
As it is known that metals have the ability to lose electrons in order to attain stability and electricity is the flow of electrons from one point to another.
Therefore, metals are good conductors of heat and electricity.
Thus, we can conclude that the statement it’s between groups 1 to 12 because it is metal best explains the probable position of the substance in the periodic table.
<span>Saturated sodium chloride
is used to transfer the product rather than water since it is not polar and
rinsing the product with water would revert any 4-methylcyclohexene back to
4-methylcyclohexanol in the Hickman Head and thus lowering the percent yield;
using water would shift the equilibrium towards the reactants. Also
sodium chloride removes the small amount of phosphoric acid and also a small
amount of water. If one were to add water, both 4-methylcyclohexene and
phosphoric acid are partially soluble making difficult to remove the water
later; sodium chloride makes the water less reactive so easier to remove by
making the aqueous later more polar.</span>
