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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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Determine the mass of 5.2 x 10 power of 21 molecules of propanol C3H7OH(l), on grams.
Alja [10]

n(\text{C}_3\text{H}_7\text{OH}) = N(\text{C}_3\text{H}_7\text{OH}) / N_A\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 5.2 \times 10^{21} / (6.02 \times 10^{23})\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 8.6 \times 10^{-3} \; \text{mol}

where N_A = 6.02 \times 10^{23}\; \text{mol}^{-1} the Avogadro's constant that relates the number of particles to their number, in the unit moles \text{mol}.

The molar mass of propanol- mass per mole propanol- can be directly deduced from its molecular formula with reference to a modern periodic table.

M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}

8.6 \times 10^{-3} \; \text{mol} of propanol molecules would thus have a mass of 8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}

5 0
3 years ago
*WILL GIVE BRAINLIEST TO CORRECT ANSWER *<br> Question in picture
sineoko [7]
The Answer for this question is : A

1 & 2
5 0
2 years ago
Read 2 more answers
Nuclear energy could come from
lianna [129]
Nuclear energy comes from splitting of Barium atom to form Krypton atom
5 0
3 years ago
A 225 g sample of an unknown solid is heated 67C and placed into a calorimeter containing 25.6 g of water at 15.6°c. If the fina
Stella [2.4K]

Answer:

= 1.271 J/g°C

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by  water.

But heat = mass × specific heat capacity × temperature change

Thus;

Heat released by the solid;

= 225 g × c ×(67 -53) , where c is the specific heat capacity of the metal

= 3150 c joules

Heat absorbed by water;

= 25.6 g × 4.18 J/g°C × (53-15.6)

= 4002.0992  joules

Therefore;

3150 c joules = 4002.0992 joules

c =4002.0992/3150

 <u> = 1.271 J/g°C</u>

6 0
3 years ago
When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
2 years ago
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