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horrorfan [7]
2 years ago
15

How is a doorbell an example of an electrical transformation

Chemistry
1 answer:
Alexeev081 [22]2 years ago
8 0
It goes from mechanical to sound I think
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The Ka1 value for oxalic acid is 5.9 x10-2 , and the Ka2 value is 4.6 x 10-5 . What are the values of Kb1 and Kb2 of the oxalate
andre [41]

Answer:

2.17x10⁻¹⁰ = Kb1

1.69x10⁻¹³ = Kb2

Explanation:

Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:

C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²

C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵

Oxalate ion, C₂O₄²⁻, has as equilibriums:

C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1

C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2

Also, you can know: KaₓKb = Kw

<em>Where Kw is 1x10⁻¹⁴</em>

Thus:

Kw = Kb2ₓKa1

1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵

<h3>2.17x10⁻¹⁰ = Kb1</h3>

And:

Kw = Kb1ₓKa2

1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²

<h3>1.69x10⁻¹³ = Kb1</h3>

<em>That is because the inverse reaction of, for example, Ka1:</em>

<em>C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1</em>

<em>+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴</em>

<em>= </em>

<em>C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1</em>

6 0
2 years ago
Which characteristic is given by the angular momentum quantum number
Marianna [84]

Answer: orbital shape.


Justification:


1) There are four quantum numbers to describe the electrons. These are:


i) Principal quantum number (n)

ii) Azimuthal quantum number (ℓ), also called angular momentum quantum number.

iii) Magnetic quantum number (m)

iv) Spin quantum number (s)


2) The principal quantum number tells the main energy level. It can be 1, 2, 3, 4, 5, 6, 7. It is related with the orbial size. 1 is a small orbital, 7 is a big orbital.


2) The Azimuthal quantum number (ℓ) or angular momentum quantum number may be a number between 0 and n - 1.


It tells the kind of orbital, which is its shape


The correspondence is:


0 = s orbital,


1 = p orbital,


2 = d orbital,


3 = f orbital.


3) Magnetic quantum number (m) tells the orientation. It can be from - ℓ to + ℓ


For example when ℓ = 1, the orbital is p, and the magnetic quantum number may be -1, 0, or +1, which correspond to px, py, pz: the orientation of the p orbital in the space.


4) Spin quantum number (s) can be either +1/2 or -1/2.

7 0
2 years ago
Read 2 more answers
I just need help with these 2 questions I accidentally answered the first one
Anni [7]

Answer:

i think the pool

Explanation:

7 0
2 years ago
The ph scale for acidity is defined by ph=−log10[h+] where [h+]is the concentration of hydrogen ions measured in moles per liter
Leokris [45]

The question is incomplete.

The complete question probably is

The pH scale for acidity is defined by pH = − log₁₀[H⁺] where [H⁺] is the concentration of hydrogen ions measured in moles per liter (M). A solution has a pH of 2.55. Find the hydrogen ion concentration.

Answer: -

0.003 M

Explanation: -

pH = − log₁₀[H⁺]

Thus the hydrogen ion concentration [H⁺] = 10^{-pH}

= 10^{-2.55}

= 0.003 M

7 0
3 years ago
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
mina [271]

<u>Answer:</u> The amount of energy released per gram of B_5H_9 is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]

Taking the standard enthalpy of formation:

\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of B_5H_9 produces -9078.57 kJ of energy.

Or,

If (2\times 63.12)g of B_5H_9 produces -9078.57 kJ of energy

Then, 1 gram of B_5H_9 will produce = \frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ of energy.

Hence, the amount of energy released per gram of B_5H_9 is -71.92 kJ

8 0
3 years ago
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