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anyanavicka [17]
3 years ago
11

A 0.125 g sample of unknown hydrocarbon is prepared for combustion analysis. After the hydrocarbon undergoes complete combustion

, 0.4225 g of CO2 and 0.0865 g of H2O are produced. What is the empirical formula of the unknown
Chemistry
1 answer:
Arturiano [62]3 years ago
8 0

Answer:

Empirical formula is CH

Explanation:

A hydrocarbon, CₐHₓ reacts with oxygen in the combustion reaction as follows:

CₐHₓ + O₂ → a CO₂ + x/2 H₂O

That means we can find the moles of C and H from the moles of CO2 and H2O. As empirical formula is defined as the simplest whole-number ratio of atoms in a molecule, we can determine it as follows:

<em>Moles CO2 = Moles C -Molar mass: 44.01g/mol:</em>

0.4225g * (1mol / 44.01g) = 0.0096 moles C

<em>Moles H2O -Molar mass: 18.01g/mol:</em>

0.0865g H2O * (1mol / 18.01g) = 0.0048moles H2O * (2mol H / 1mol H2O) =

0.0096 moles H

The ratio of atoms of C:H is:

0.0096moles C / 0.0096moles H = 1

That means empirical formula is CH

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A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128
alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is 50.26^{o}C.

Explanation:

The given data is as follows.

mass = 7.62 g,           T_{2} = 10.8^{o}C

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

       q = mC \times \Delta T    

          = 7.62 g \times 4.184 J/^{o}C \times (52.3 - T)

Now, heat gained by lead will be calculated as follows.

       q = mC \times \text{Temperature change of lead}  

           = 2.04 \times 0.128 \times (T - 11.0)

According to the given situation,

     Heat lost = Heat gained

7.62 g \times 4.184 J/^{o}C \times (52.3 - T) = 2.04 \times 0.128 \times (T - 11.0)

        T = 50.26^{o}C

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is 50.26^{o}C.

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3 years ago
Consider the reversible reaction, A+B⇌C+D. If the concentration of product D is increased, the rate of the reverse reaction woul
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Answer:

If the concentration of product D is increased, the rate of the reverse reaction would increase.

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A + B ⇄ C + D

In given condition the equilibrium is disturb by increasing the concentration of product.

When the concentration of product D is increased the system will proceed in backward direction in order to regain the equilibrium. Because when the product concentration is high it means reaction is not on equilibrium state the reaction will proceed backward direction to regain the equilibrium state.

According to the Le- Chatelier principle,

At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.

The equilibrium can be disturb,

By changing the concentration

By changing the volume

By changing the pressure

By changing the temperature

4 0
3 years ago
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