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Natalija [7]
3 years ago
5

Since water can be prepared in the laboratory

Chemistry
1 answer:
Arturiano [62]3 years ago
7 0

Answer:

# In a familiar high-school chemistry demonstration, an instructor first uses electricity to split water into its constituent gases, Hydrogen and Oxygen. Then, by combining the two gases and igniting them with a spark, the instructor changes the gases back into water with a loud pop (That means the energy is released in the process).

# There are new other ways to produce water in laboratory, however, the scientists can not produce water in large quantity for the masses, because of some reasons.

1- Theoretically, this is possible, but it would be an extremely dangerous process. Since Hydrogen is extremely flammable and Oxygen supports combustion, it wouldn’t take much to create this force, but we also have an explosion. That’s why this process can be a deadly one if our experiment is big enough.

2- Personally, I think that it makes no sense to produce water in a laboratory ( or in a large plant) for people to use as daily water. The much more important thing we need to do is to save our environment, our planet Earth. Because the daily water people drink contains not just water molecules but other minerals, the marine life is depend not just in water molecules but diferent factors, etc.

Explanation:

This is just my personal opinion. Hope that can help you a little. Have a nice day

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( a ) Filtration

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( a ) Filtration

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( b ) sugar refining

The raw sugar can be purified from the process of sugar refining .

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A chemical reaction happens
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Where does the equilibrium point occur in a reaction system?
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Equilibrium occurs when the rate of the forward reaction is the same as the rate of the reverse reaction. This doesn't necessarily mean the concentrations or pressure are the same on both sides of the equation, only the rates are the same
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3 years ago
The first-order decay of radon has a half-life of 3.823 days. How many grams of radon remain after 7.22 days if the sample initi
Nesterboy [21]

substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span> Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span> <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span> <span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>

6 0
3 years ago
With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?
mina [271]

Answer: 7.07 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

\text{Moles of} zinc=\frac{21g}{65g/mol}=0.32moles

\text{Moles of} CuCl_2=\frac{7g}{134g/mol}=0.052moles

Zn+CuCl_2\rightarrow Cu+ZnCl_2

According to stoichiometry :

1 mole of CuCl_2 require 1 mole of Zn

Thus 0.052 moles of CuCl_2 will require=\frac{1}{1}\times 0.052=0.052moles  of Zn

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Zn is the excess reagent.

As 1 mole of CuCl_2 give = 1 mole of ZnCl_2

Thus 0.052 moles of CuCl_2 give =\frac{1}{1}\times 0.052=0.052moles  of ZnCl_2

Mass of ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g

Thus 7.07 g of ZnCl_2 will be produced from the given masses of both reactants.

8 0
3 years ago
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