Ask question

Ask question

All categories

3 years ago

8

Combustion of 6.90g of this compound produced 13.8g of CO2 and 5.64g of H2O. What is the empirical formula of the unknown compou

nd?

1 answer:

Alla [95]3 years ago

4 0

The empirical formula for the unknown compound would be: C2H4O (2 molecules of Carbon, 4 molecules of Hydrogen, and 1 molecule of Oxygen)

You might be interested in

An article in a magazine denies that Earth is warming and climate change is occurring. The writer reports that this season was t

ivann1987 [24]

First off, you should be looking at the whole world, not just the northeastern US. Are other regions warming? Can you provide proof that simply because the northeast received a lot of snow and low temperatures, climate change is not a viable theory?

6 0

3 years ago

Natation TT mean to genetics

neonofarm [45]

I am not sure what you are asking, but TT likely refers to the pure tall peas Mendel worked with.

4 0

3 years ago

What cellular macromolecules make up the complement pathway?

never [62]

The cellular macromolecules that are referred by the description above are the LIPIDS. The complement pathway system is a constituent of the natural immunity and this provides us a natural defense against infection. In the complement pathway of the immune system, the macromolecules that make this up are the lipids.

5 0

3 years ago

Problem #6 If you dissolve lead (II) nitrate and potassium iodide in water they will react to form lead (II) iodide and potassiu

LenaWriter [7]

$\implies \huge \tt \underline \red{answer}$

Should ever come back

8 0

3 years ago

Use the ΔHrxn values of the following reactions: 2SO2(g) + O2(g) → 2SO3(g) ΔHrxn = –196 kJ 2S(s) + 3O2(g) → 2SO3(g) ΔHrxn = –790

sesenic [268]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -297 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$S(s)+O_2(g)\rightarrow SO_2(g)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$ $\Delta H_1=-196kJ$

(2) $2S(g)+3O_2(g)\rightarrow 2SO_3(g)$ $\Delta H_2=-790kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=\frac{[1\times (-\Delta H_1)]+[1\times \Delta H_2]}{2}$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=\frac{[(1\times -(-196))+(1\times (-790))}{2}=-297kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -297 kJ.

8 0

3 years ago