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Karolina [17]
2 years ago
10

MOLE CONVERSIONS

Chemistry
1 answer:
madam [21]2 years ago
4 0

Answer:

1.    6.116x1024 Molecules of H2O

2.    13400 L

3.    8.001x1024 Molecules of Mg3(PO4)2

4.    572 g.

5.    1.017x1025 Molecules of N2

6.    7.24 g

.7.     6980 g. of Al(OH)3

8.    3H2 + N2    => 2NH3

9.    S8 + 8O2   => 8SO2

10.  Ni(ClO3)2→     NiCl2   +  3O2

11.  C2H4      +      3O2→      2CO2      +     2H2O

12.  2KClO3→      2KCl      +      3O2

13.  Cu(OH)2   +   2HC2H3O2→     Cu(C2H3O2)2      +     2H2O

14.  C3H8      +      5O2→      3CO2      +      4H2O

15.  191 g of CO

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Answer:

Forming a problem requires the scientist to  use creativity to imagine new solutions.

Explanation:

Albert Einstein remains a critically prominent figure who conducted remarkable, ground-breaking research that not only formed the foundations of modern physics but also strongly affected the scientific world. It is difficult to teach imagination but it can be harnessed and accepted. Nothing incites our imaginative impulses we love more than the prospect of immediate creative inspiration. And creativity hits its full potential when paired with the experience, insights, and skills people gained by questioning the real-life problems.

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2 years ago
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Chemistry help please! I just need to make sure the answers correct.... Thank you!
Veronika [31]
1. At constant tempaerature and pressure, 3 tablets produce 600cm^3 of gas
Thus calculating for 1 tablet that produces 600 / 3 = 200 cm^3
So now two tablets produce 200 x 2 = 400 cm^3
2. We have the equation PV = nRT, n being the number of moles
Pressure P = 1,000 kPa
Volume V = 3 L
R = 8.31 L kPa/mol-K
Temperature T = 298 K
n = PV / RT = (1000 x 3) / (8.31 x 298) = 3000 / 2476.38 = 1.21 moles
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8 0
3 years ago
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!

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Answer:267 g

Explanation:

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Answer:

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