Answer:
You may be referring to the gas that makes up 21% of the earth's atmosphere, which is oxygen.
Explanation:
According to NASA, the gases in Earth's atmosphere include:
Nitrogen — 78 percent
Oxygen — 21 percent
Argon — 0.93 percent
Carbon dioxide — 0.04 percent
(Trace amounts of neon, helium, methane, krypton and hydrogen, as well as water vapor)
Answer:
A catalyst speeds up a chemical reaction, without being consumed by the reaction. It increases the reaction rate by lowering the activation energy for a reaction. ... Remember that with a catalyst, the average kinetic energy of the molecules remains the same but the required energy decreases
Explanation:
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hope that is helpful
Answer:
The sun radiates energy in all directions.
Explanation:
Most of it dissipates into space, but the tiny fraction of the sun's energy that reaches Earth is enough to heat the planet and drive the global weather system by warming the atmosphere and oceans
It's not really possible to tell longitudinal vs. transverse in this image as given. However, we can say that the waves labeled A are high-frequency (short wavelengths) while the waves labeled B are low-frequency (long wavelengths). So, this third answer choice would be correct here.
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
