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3 years ago

Oily vegetable or animal substances are _ (choose only one answer)

Chemistry

1 answer:

Vlad [161]3 years ago

6 0

The answer is 'lipids'

hopefully this helps

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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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2 years ago

P + O2 → P4O10 balance the equation

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4P + 502 -> P4O10 this is the answer

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2 years ago

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Taya2010 [7]

Answer:

Explanation:

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3 years ago

A pure gold bar is made up of only particles. A.nickel B.copper C.god D.gold

Minchanka [31]

Answer: gold

Explanation:

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3 years ago

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

scZoUnD [109]

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
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2 years ago