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yKpoI14uk [10]
3 years ago
13

A molecule of alkene has the chemical formula C3H4. How many carbon-carbon double bonds are present in the molecule?Choices:123

Chemistry
1 answer:
alexdok [17]3 years ago
4 0
2 Carbon double bonds
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Balance the following chemical equation <br> __Mg+__O2–&gt;__MgO
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Answer:

2Mg^+ +O2 right arrow 2MgO

Explanation:

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Name each subatomic particle, its charge, and its location in an atom.
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Protons: charge +1, have a mass of 1 and are found in the nucleus 

Neutrons: charge 0, have a mass of 1 and are found in the nucleus

Electrons: charge -1, have a mass of 1/840 and are found on the outside of the nucleus 

hope that helps 
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3 years ago
Lucy wants to attach a goal cost to each of her life goals. Why might she do this?
Natasha2012 [34]

Answer:

So she can have something to reach or look forward to.

Explanation:

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3 years ago
What concentration of monosodium phosphate would be required to make a solution of pH 7.4 with 0.2 M disodium phosphate ( pKa
denis-greek [22]

Answer:

The concentration of monosodium phosphate is 0.1262M

Explanation:

The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2

To determine the pH you must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>

For H₂PO₄⁻ / HPO₄⁻ buffer:

pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]

As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:

7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]

0.2 = log [0.2] / [H₂PO₄⁻]

1.58489 = [0.2] / [H₂PO₄⁻]

[H₂PO₄⁻] = 0.1262M

<h3>The concentration of monosodium phosphate is 0.1262M</h3>

<em />

8 0
3 years ago
Having been heated to 800 K , at some point the tank starts to leak. By the time the leak is repaired, the tank contains only ha
Liula [17]

Answer: The temperature of the gas reduced to 400K.

Explanation:

Stated that ; The pressure remains the same, that is initial and final pressure equals 1atm.

Applying Charles Law

V1/T1 = V2/T2

Initial volume V1 = 1

Final volume V2 = 1/2 (halved)

Initial temperature T1 =800K

Final temperature T2 = ?

(1/800) = (1/2)/T2

T2 = 800/2

T= 400K

Therefore, when the volume is halved, the temperature reduced also to half ( 400K)

3 0
3 years ago
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