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Korvikt [17]
2 years ago
5

What is the empirical formula of a compound that contains 10.0g aluminum and 8.88g oxygen?

Chemistry
1 answer:
Fiesta28 [93]2 years ago
6 0

A s the empirical formula of a compound that contains 10.0g aluminum and 8.88g oxygen assessing the effect of anxiety (low vs. high) and stress (low vs. moderate vs. high) on test.

Everyone experiences anxiety occasionally, but persistent anxiety can reduce your quality of life. Though likely best known for altering behavior, worry can have negative effects on our physical health. niety speeds up our heartbeat and breathing, concentrating blood flow aluminum the parts of our brains that need it. You are getting ready for a challenging situation by having this aluminum bodily reaction. Test performance may be impacted by anxiety. According to studies, pupils with low levels of test anxiety perform better on multiple-choice question (MCQ) exams than pupils with high levels of anxiety. Studies have indicated that female students have greater levels of test anxiety than male students.

Learn more about anxiety here:

brainly.com/question/4913240

#SPJ4

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal
bonufazy [111]

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = \frac{0.840}{148.2}moles of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release (5539\times 0.00567)kJ of heat or 31.41 kJ of heat

6.60 kJ of heat increases 1^{0}\textrm{C} temperature of calorimeter.

So, 31.41 kJ of heat increases (\frac{1}{6.60}\times 31.41)^{0}\textrm{C} or 4.76^{0}\textrm{C} temperature of calorimeter

So, the final temperature of calorimeter = (20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}

3 0
3 years ago
B) Quelle est la masse de tétraoxyde de trifer (Fe3O4) produite si 3,60 moles de trioxyde de
hram777 [196]

Answer:

626,4 g de Fe₃O₄

Explanation:

Nous commencerons par écrire l'équation équilibrée de la réaction entre le fer (Fe) et le trioxyde d'aluminium. Ceci est donné ci-dessous:

9Fe + 4Al₂O₃ -> 3Fe₃O₄ + 8Al

De l'équation équilibrée ci-dessus,

4 moles d'Al₂O₃ ont réagi pour produire 3 moles de Fe₃O₄.

Par conséquent, 3,6 moles d'Al₂O₃ réagiront pour produire = (3,6 × 3) / 4 = 2,7 moles de Fe₃O₄

Ainsi, 2,7 moles de Fe₃O₄ sont produites à partir de la réaction.

Enfin, nous déterminerons la masse massique de Fe₃O₄ produite par la réaction. Ceci peut être obtenu comme suit:

Mole de Fe₃O₄ = 2,7 moles

Masse molaire de Fe₃O₄ = (3 × 56) + (4 × 16)

= 168 + 64

= 232 g / mol

Masse de Fe₃O₄ =?

Mole = masse / masse molaire

2,7 = Masse de Fe₃O₄ / 232

Croiser multiplier

Masse de Fe₃O₄ = 2,7 × 232

Masse de Fe₃O₄ = 626,4 g

Par conséquent, 626,4 g de Fe₃O₄ sont produits à partir de la réaction

7 0
2 years ago
What are the signs that a chemical reaction has occurred
Finger [1]
Well. There is the color change and the formation of bubbles. 
8 0
3 years ago
Read 2 more answers
Write the ions present in a solution of K2CO3. Express your answers as chemical formulas separated by a comma. Offset subscripts
meriva

Answer:

K ^+ , CO3 ^2-

Explanation:

The compound is potassium trioxocarbonate(IV).

It contains cation (potassium ion) and acid radical ( trioxocarbonate (IV) ion).

Since K is in group 1 of the periodic table, it loses one electron to form ion i.e K^1. trioxocarbonate IV ion has a charge of 2-.and so the ions of the compound are as shown in the answer above.

8 0
2 years ago
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