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Aloiza [94]
3 years ago
10

Devices that mix air or oxygen with flammable gasses cannot be used unless approved by

Chemistry
1 answer:
Feliz [49]3 years ago
8 0

Flammable gases are gases that have the tendency to <u>explode (burst into flames)</u> when they come in contact with the <u>appropriate quantities of air, oxygen, or any suitable oxidant.</u>

<u />

<u />

Devices that mix air or oxygen with flammable gasses cannot be used unless approved by an <u>authorized or approved personnel.</u>

<u></u>

<u></u>

  • Examples of flammable gases are hydrocarbons such as <u>Propane, Acetylene</u>, e.t.c.

  • Flammable gases are dangerous to humans and the environment and if they are not handled properly, can cause severe damages and even death.

  • During the preparation and usage of flammable gases, safety precautions should be properly enforced such as the <u>wearing of appropriate safety gear and goggles</u>.

  • Devices that mix air or oxygen with flammable gases cannot be used unless approved by an <u>authorized or approved personnel.</u>

<u></u>

<u></u>

To learn more, visit the link below:

brainly.com/question/3702349

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Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu
erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
How many electrons can be held in the second orbital
mariarad [96]
Second orbital can hold 8 electrons

3 0
3 years ago
Read 2 more answers
Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.
bonufazy [111]

Answer:

a) 72 °F= 22.22 °C

b)  213.8 °C=  416.84°F

c) 180 °C= 453.15 °K

d) 315 °K=  107.33 °F

e) 1750 °F= 1227.594 °K

f) 0 °K=  -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

  • Fahrenheit a Celsius: C=\frac{F-32}{1.8}
  • Celsius a Fahrenheit: °F= °C*1.8 + 32
  • Celsius a Kelvin: °K= °C + 273.15
  • Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
  • Fahrenheit a Kelvin:K=\frac{F-32}{1.8} + 273.15

Entonces se obtiene:

a) 72 °F= \frac{72-32}{1.8}=22.22 °C

b)  213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= \frac{1750-32}{1.8} + 273.15= 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

7 0
3 years ago
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Tpy6a [65]
This is a lab assignment, you need to do something physically to find your answers
5 0
3 years ago
An ion with a charge of +3 would most likely be an element in Group ____.
Dmitrij [34]
1) D
2) B
3) A
Hope this helps! :)
5 0
3 years ago
Read 2 more answers
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