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kozerog [31]
3 years ago
9

A strong acid like hcl _____.a strong acid like hcl _____.dissociates completely in an aqueous solutionis a strong buffer at low

phincreases the ph when added to an aqueous solutionreacts with strong bases to create a buffered solution
Chemistry
1 answer:
vladimir2022 [97]3 years ago
4 0
Answer: A strong acid like HCl dissociates completely in an aqueous solution.

By definition a strong acid is that that dissociates completely in aqueous solutions. That means that all the molecules of the acid will be inoized into hydronium cation (H3O+) and anion (the negative radical).

For expample, HCl is a strong acid because

HCl + H2O----> H3O(+) + Cl-

The forwar arrow indicates that all the molecules of HCl reacted with the water for form the ions.
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A 4.305-g sample of a nonelectrolyte is dissolved in 105 g of water. the solution freezes at -1.23c. calculate the molar mass of
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The answer is 62.00 g/mol. 
Solution: 
Knowing that the freezing point of water is 0°C, temperature change Δt is 
     Δt = 0C - (-1.23°C) = 1.23°C 
Since the van 't Hoff factor i is essentially 1 for non-electrolytes dissolved in water, we calculate for the number of moles x of the compound dissolved from the equation 
     Δt = i Kf m 
     1.23°C = (1) (1.86°C kg mol-1) (x / 0.105 kg) 
     x = 0.069435 mol 
Therefore, the molar mass of the solute is  
     molar mass = 4.305g / 0.069435mol = 62.00 g/mol
6 0
3 years ago
15 POINTS RIGHT NOW!!!!
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Answer:

It is valid because his experiment had clear variables.

Explanation:

He had clear variables and did everyhting the same for both plants, even though he had calcium in one and not in the other, which gives him a true and good result.

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3 years ago
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How many electrons would be in an atom of carbon
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12

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Identify the Lewis acid in this balanced equation: SnCl4 + 2Cl− → SnCl62−
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Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr
Len [333]

Answer:

a. 1875 nm

b. 4051 nm

c. 1282 nm

These all are infrared electromagnetic radiation.

Explanation:

Our strategy here is to utilize the Rydberg equation for hydrogen atom electronic transition.

1/λ = Rh x (1/n₁² - 1/n₂²)   where  λ is the wavelength

                                                   Rh is Rydberg constant

                                                   n₁ and n ₂ are the energy levels ( n₁ < n₂ )

Now lets star the calculations.

a.  n₁  = 3, n₂ = 4

1/λ = 1.097 x 10⁷ /m x (1/3² - 1/4²) = 5.333 x 10⁵/m

λ  = 1/(5.333 x 10⁵ /m) = 1.875 x 10⁻⁶ m

Converting λ to nanometers:

1.875 x 10⁻⁶ m x (1 x10⁹ nm/m) = 1875 nm

b.  n₁  = 4, n₂ = 5

1/λ = 1.097 x 10⁷ /m x  (1/4² - 1/5²) = 2.468 x 10⁵/m

λ  = 1/(2.468 x 10⁵/m) = 4.051 x 10⁻⁶ m

4.051 x 10⁻⁶ m x  (1 x10⁹ nm/m)  = 4051 nm

c.  n₁  = 3, n₂ = 5

1/λ = 1.097 x 10⁷ /m x  (1/3² - 1/5²) = 7801 x 10⁵/m

λ  = 1/(7801 x 10⁵/m) = 1282 x 10⁻⁶ m

1282 x 10⁻⁶ m x  (1 x10⁹ nm/m)  = 1282 nm

All of these transitions fall in the infrared region of the spectrum.

8 0
3 years ago
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