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emmainna [20.7K]
3 years ago
5

50 points!! i need 3 answers!!!

Chemistry
2 answers:
iris [78.8K]3 years ago
5 0

Answer:

Real gas particles have significant volume

Real gas particles have more complex interactions than ideal gas particles

tester [92]3 years ago
3 0
Option E, Real gas particles have more complex interactions than ideal gas particles.

In ideal gases, there is absolutely no interaction between any atoms. At all. Atoms simply don't bump into each other in ideal gases.

Obviously, you know that's unrealistic. In real gases, atoms collide into each other all the time.

-T.B.
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a lithium atom has three protons three neutrons and three wletrons.what is the overall charge on this atom
CaHeK987 [17]
The overall charge is neutral as three electrons cancels out the 3 proton
3 0
4 years ago
For the reaction: CH3NH2(aq) + H2O(aq) ⇌ CH3NH3 +(aq) + OH- Determine the change in the pH (ΔpH) for the addition of 6.7 M CH3NH
Korolek [52]

Answer:

The change in the pH (ΔpH) is 2,17

Explanation:

The reaction:

CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻

kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]} <em>(1)</em>

In equilibrium, a solution of CH₃NH₂ 4,7M produces:

[CH₃NH₂] = 4,7 - x

[CH₃NH₃⁺] = x

[OH⁻] = x

Replacing in (1):

4,38x10^{-4} = \frac{x^2}{4,7-x}

x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0

The solutions are:

x = -0,0456 No physical sense. There are not negative concentrations.

x = 0,04515 Real answer.

The concentration of [OH⁻] is 0,04515 M.

As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:

<em>pH = 12,65</em>

The addition of 6,7M produce this changes in concentrations:

[CH₃NH₂] = 4,656 + x

[CH₃NH₃⁺] = 6,74515 - x

[OH⁻] = 0,04515 - x

Replacing in (1) you will obtain:

x² - 6,7907x + 0,3025 = 0

Solving for x:

x = 6,74586 No physical sense

x = 0,04484 Real answer.

Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M

pOH = 3,51.

<em>pH = 10,49</em>

Thus ΔpH is 12,65 - 10,49 = <em>2,16 ≈ 2,17</em>

I hope it helps!

4 0
4 years ago
How many molecules are in the substance formula of 2C6H1206 (use coefficients)
FrozenT [24]

<u>Answer:</u> The number of formula units present in 2 moles of C_6H_{12}O_6 are 1.2044\times 10^{24}

<u>Explanation:</u>

Formula units are defined as the number of molecules or atoms present in 1 mole of a compound or element respectively.

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of formula units

Here, 2 represents the number of moles of C_6H_{12}O_6

We are given:

Moles of C_6H_{12}O_6 (glucose) = 2 moles

Number of formula units of C_6H_{12}O_6=(2\times 6.022\times 10^{23})=1.2044\times 10^{24}

Hence, the number of formula units present in 2 moles of C_6H_{12}O_6 are 1.2044\times 10^{24}

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