Answer : The molar mass of the solute would be low.
Explanation :
Formula used for depression in freezing point is:

where,
= change in freezing point
= freezing point of solution
= freezing point of water
i = Van't Hoff factor
= freezing point constant
m = molality
= mass of solute
= mass of solvent
= molar mass of solute
From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.
Hence, the molar mass of the solute would be low.
Answer:
28
Explanation:
28 is correct. (NH4)2 is equal to 10 atoms. (8 H and 2 N ) CO3 is equal to 4 atoms. (1 C and 3 O) That's 14 atoms. But it's two molecules of this. So you just multiply by the number of molecules (2), and you get 28.
Juan is taking out a loan for $2,500 with an annual compound interest rate of 7% for 3 years Juan will not make any additional deposits or withdrawals how much interest will he have to pay at the end of the loan period.
Answer:
Hi
Williamson's ether reactions imply that an alkoxide reacts with a primary haloalkane. Alkoxides consisting of the conjugate base of an alcohol and are formed by a group R attached to an oxygen atom. They are often written as RO–, where R is the organic substituent (Step 1).
Sn2 reactions are characterized by the reversal of stereochemistry at the site of the leaving group. Williamson's synthesis is no exception and the reaction is initiated by the subsequent attack of the nucleophile. This requires that the nucleophile and electrophile be in anti-configuration (Step 2).
As an example (figure 3).
In the attached file are each of the steps of Williamson's synthesis.
Explanation:
Answer:
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