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2 years ago

13

PLEASE HELP i’ll mark brainliest ( reporting fake answers)

1 answer:

kherson [118]2 years ago

6 0

Delta S reaction= Delta S products- Delta S reactants

don't forget to mulitiply by coefficients

also

here is a really slow way to do it

you know the moles of gas increased

so Delta S is positive

so its B or D

then just do the units digit to see which one match up

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Describe a simple chemical test that could be used to distinguish between members of each of the following pairs of compounds (a

MA_775_DIABLO [31]

Answer:

Explanation:

(A) Phenol gives violet color by complexation with Fe3+ solution . It is best identification...

(B) For b , same phenolic test can be done ...But other esterification is also possibility...

(C) Picric acid has a particular identification test ,

(D) Here , like the first one , phenolic test with FeCl3 gives violet color for 4-ethylphenol and no color for ethyl phenyl ether...

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3 years ago

A _____________reaction mechanism involves loss of a leaving group, formation of a ________ followed by the removal of a proton

Molodets [167]

Answer:

The correct answer is Option C (E1) and Option B (carbocation).

Explanation:

Intramolecular immunity idols are considered as that of the formation mechanism with E1 responses or reactivity.
Reactants with E1 were indeed obligations of both parties, meaning that an E1 reaction was conducted thru all the two stages known as ionization but rather deprotonation. Involves the absence of either an aromatic ring, a carbocation has been generated throughout the ionization solution.

Some other possibilities offered aren't relevant to the procedure outlined. So the above alternative is accurate.

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2 years ago

A saline solution contains 1.6 % NaCl by mass.How much NaCl is present in 57.2 g of this solution?

nekit [7.7K]

Answer:

0.92 g

Explanation:

Given data:

Percentage of NaCl = 1.6%

Amount of NaCl present in 57.2 g = ?

Solution:

1.6% of 57.2 g NaCl:

1.6/100× 57.2 g

0.92 g

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6 0

3 years ago

What happens to an elements property when he doesnt pay his taxes?

Karolina [17]

Answer:

I've got a feeling...

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3 years ago

A 20.7167 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation mgco3(s) → mgo(s

marishachu [46]

The mass of impure $MgCO_{3}$ is 20.7167 g. The decomposition reaction is as follows:

$MgCO_{3}(s)\rightarrow MgO(s)+CO_{2}(g)$

Here, 1 mole of $MgCO_{3}$ gives 1 mole of MgO.

Molar mass of $MgCO_{3}$ and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

$n=\frac{m}{M}$

Here, M is molar mass.

Since, $n_{MgCO_{3}}=n_{MgO}$

Thus, $\frac{m_{MgCO_{3}}}{M_{MgCO_{3}}}=\frac{m_{MgO}}{M_{MgO}}$

On putting the values,

$\frac{m_{MgCO_{3}}}{84.31}=\frac{m_{MgO}}{40.3044}$

Rearranging,

$m_{MgO}=\frac{m_{MgCO_{3}}}{84.31}\times 40.3044$

Or,

$m_{MgCO_{3}}=\frac{m_{MgO}}{0.4780}$ ...... (1)

Let the mass of impurity be $M_{I}$ and mass of impure $MgCO_{3}$ is 20.7167 g thus,

$M_{MgCO_{3}}=(20.7167-M_{I})g$ ...... (2)

Also, mass of impure MgO is 16.8817 g thus,

$M_{MgO}=(16.8817-M_{I})g$ ...... (3)

On comparing equations (2) and (3),

$M_{MgO}=M_{MgCO_{3}}-3.835$

Putting the value of $M_{MgO}$ in equation (1),

$m_{MgCO_{3}}=\frac{M_{MgCO_{3}}-3.835}{0.4780}$

Or,

$0.4780 M_{MgCO_{3}}=M_{MgCO_{3}}-3.835$

Or,

$M_{MgCO_{3}}-0.4780M_{MgCO_{3}}=3.835$

Or,

$0.522 M_{MgCO_{3}}=3.835$

Or,

$M_{MgCO_{3}}=\frac{3.835}{0.522}=7.35 g$

Thus, magnesium carbonate present in the original sample is 7.35 g.

5 0

3 years ago