Molarity=moles/liter
molarity=43/0.64
molarity=67.19moles/litre
c. a tertiary alcohol; when a ketone reacts with a grignard reagent followed by protonation a tertiary alcohol is formed.
More about tertiary alcohol:
No hydrogen atoms are bonded to the functional group's carbon in a tertiary alcohol. Alcohols that have a hydroxyl group bonded to the carbon atom and are linked to three alkyl groups are referred to as tertiary alcohols. These alcohols' structural makeup largely determines their physical characteristics.
This -OH group's existence enables alcohols to create hydrogen bonds with the atoms next to them. Because of this weak connection, alcohols have higher boiling points than their alkane counterparts.
The alcohol is referred to as a tertiary (3°) alcohol if the carbon atom carrying the alcohol group is connected to three other carbon atoms in the alcohol molecule.
Learn more about tertiary alcohol here:
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Answer:
Yes it would
Explanation: Well it kinda depend on the voltage and how the battery has been in use or based on the condition
Answer:
0.1357 M
Explanation:
(a) The balanced reaction is shown below as:

(b) Moles of
can be calculated as:
Or,
Given :
For
:
Molarity = 0.1450 M
Volume = 10.00 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 10×10⁻³ L
Thus, moles of
:
Moles of
= 0.00145 moles
From the reaction,
1 mole of
react with 2 moles of NaOH
0.00145 mole of
react with 2*0.00145 mole of NaOH
Moles of NaOH = 0.0029 moles
Volume = 21.37 mL = 21.37×10⁻³ L
Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M
Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation: