1. Explain Hydrogen peroxide, H 2 O 2 properties and decomposition reaction.

Chemistry

1 answer:

Len [333]2 years ago

3 0

Answer:

1) Hydrogen Peroxide. Hydrogen Peroxide (H2O2) is a colourless liquid that resembles water in many respects. Its physical properties are very similar to those of water, except that it is 40% denser. ... The single bond between the two oxygen atoms is weak, so that H2O2 readily fragments into either H and HO2 or two OHs.

2) Is human activity warming the Earth or do recent signs of climate change signal natural variations? In this feature article, scientists discuss the vexing ambiguities of our planet's complex and unwieldy climate.

a) The Water Cycle. ... The water cycle describes how water evaporates from the surface of the earth, rises into the atmosphere, cools and condenses into rain or snow in clouds, and falls again to the surface as precipitation.

b) Carbon is an essential element for life on Earth. Every living organism has carbon compounds inside each of its cells, such as fats and proteins. The carbon cycle shows how atoms of carbon can exist within different compounds at different times and be recycled between living organisms and the environment.

Explanation:

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What reaction might we use to synthesize nickel sulfate,niso4? Write equations for acid base neutralization. Include states in y

il63 [147K]

Answer:

Here's what I get

Explanation:

1. Nickel sulfate

base + acid ⟶ salt + water

NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.

Hydroxides of transition metals are insoluble; most sulfates are soluble.

$\underbrace{\hbox{Ni(OH)$_{2}$(s)}}_{\hbox{base}} + \underbrace{\hbox{H$_{2}$SO$_{4}$(aq)}}_{\hbox{acid}} \longrightarrow \, \underbrace{\hbox{NiSO$_{4}$(aq)}}_{\hbox{salt}} + \underbrace{\hbox{2H$_{2}$O(l)}}_{\hbox{water}}$

2. Carbonate + acid

Most carbonates are insoluble.

They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.

$\rm NiCO_{3}(s) + H_{2}SO_{4}(aq) \longrightarrow \, NiSO_{4}(aq) + H_{2}O(l) + CO_{2}(g)$

5 0

2 years ago

Using the equations and enthalpy values provided, which mathematical expression can be used to determine the unknown enthalpy ch

vlabodo [156]

In order to solve this, we need to make use of Hess' Law.

We are already given the equations and their corresponding deltaH. Using Hess' Law, we can generate this equation:
104 kJ = x - (-1182 kJ) - (-1144 kJ)

Among the choices, the answer is
<span>B.104 = x - [(-1182) + (-1144)]
</span>

3 0

2 years ago

Read 2 more answers

stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,

Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.