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cricket20 [7]
3 years ago
8

What is the mass of 2.2 moles of Beryllium chloride

Chemistry
1 answer:
Fiesta28 [93]3 years ago
5 0

Answer:

the mass of 2.2 moles of Beryllium chloride is

Explanation:

175.8200004 gram

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Determine the number of grams of carbon dioxide that can be formed from 0.500 grams of iron oxide and an excess of carbon.
Nimfa-mama [501]

Taking into account the reaction stoichiometry, 0.2066 grams of CO₂ are formed from 0.500 grams of iron oxide and an excess of carbon.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Fe₂O₃ + 3 C → 4 Fe + 3 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Fe₂O₃: 2 moles
  • C: 3 moles
  • Fe:4 moles
  • CO₂: 3 moles

The molar mass of the compounds is:

  • Fe₂O₃: 159.7 g/mole
  • C: 12 g/mole
  • Fe: 55.85 g/mole
  • CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Fe₂O₃: 2 moles ×159.7 g/mole= 319.4 grams
  • C: 3 moles ×12 g/mole= 36 grams
  • Fe: 4 moles ×55.85 g/mole= 223.4 grams
  • CO₂: 3 moles ×44 g/mole= 132 grams

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 319.4 grams of Fe₂O₃ form 132 grams of CO₂, 0.500 grams of Fe₂O₃ form how much mass of CO₂?

mass of CO_{2} =\frac{0.500 grams of Fe_{2}O_{3}x132 grams of CO_{2} }{319.4 grams of Fe_{2}O_{3}}

<u><em>mass of CO₂= 0.2066 grams</em></u>

Then, 0.2066 grams of CO₂ are formed from 0.500 grams of iron oxide and an excess of carbon.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

7 0
2 years ago
What is a solution called when it contains as much solute as it can possibly hold at a given temperature?
8090 [49]
Saturated solution is when a solution has dissolved as much as it can
4 0
4 years ago
Read 2 more answers
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
tekilochka [14]

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>

6 0
4 years ago
What is catastrophism
BigorU [14]
<span>the theory that changes in the earth's crust during geological history have resulted chiefly from sudden violent and unusual events.

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6 0
4 years ago
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In the periodic table hydrogen is placed in Group 1A (group 1) and helium is placed in Group 8A (group 18). The most likely reas
Alexus [3.1K]

Answer:

This question appears incomplete

Explanation:

This question appears incomplete because of the absence of options. However, hydrogen is placed in group 1 because it has just one electron in it's outermost shell (which happens to be the only shell it has) just like every other group 1A/group 1 element. While helium is placed in group 8A/group 18 because it has a completely filled outermost shell (which is also the only shell it has) just like every other element in group 8A/group 18.

8 0
3 years ago
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