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aleksandr82 [10.1K]
2 years ago
10

Suggest a name for a compound containing potassium, carbon and oxygen.

Chemistry
1 answer:
spin [16.1K]2 years ago
4 0

Answer:

Oxidation of potassium amalgam with carbon dioxide results in the formation of potassium oxalate. Potassium is not reactive with benzene, although heavier alkali metals such as cesium react to give organometallic products.

Please Mark Brainliest If This Helped!

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When iron, Fe, react with chlorine, Cl, the product is iron(II) chloride, FeCl. If 56.8 g of
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3 years ago
Which graphic below shows the correct orientation of each water molecule when it is near the cl- ion in the aqueous solution?
andrew-mc [135]
The question is missing the graphics required to answer which I have attached as an image.

There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.

We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.

The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.

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8 0
4 years ago
For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (
vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g)  ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0
3 years ago
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