It is deformation of heat and pressure because igneous rock is made by magma when it cools. After a long time and pressure, it will turn into sedimentary rock.

6 0

3 years ago

What types of hydrocarbons are usually liquid at room temperature?

marishachu [46]

<span>These hydrocarbons are many and this depends on the phase diagram of the substance. In a phase diagram, the phase can be determined by looking at a certain temperature and pressure. For this case, at a temperature of 20 degrees Celsius and at 1 atmosphere.</span>

8 0

3 years ago

Use the thermochemical equations shown below to determine the enthalpy for the reaction: COCl2(g) + H2O(l) --> CH2Cl2(l) + O2

tatiyna

Answer:

ΔH° (Enthalpy change) for the reaction is equal to (-151.5 KJ)

Explanation:

Determine the enthalpy for the reaction

COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)

It becomes easier to find the unknown value of enthalpy of a particular reaction by applying the Hess's law on the given reaction with the known value of standard enthalpy change.

$\frac{1}{2}$ H₂(g) + $\frac{1}{2}$ Cl₂(g) --> HCl(g) ΔH=-46 kJ -----------------------------(1)

H₂O(g) + Cl₂(g) --> 2 HCl(g) + $\frac{1}{2}$ O₂ (g) ΔH=-21 KJ-------------------(2)

CH₂Cl₂(l) + H₂(g) + $\frac{3}{2}$ O₂(g) --> COCl₂(g) + 2 H₂O(l) ΔH = 80.5 kJ ----------(3)

Equation (1) Multiply by factor 2 we get

H₂(g) + Cl₂(g) ---> 2 HCl (g) ΔH = - 46 x 2 = - 92 KJ ------------(4)

Revers the equation (2) and (3) we get

2 HCl(g) + $\frac{1}{2}$ O₂ (g) --> H₂O(g) + Cl₂(g) ΔH=+21 KJ ----------------(5)

COCl₂(g) + 2 H₂O(l) --> CH₂Cl₂(l) + H₂(g) + $\frac{3}{2}$ O₂(g) ΔH = - 80.5 kJ -----------(6)

Now add these three equations i.e., (4) + (5) + (6)

we can get these equation

COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)

The enthalpy change during the reaction = -92 + 21 -80.5 = - 151.5 KJ

6 0

4 years ago