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marysya [2.9K]
3 years ago
15

Alpha and beta particles and gamma rays are examples of:

Chemistry
1 answer:
elixir [45]3 years ago
6 0
B. They are all types of radiation.
For example helium has 2 protons and 2 neutrons in its nucleus. If both of its electrons were removed it would become an alpha particle and so on.
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What’s “the ring of fire” meaning ( no answers from goo gle please I really wanna understand this because it’s coming in my test
klemol [59]

Answer:

"Ring of Fire" sounds ominous, the term refers to falling in love

Explanation:

Brainliest???

8 0
2 years ago
If a flask maintained at 674 K contains 0.138 moles of NH4I(s) in equilibrium with 4.34×10-2 M NH3(g) and 9.39×10-2 M HI(g), wha
r-ruslan [8.4K]

Answer:

K = 4.07x10⁻³

Explanation:

Based on the reaction:

NH₄I(s) ⇄ NH₃(g) + HI(g)

You can define K of equilibrium as the ratio of concentrations of reactants and products, thus:

K = [NH₃] [HI] / [NH₄I]

But, as NH₄I is a solid, is not taken into account in the equilibrium, that means K expression is:

K = [NH₃] [HI]

As the concentrations in equilibrium of the gases is:

[NH₃] = 4.34x10⁻²M

[HI] = 9.39x10⁻²M

Equilibrium constant, K, is:

K = 4.34x10⁻²M * 9.39x10⁻²M

<h3>K = 4.07x10⁻³</h3>

4 0
3 years ago
What is the dewpoint when the air temperature is 26c and the relative humidity is 77%?
Zielflug [23.3K]
The dew point would be about 22 . (this is C )  
7 0
3 years ago
Read 2 more answers
Generic Formula &amp; Molecular Geometry of o3
ella [17]

Answer:

We use the following formula as given below Use the formula below to find the lone pair on the oxygen atom of the SO3 molecule. L.P (O) = V.E (O) – N.A (S-O) Lone pair on the terminal oxygen atom in SO3 = L.P (O)

Explanation:

5 0
2 years ago
How many grams of oxygen are required to react with 17.0 grams of octane (c8h18) in the combustion of octane in gasoline?
Keith_Richards [23]
The balanced equation for the combustion of octane is as follows
2C₈H₁₈ + 25O₂ ---> 16CO₂  + 18H₂O
stoichiometry of C₈H₁₈  to O₂ is 2:25
number of octane moles reacted - 17.0 g / 114.2 g/mol = 0.149 mol 
according to molar ratio 
if 2 mol of octane reacts with 25 mol of O₂
then 0.149 mol of octane reacts with - 25 /2 x 0.149 mol = 1.86 mol of O₂
 mass of O₂  - 1.86 mol x 32 g/mol = 59.5 g
59.5 g of O₂ is required to react with 

8 0
3 years ago
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