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Dennis_Churaev [7]
3 years ago
6

4.50 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 1

28. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product carbon dioxide water mass 15.47 g 2.53 g Use this information to find the molecular formula of X.
Chemistry
1 answer:
Anna11 [10]3 years ago
5 0

Answer:

\mathbf{C_{10}H_8}   ( Naphthalene )

Explanation:

Given that:

4.50 g of a Compound X is made up of Carbon , Hydron and Oxygen

It's molecular molar mass = 128 g/mol

Compound X undergoes combustion reaction and the product yield :

CO_2 with mass 15.47g and :

H_2O with mass 2.53 g

The objective is to use this information to determine the molecular formula of X.

We all know that ; number of moles = mass/molar mass

where the molar mass of H_2O is 18 g/mol

number of moles of H_2O product = 2.53 g/18 g/mol

number of moles of H_2O product = 0.1406 moles

Also; the molar mass of CO_2 product = 44 g/mol

number of moles of CO_2 product = 15.47g/ 44 g/mol

number of moles of CO_2 product =  0.3516 moles

number of moles of Compound X in the reactant side= 4.50 g /128 g/mol

number of moles of Compound X n the reactant side= 0.03516 moles

Now; number number of moles of CO_2 in reactant = 0.3516 moles/0.03516 moles

Now; number number of moles of CO_2 in reactant = 10

number of moles of H_2O reactant = 0.1406 moles × 2/0.03516

number of moles of H_2O reactant = 7.997 ≅ 8

Since we said the Compound X is known to be made of Carbon C , Hydrogen H and Oxygen O

Then the molecular formula can be written as :

\mathbf{C_{10}H_8O_{x}}

In order to find the x; we have

128  = (12 × 10 + 1 × 8 + 16 × x)

128 = 120 + 8 + 16x)

128  =  128 + 16 x

128 - 128 = 16 x

0 = 16 x

x = 0/16

x = 0

As x = 0 ; hence there are no oxygen present in the reaction

Thus; the molecular formula of Compound X = \mathbf{C_{10}H_8} which is also known as Naphthalene

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2 years ago
[H] = 1.00 x 10^-7 M <br><br> a. Acidic <br> b. Basic <br> c. Neutral
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8 0
3 years ago
What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

7 0
2 years ago
The kb of hypochlorous acid is 3. 0×10^–8 at 26. 0 C. What is the percent of ionization of hypochlorous?
astraxan [27]

The acid dissociation constant (Ka) defines the difference between a weak and a strong acid. The % ionization of hypochlorous acid is 0.14%.

<h3>What is the acid dissociation constant?</h3>

The acid dissociation constant is used to define the ionization constant of an acidic substance. It gives the quantitative measurement of the strength.

The ICE table is attached to the image below.

The acid dissociation constant (Ka) for the reaction is,

Ka = [H⁺][ClO⁻] ÷ [HClO]

= a² ÷ (0.015 - a)

= 3.0 x 10⁻⁸

Now, a² + 3.0 x 10⁻⁸ a - 4.5 × 10⁻¹⁰ = 0

So, a = 2.210 × 10⁻⁵

Solving further,

[H+] = a = 2.210 × 10⁻⁵ M

The percent ionization is calculated as,

[H+] ÷ [HClO] × 100

= 2.210 × 10⁻⁵ M ÷ 0.015 × 100

= 0.14 %

Therefore, 0.14 % is the percentage of hypochlorous ionization.

Learn more about acid dissociation constant here:

brainly.com/question/22668939

#SPJ4

Your question is incomplete, but most probably your full question was, The ka of hypochlorous acid (HClO) is 3.0 x 10⁻⁸ at 25.0°C. What is the % of ionization of hypochlorous acid in a 0.015 aqueous solution of HClO at 25.0C?

6 0
1 year ago
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