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2 years ago

Decribe how hydrogen and oxygen form water

Chemistry

1 answer:

rewona [7]2 years ago

8 0

Well, its a very simple process actually. Here is how it works;

When molecular hydrogen, also known as H2, and oxygen, also known as O2, are combined and are able to react together, energy is released and the molecules of the hydrogen and the oxygen. These can combine to form either water or hydrogen peroxide, depending on how the chemicals mix.

I hope I made your day a little easier,

Have a great day and happy Halloween!

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Write a method that could be used to produce pure crystals of copper chloride from copper oxide and hydrochloric acid.

joja [24]

- Describe how you would make the salt from the reactants.

CuO + 2 HCl → CuCl₂ + H₂O

- Describe how you would purify the salt from the reaction mixture.

Filter the solution and let the product crystallize.

Explanation:

The reaction between copper oxide (CuO) and hydrochloric acid (HCl) will produce copper chloride (CuCl₂) and water:

CuO + 2 HCl → CuCl₂ + H₂O

-Describe how you would make the salt from the reactants

In a beaker which contain cooper oxide add the hydrochloric acid. To avoid working with concentrated hydrochloric acid, the acid may be diluted with water but make sure the add the stoechiometric amount (and a little bit of excess) in respect with the copper oxide.

-Describe how you would purify the salt from the reaction mixture.

Let the reaction proceed and then filter the solution to remove the unreacted cooper oxide.

Let the filtered solution, which contain copper chloride, water and unreacted hydrochloric acid, to stand undisturbed for several days. You may see the crystals growing from the solution. To speed up the process you may reduce the temperature.

Learn more about:

purifying compounds

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3 0

3 years ago

A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad

Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -Weak acid-

1.0L × (0.050 mol / L) = 0.050 moles of NaF -Conjugate base-

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

(a) adding 0.050 mol of HCl: The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid destroying the buffer.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer wouldn't be destroyed.

3 0

3 years ago

What constitutes a positive test for ammonia?

klemol [59]

Ammonia has a characteristic sharp, choking smell. It also makes damp red litmus paper turn blue. Ammonia forms a white smoke of ammonium chloride when hydrogen chloride gas, from concentrated hydrochloric acid, is held near it.

7 0

2 years ago

For the overall reaction below, which of the following is the correctly written rate law? Overall reaction: O3(g)+2NO2(g)→N2O5(g

zavuch27 [327]

Answer: $Rate=k[O_3][NO_2]^2$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$O_3(g)+NO_2(g)\rightarrow NO_3(g)+O_2(g)$ slow

$NO_3(g)+NO_2(g)\rightarrow N_2O_5(g)$ fast

To determine the net chemical equation, we will simply add the above two equations, we get:

$O_3(g)+2NO_2(g)\rightarrow N_2O_5(g)+O_2(g)$

$Rate=k[O_3][NO_2]^2$

Order with respect to $O_3$ is 1 and Order with respect to $NO_2$ is 2.

Thus the rate law will be: $Rate=k[O_3][NO_2]^2$

6 0

3 years ago

How do you show the formation of an ionic bond using a Lewis dot diagram ?

Luden [163]

An x would represent the gained electrons
A . Would represent the valence electrons
You would just draw [ ] around the diagram
And the charge should be placed outside the brackets

6 0

2 years ago