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Rudik [331]
2 years ago
5

When a bunsen burner is properly adjusted, what should the flame look like?.

Chemistry
1 answer:
DochEvi [55]2 years ago
8 0

When a bunsen burner is properly adjusted, the flame look like outer violet and inner blue colour.

<h3>What is the clolur of flame?</h3>

When a burner is properly adjusted the flame should have an outer violet oxidizing flame and an inner blue reducing flame. The temperature of blue colour is 15000 degree Celsius whereas the cool region outside the blue has 3000C temperature.

So we can conclude that When a bunsen burner is properly adjusted, the flame look like outer violet and inner blue colour.

Learn more about flame here: brainly.com/question/448102

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Answer:

2Fe +N_{2} -> 2FeN

Explanation:

Iron is Fe, nitrogen is N. Nitrogen is diatomic, which means it occurs as a molecular pair by itself. Iron III nitride has a chemical formula of FeN because nitrogen has a charge of 3-, and iron III tells us the iron has a charge of 3+ so you just need one of each to make the charges balance and the compound neutral.

3 0
2 years ago
Which is a characteristic of a strong base
Lubov Fominskaja [6]

Answer:

Explanation:

The strong bases have following properties:

1. In solution, strong bases ionize fully.

2. On dissolving the strong bases in water they produce all hydroxide ion which they have.  

3. For strong bases the value of equilibrium constant (Kb  ) is large.

4. In general the strong base ionizes completely means concentration of ions are greater means conductivity also greater.

5. For strong bases the value of equilibrium constant (Kb) is large, thus the value of dG0 is very large negative number.

6 0
3 years ago
What will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
Marina86 [1]

Answer:

THE VOLUME OF THE NITROGEN GAS AT 2.5  MOLES , 1.75 ATM AND 475 K IS 55.64 L

Explanation:

Using the ideal gas equation

PV = nRT

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T = 475 K

R = 0.082 L atm/mol K

V = unknown

Substituting the variables into the equation we have:

V = nRT / P

V = 2.5 * 0.082 * 475 / 1.75

V = 97.375 / 1.75

V = 55.64 L

The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L

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