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dimaraw [331]
2 years ago
9

Pls help I don’t understand at all

Chemistry
1 answer:
galina1969 [7]2 years ago
4 0

Answer:

Mass = 12.48 g

Explanation:

Given data:

Mass of sulfur dioxide = 25.0 g

Mass of sulfur formed = ?

Solution:

Chemical equation:

SO₂      →        S   + O₂

Number of moles of SO₂:

Number of moles = mass/molar mass

Number of moles = 25.0 g / 64.07 g/mol

Number of moles = 0.39 mol

Now we will compare the moles of SO₂  with S.

                     SO₂        :          S

                        1          :           1

                       0.39     :         0.39

Mass of sulfur:

Mass = number of moles × molar mass

Mass = 0.39 mol × 32 g/mol

Mass = 12.48 g

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jok3333 [9.3K]
The amount of grams that are in 2.3 moles of N = 32.223 or 32/100
Because there are 14.01 grams per mile of nitrogen atoms.
So…
14.01 x 2.3= 32.223
Hope this helps :)
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NARA [144]

Answer:

The answer will be listed below.

Explanation:

Kinetic Energy- Energy of motion, increases with mass

Potential Energy- Stored energy, increases with height

Both- Increases with velocity

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The nucleus is like the____of the cell
Tpy6a [65]

Answer:

Brain

Explanation:

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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
PO3−4 what's the formula name?
masha68 [24]

Answer:

Phosphate ion

Explanation:

4 0
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