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3 years ago

Can someone please help me!! What are the parts of a Bunsen burner if possible add a picture and name please thanks

Chemistry

1 answer:

borishaifa [10]3 years ago

3 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>BUNSEN BURNER:</h2>

Bunsen burner is an instrument which is used as a source of heat using natural gas or petrol. It is used for the heating, sterlization, and combustion. Bunsen burner is named after its inventor Robert bunsen.

<h2>____________________________________</h2><h3>Mechanism:</h3>

Burner provide a single, continuous flame by mixing gas with air in a controlled fashion. The ratio of gas to air that is mixed together can be manually adjusted, allowing the user to control the intensity, temperature, and size of the flame.

<h2>____________________________________</h2><h3>PICTURE IS ATTACHED BELOW</h3><h2>____________________________________</h2><h2>Best Regards!</h2><h2>'Borz'</h2>

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4 years ago

What is the energy of light with a wavelength of 589 nm? (The speed of light in a vacuum is 3.00 x 10^8 m/s, and Planck's consta

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Answer:

Explanation:

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3 years ago

To show the electron configuration for an atom, what is the advantage of using an orbital notation compared to a dot structure?

stich3 [128]

The advantage of using an orbital notation is that it shows the electron distribution in shells.

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3 years ago

Read 2 more answers

How would you separate food coloring and water?

mr_godi [17]

Heat the water on an oven burner or place it in sunlight for several days to evaporate the water out of the pan. Allow the water to evaporate away and you will be left with the food coloring remaining in the pan.

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3 years ago

A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o

Llana [10]

Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

Substitute and solve

1.3 x 10-5 = (x)(0.206)

(0.0281)

X= 1.77 x 10-6M => pH = 5.75

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3 years ago