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3 years ago

The limiting reactants appeared to be _ Because_____

Chemistry

1 answer:

Igoryamba3 years ago

8 0

Answer:

The limiting reagents seemed to be those that were consumed first .

Because when this reagent is consumed, the reaction stops. The quantity of this determines the total quantity of the product formed.

Explanation:

A limiting reagent is one that is consumed in its entirety. In this way, it delimits the amount of product that can be formed.

Take the case of two substances that interact and produce a chemical reaction. If one of the substances runs out as it is consumed during the process, the reaction will stop. The reagent consumed acts as a limiting reagent, that is, it limits the possibility of the reaction proceeding, and therefore it also limits the amount of the product generated by the reaction.

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A lab technician mixed a 550 ml solution of water and alcohol. if 3% of the solution is alcohol, how many milliliters of alcoh

kupik [55]

The solution 550 ml total and first we will find the amount of alcohol. 3% = 0.03 550 ml x .03 = 16.5 ml alcohol
Then to find the amount of water used, we just have to subtract the amount of alcohol from the total volume
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3 years ago

Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

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Explanation:

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Read 2 more answers

Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:

Andru [333]

Answer:

gde

Explanation:

We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.

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The limiting reactants appeared to be _______ Because___________

The limiting reactants appeared to be _ Because_____