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Kamila [148]
3 years ago
11

How does a student need to adjust the bunsen burner in order to change a luminous yellow flame into a non luminous blue flame?

Chemistry
2 answers:
11Alexandr11 [23.1K]3 years ago
6 0
A student can adjust the bunsen burner in order to change a luminous yellow flame into a non luminous blue flame by opening the vents of the burner allowing more oxygen from air to combust the fuel. Allowing more oxygen would lead to a more complete combustion resulting to a blue flame.
WINSTONCH [101]3 years ago
4 0

Answer:

Open the air holes

Explanation:

Hello,

In this case, by opening the air holes we allow more oxygen to enter in contact with the fuel so a hotter blue nonluminous flame appears as a product of the complete combustion of the fuel. On the other hand, by closing the air holes, an incomplete reaction undergoes, so a luminous yellow flame comes up.

Best regards.

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aev [14]

Answer:

Copper(II) sulphate – sodium hydroxide reaction

The reaction between copper(Il) sulphate and sodium hydroxide solutions is a good place to start. If you slowly add one to the other while stirring, you will get a precipitate of copper(II) hydroxide, Cu(OH)2.

5 0
3 years ago
Question 1 (1 point)
Dima020 [189]

Answer:

1. Both are made up of two substances that are chemically combined.                   2. oxygen(O2) , 3. CARBON (C) , 4. AIR(N2 nixed with O2 AND CO2), 5. CANNOT BE SEPARATED BY PHYSICAL MEANS

Explanation:

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4 0
3 years ago
If an atom has 5 protons, how many electrons does it have?
IRISSAK [1]

Answer:

5

Explanation:

An atom has the same number of protons and electrons

8 0
3 years ago
Read 2 more answers
Doubling the amplitude of a wave means that you increase the ___ of a wave factor of _______
Nesterboy [21]
Doubling the amplitude of a wave means that you increase the energy of a wave factor of quadrupling
5 0
3 years ago
What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)
My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ →  2KCl  +  3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1  .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol . 32 g/ 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

(3.2g / 4.8g) . 100 = 66.7 %

8 0
3 years ago
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