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barxatty [35]
2 years ago
5

10. i really really need help plz don't waste time, thx :)

Chemistry
2 answers:
egoroff_w [7]2 years ago
6 0
Lollll lollll lolll lolll
Serjik [45]2 years ago
3 0
Dude use a calculator app
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Which of the following is the balanced equation for P + O2 → P2O5? Question 3 options: A) 4P + O2 → 2P2O B) P + 5O2 → 2P2O5 C) 4
Luden [163]

Given equation:

P + O2 → P2O5

In order for the equation to be balanced, the stoichiometry of the atoms of one kind on the reactant side must be equal to that on the product

Reactants                                            Products

P = 1                                                       P = 2

O = 2                                                     O = 5

The balanced equation would be:

4P + 5O2 → 2P2O5

Reactants                                            Products

P = 4                                                      P = 4

O = 10                                                    O = 10

Ans: D)


4 0
3 years ago
Read 2 more answers
During which step in the Can Crush Lab did water vapor force air from the can?
irina [24]
When you heated the can with the bit of water inside and you boiled it over a flame, the water turned to vapor (gas) and the pressure in the inside of the can is different from the pressure on the outside of the can. When you placed the can into a ice water beaker or a container, the can shrunk it's size, decreasing it's mass and density. The can shrunk as a result of the inside pressure being equalized with the outside pressure.
The part where you placed it in the ice bath or container was when the water vapor was forced out of the can.
3 0
3 years ago
PLEASE HELP AGAIN lol thank you :))
Annette [7]
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8 0
2 years ago
Please help
Kay [80]

Answer:

Can you post the pic to it so I can give you more info?

Explanation:

5 0
3 years ago
A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

7 0
3 years ago
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