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2 years ago

How many total atoms are in C3H-NO3?

Chemistry

1 answer:

Alik [6]2 years ago

5 0

Answer:

Percent composition by element

Element Symbol # of Atoms

Hydrogen H 5

Carbon C 3

Nitrogen N 3

Oxygen O 9

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LolnnrjejahaushanakKababab

snow_lady [41]

Answer:

LolnnrjejahaushanakKababab

Explanation:

Source: Trust me bro

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2 years ago

Which statement is true of a

soldier1979 [14.2K]

I think the awnser to your question is C

6 0

2 years ago

BRAINLIESTT ASAP! PLEASE HELP ME :)

ruslelena [56]

Answer:

See below

Step-by-step explanation:

Matter is either a <em>pure substance</em> or a <em>mixture. </em>

Pure substances

Are composed of one type of atom or molecule.
Have a constant chemical composition
Have fixed chemical properties
Have fixed physical properties

• For example, melting point, boiling point, density, solubility

Mixtures:

Consist of two or more substances not chemically combined
Have a variable composition
Can be separated into two or more components by physical means

• For example, filtration, distillation, centrifugation

Each component retains its own properties

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2 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

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2 years ago

What type of scientific inquiry is friction ridge analysis

Ilia_Sergeevich [38]

Answer:

Towards this goal, this project aims to develop a statistical measure of the uncertainty of the decisions made on the friction ridge evidence (i.e., evidential value of fingerprint comparison), which ultimately can be referred to as a scientific basis of the identification decisions made in friction ridge analysis.

Explanation:

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2 years ago

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