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3 years ago

How does adding a non-volatile solute to a pure solvent affect the boiling point of the pure solvent?

1 answer:

7 0

The correct answer is the second statement. The solvent will have a higher boiling point. Adding a non-volatile solute to a pure solvent will increase the boiling point of the solvent. This solution exhibit colligative properties. Colligative properties depend on the amount of solute dissolved in a solvent. These set of properties do not depend on the type of species present.

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Use the drop-down menus to select the names of the labeled structures.<br><br> A: B: C:

Naya [18.7K]

Answer:

A Petal B Stigma C Stamen

Explanation:

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3 years ago

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An alien civilization has different names and symbols for six elements. Place each element in the correct location in the period

Margarita [4]

Answer:

why whould I know that!!??

Explanation:

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3 years ago

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Element x reacts with oxygen to produce x2o3 in an experiment it is found that 1.0000 g of x produces 1.1xxx g of x2o3 what is t

8_murik_8 [283]

Since X is 1 g, therefore O must be 0.1 g. Therefore:

moles O = 0.1 g / (16 g / mol) = 0.00625 mol

We can see that for every 3 moles of O, there are 2 moles of X, therefore:

moles X = 0.00625 mol O (3 moles X / 2 moles O) = 0.009375 mol

Molar mass X = 1 g / 0.009375 mol

<span>Molar mass X = 106.67 g/mol</span>

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3 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

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3 years ago

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Which of the following is an incorrect name for an acid? A. acetic acid B. hydrocarbonate acid C. hydrocyanic acid D. sulfurous

just olya [345]

Answer:

The answer to your question is letter B

Explanation:

Incorrect name

A. acetic acid This name is correct for the acid with formula CH₃COOH

B. hydrocarbonate acid This is not the name for acid but for a molecule that has hydrogen and a metal.

C. hydrocyanic acid This name is correct for the inorganic molecule with formula HCN

D. sulfurous acid This name is correct and is the name of the inorganic molecule with formula H₂SO₃.

E. phosphoric acid This name is correct for the acid with formula H₃PO₄.

5 0

3 years ago