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murzikaleks [220]
2 years ago
6

I need chemistry help text me at (951) 897-8325 PLEASE

Chemistry
1 answer:
user100 [1]2 years ago
7 0
Just post the question on here
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if you burn a piece of wood you will feel heat see water vapor and be left with ashes how come the matter was not destroyed
Elanso [62]
The matter wasn't destroyed because albert einsteins theory of the black hole says that matter cannot be destroyed if it is burned into the same color of a black hole. this eqation proves my thesis:

²²↑↑Ф\left \{ {{y=2} \atop {x=2}} \right.  \left \{ {{y=2} \atop {x=2}} \right.  \left \{ {{y=2} \atop {x=2}} \right.  \left \{ {{y=2} \atop {x=2}} \right.
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Which subatomic particle has negligible mass and travels around outside the nucleus?
allsm [11]
The electron is a subatomic particle that has a negative charge and a negligible mass. The electron travels around out side the nucleus.

The subatomic particles that are inside the nucleus are protons and neutrons.

The answer to your question is electron. 
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to form brass, copper and zinc combine but do not chemically bond. brass is a(n) ________. a atom b element c compound d
umka21 [38]

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2 years ago
When water reacts with potassium metal the hydrogen produced ignites explosively on the surface of water .What cause the ignitio
Julli [10]

Answer:

Pottasium reacts with water vigorously and the reation is exothermic. The heat released causes the hydrogen released to ignite

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7 0
3 years ago
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
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