All categories

3 years ago

How Do Glow Sticks Glow?

Chemistry

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer:

All liquid glow products depend on a chemical process known as CHEMILUMINESCENCE to produce their light. Chemiluminesence is a chemical reaction that causes a release of energy in the form of light. To produce this light the electrons in the chemicals become excited and rise to a higher energy level.

To utilise this process glowsticks contain two liquids; hydrogen peroxide and tert-butyl alcohol. When mixed together it is these liquids that create the glow. Fluorescent dyes are also used in the alcohol to alter the colour of the light emitted.

Explanation:

You might be interested in

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Determine the mass in grams of 5.15 × 10²¹ atoms of chromium. (The mass of one mole of chromium is 52.00 g.)

Dennis_Churaev [7]

The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Cr

But:

1 mole of Cr = 52 g

Thus, we can say that:

6.02×10²³ atoms = 52 g of Cr

With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:

6.02×10²³ atoms = 52 g of Cr

Therefore,

5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³

5.15×10²¹ atoms = 0.44 g of Cr

Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

Learn more: brainly.com/question/15488332

5 0

2 years ago

Natural gas burns in air to form carbon dioxide and water, releasing heat. CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔHrxn = -802.3 kJ.

Olenka [21]

Answer:

1) Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Explanation:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g) ,\Delta H_{rxn} =-802.3 kJ$

1) Minimum mass of methane required to raise the temperature of water by 21.0°C.

Mass of water = m = 45.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 21.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 45.0 g\times 4.18 J/g^oC\times 21.0^oC$

Q = 3,950.1 J = 3.9501 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 3.950.1 kJ of heat will be given by:

$=\frac{3.950.1 kJ}{802.3 kJ}=0.004923 mol$

Mass of 0.004923 moles of methane :

0.004923 mol × 16 g/mol=0.0788 g

Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to raise the temperature of water by 26.0°C.

Mass of water = m = 50.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 26.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 50.0 g\times 4.18 J/g^oC\times 26.0^oC$

Q = 5,434 J= 5.434 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 5.434 kJ of heat will be given by:

$=\frac{5.434 kJ}{802.3 kJ}=0.006773 mol$

Mass of 0.006773 moles of methane :

0.006773 mol × 16 g/mol= 0.108 g

Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

6 0

4 years ago

Complete the sentence using the word distance, elevation, or relief. Scientists use mean sea level to make sure they have a stan

alexandr402 [8]

Answer:

ELEVATION

Explanation:

4 0

3 years ago

If Steve throws a football 57 meters in 3 seconds, what is the average speed of the football?

sattari [20]

Answer:

19 m/s

Explanation:

57/3

7 0

3 years ago