Answer:
the answer is destructive interference
Uhh add a picture so I can help
The option which gives the correct mole ratios is H₂S : SO₂ = 2 : 2 and O₂ : H₂O = 3 : 2
<h3 /><h3>What is Mole ratio ?</h3>
It is a conversion factor between compounds in a chemical reaction, that is derived from the coefficients of the compounds in a balanced equation
Molar ratio also known as stoichiometry is the ratio in which the reactants and products are either formed or reacted in the given equation
The balanced equation for given reaction is as follows ;
2H₂S + 3O₂ --> 2SO₂ + 2H₂O
Molar ratio can be determined by the coefficients of the compounds in the balanced reaction
Coefficient is the number in front of the chemical compound and they are as follows
- H₂S - 2
- O₂ - 3
- SO₂ - 2
- H₂O - 2
Therefore, correct option is H₂S : SO₂ = 2 : 2 and O₂ : H₂O = 3 : 2
Learn more about mole ratio here ;
https://brainly.in/question/32799056
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Answer:
The correct answer is option B.
Explanation:
![3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2](https://tex.z-dn.net/?f=3NaOH%2B2NBr_3%5Crightarrow%203HOBr%2B3NaBr%2BN_2)
Moles of
= 40 mol
Moles of NaOH = 48 mol
According to reaction, 3 moles of NaOH reacts with 2 moles ![NBr_3](https://tex.z-dn.net/?f=NBr_3)
Then ,48 moles of NaOH will reacts with:
of ![NBr_3](https://tex.z-dn.net/?f=NBr_3)
Then ,40 moles of
will reacts with:
of NaOH
As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.
Moles left after reaction = 40 mol - 32 mol = 8 mol
Hence, the
is an excessive reagent.
Answer:
Explanation:
Matter is anything that has weight and occupies space.
To prove that a bicycle is a matter, we need to show that it has weight and will occupy space.
When you put the bicycle on a weighing scale, you will see the weight of the bicycle. This a proof that bicycle is a matter.
To show that the bicycle can occupy space, place the device in a tank full of water. From the tank, the bicycle will displace some water. Substances that cannot occupy space will not behave in such manner.