Use the Periodic Table of the Elements to answer

What is the empirical formula for C14H28O7?

mrs_skeptik [129]

Answer:

Empirical formula of C4H8O2, is the answer

Explanation:

7 0

3 years ago

The student is now told that the four solids, in no particular order, are aluminum chloride (AlCl3), sugar (C6H12O6), benzoic ac

zaharov [31]

Answer:

AlCl₃, NaBr, C₆H₅COOH, C₆H₁₂O₆.

Explanation:

The more ions in solution, the greater the conductivity of a solution because these charged particles can carry electrons in the solution.

In the first place, when AlCl₃ dissolves in water it produces 4 ions:

AlCl₃(aq) ⇄ Al³⁺(aq) + 3 Cl⁻(aq)

If the solution is 0.20M in AlCl₃, it will be 4 x 0.20M = 0.80M in ions.

Secondly, NaBr is a strong electrolyte (complete ionization) so it produces 2 moles of ions per each mole of NaBr dissolved:

NaBr(aq) ⇄ Na⁺(aq) + Br⁻(aq)

If the solution is 0.20M in NaBr, it will be 2 x 0.20M = 0.40M in ions.

Then, benzoic acid is a weak electrolyte (partial ionization) so it will produce less than 2 moles of ions per each mole of benzoic acid dissolved:

C₆H₅COOH(aq) ⇄ C₆H₅COO⁻(aq) + H⁺(aq)

If the solution is 0.20M in benzoic acid, the solution will be far below 0.40M in ions.

Finally, sugar has only nonpolar covalent bonds so it will produce no ions in solution, thus being a poor electricity conductor.

4 0

3 years ago

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g CO2 and 0.6551 g

Ede4ka [16]

Answer:

C2H4O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.6004/44 = 0.0364

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0364= 0.4368g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.6551/18 = 0.0364 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0364= 0.0728

The mass of hydrogen is thus 0.0728* 1 = 0.0728g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.8009 - 0.0728 - 0.4368 = 0.2913 mole

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.2913/16 = 0.0182 mole

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.0182

H = 0.0728/0.0182 = 4

C= 0.0364/0.0182 = 2

O= 0.0182/0.0182= 1

The empirical formula is thus C2H4O

3 0

3 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago

Is chemistry hard in grade 11 explain to me everything plsss thanks

damaskus [11]

Answer: yes it’s hard

Explanation:

8 0

2 years ago

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