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3 years ago

The study of all chemicals containing carbon

Chemistry

2 answers:

Klio2033 [76]3 years ago

8 0

Answer:

Organic chemistry

Explanation:

laila [671]3 years ago

6 0

The correct answer is
3. Organic Chemistry.....

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When an object such a a stone is dropped into water, it disturbs the surface of the water. Waves form at the surface of the wate

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Answer: sorry i’m late but the answer is D- wavelength :)

Explanation:

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3 years ago

By what method do we measure the distance to objects that are a few thousand light-years away?

Zigmanuir [339]

Answer:

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Explanation:

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3 years ago

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Metal is a conductor of heat

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3 years ago

If the relative rate of diffusion of ozone as compared to chlorine is 6:3 and further if the density of chlorine is 36 Find out

Alexandra [31]

Answer:

The density of ozone is 4.24.

Explanation:

The relation between the relative rate of diffusion and density is given by :

$r\propto \dfrac{1}{\sqrt d}$

The given ratio of the relative rate of diffusion of ozone as compared to chlorine is 6:3.

Let the density of ozone is d₂.

$\dfrac{r_1}{r_2}=\sqrt{\dfrac{d_2}{d_1}} \\\\\dfrac{6}{3}=\sqrt{\dfrac{d_2}{36}} \\\\3=\dfrac{\sqrt{d_2}}{6}\\\\d_2=\sqrt{18} \\\\d_2=4.24$

So, the density of ozone is 4.24.

6 0

3 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

3 years ago