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yulyashka [42]
3 years ago
15

Kinetic energy is the energy an object has as a result of its    

Chemistry
2 answers:
mestny [16]3 years ago
7 0
Kinetic energy is the energy and object has as a result of its motion
arlik [135]3 years ago
4 0

Answer : The kinetic energy is the energy an object has as a result of its motion.

Explanation :

Kinetic energy : It is defined as the energy an object has due to its motion.

All the moving object have the kinetic energy. Not all moving object have the same amount kinetic energy.

That means the amount of kinetic energy depends on the mass of an object and the speed of the object.

The formula used for the kinetic energy :

K.E=\frac{1}{2}\times (m\times v^2)

where,

K.E = kinetic energy of an object

m = mass of an object

v = speed of an object

Hence, the kinetic energy is the energy an object has as a result of its motion.

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The burning of fossil fuels produces what kind of energy​
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Answer:

When fossil fuels are burned carbon and hydrogen react with oxygen in air to carbon dioxide (CO2) and water (H2O). During this reaction heat is released which further amplifies the reaction. Electricity is generated by transforming mechanical energy (heat) to electrical energy in a turbine or generator.

Explanation:

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Why is Potassium not used in school laboratory
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<h3>Because it is harmful for school environment.</h3>

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3 years ago
Methods: Part A: Preparation of Buffers Make two buffers starting with solid material, which is the most common way to make buff
Alecsey [184]

Answer:

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀ \frac{A^{-} }{HA}

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; <em>pka=7,21</em>

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀ \frac{HPO4^{2-} }{H2PO4^{-} }

If desire pH is 7,0 you will obtain:

<em>0,617 =  \frac{HPO4^{2-} }{H2PO4^{-} } </em><em>(1)</em>

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] <em>(2)</em>

Replacing (1) in (2) you will obtain:

<em>[H₂PO₄⁻] = 0,0618 M</em>

And with this value:

<em>[HPO₄²⁻] = 0,0382 M</em>

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× \frac{0,0382mol}{1L}× \frac{141,96g}{1mol} = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× \frac{0,0618mol}{1L}× \frac{119,96g}{1mol} = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀ \frac{Tris-base }{Tris-H^{+} }

If desire pH is 8,0 you will obtain:

<em>0,841 =  \frac{Tris-base }{TrisH^{+} } </em><em>(3)</em>

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] <em>(4)</em>

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× \frac{0,0457mol}{1L}× \frac{121,1g}{1mol} = 0,553 g of Tris-base

Tris-HCl = 0,1 L× \frac{0,0543mol}{1L}× \frac{157,6g}{1mol} = 0,856 g of Tris-HCl

I hope it helps!

8 0
3 years ago
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