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nikklg [1K]
2 years ago
10

I need help with Hess law

Chemistry
1 answer:
ICE Princess25 [194]2 years ago
6 0

Answer:

alright

Explanation:

Hess law is  the enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps. Consistent with the law of conservation of energy. Starting and final conditions must be the same.

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Ionic or covalent C2H5OH
Afina-wow [57]

Answer:

Covalent  

Explanation:

A molecule of C₂H₅OH has C-C, C-H, C-O, and O-H bonds.

A bond between A and B will be ionic if the difference between their electronegativities (ΔEN) is greater than 1.6.

\begin{array}{ccc}\textbf{Bond} & \textbf{$\Delta$EN} & \textbf{Polarity}\\\text{C-C} & 2.55 - 2.55 = 0.00 & \text{Nonpolar covalent}\\\text{C-H} & 2.55 - 2.20 = 0.35 & \text{Nonpolar covalent}\\\text{C-O} & 3.44 - 2.55 = 0.89 & \text{Polar covalent}\\\text{O-H} & 3.44 - 2.20 = 1.24 & \text{Polar covalent}\\\end{array}

No bond has a large enough ΔEN to be ionic.

C₂H₅OH is a covalent molecule.

6 0
3 years ago
If we know the specific heat of a material, can we determine how much heat is released under a given set of circumstances?
Alekssandra [29.7K]

Answer:

Yes

Explanation:

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6 0
2 years ago
Unsur senyawa CaCO3 adalah
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7 0
2 years ago
Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w
jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, COCl_2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:  

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The value of K_c for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which p_{CO}=p_{Cl_2}=0.265atm and p_{COCl_2}=0.000atm ?

<u>Answer:</u> The equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of K_c\text{ and }K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p = Equilibrium constant in terms of partial pressure

K_c = Equilibrium constant in terms of concentration = 5.79

\Delta n_g = Difference between gaseous moles on product side and reactant side = n_{g,p}-n_{g,r}=1-2=-1

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

T = Temperature = 570 K

Putting values in above equation, we get:

K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

                      CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

<u>Initial:</u>            0.265      0.265

<u>At eqllm:</u>        0.265-x    0.265-x        x

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}

Putting values in above equation, we get:

0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = (0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of Cl_2=(0.265-x)=(0.265-0.008)=0.257atm

The equilibrium partial pressure of COCl_2=x=0.008atm

Hence, the equilibrium partial pressure of CO, Cl_2\text{ and }COCl_2 is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0
3 years ago
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