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stiks02 [169]
2 years ago
6

2. Water is an example of a(n)

Chemistry
1 answer:
Alexus [3.1K]2 years ago
6 0

it would be A ,inorganic Compound

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What is the predominant intermolecular force in CBr4
vovikov84 [41]
CBr4 is a symmetric tetrahedral molecule so it will be non-polar.
3 0
3 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
3 years ago
How many molecules are there in 3 moles of aluminum chloride ?
Svetradugi [14.3K]

there are 4 atoms in aluminum chloride

6 0
2 years ago
. One of the essential minerals in the human body is salt. How much salt (NaCl) is in the average adult human body?
son4ous [18]

Answer:

200g or 40 teaspoons

Explanation:

An average human, weighing about 50 pounds, has about 200 g or 40 tps of NACl

4 0
3 years ago
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

  • 1.67 mol * 65 g/mol = 108.6 g
6 0
2 years ago
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