Following are the advantages of corrosion: A layer of protection: A layer of oxide is formed in surface corrosion, which protects the inner metal from corrosion. Sacrificial anodes such as zinc is used as a preventive measure to stop corrosion of other metals.
Answer:
65 g (C₂H₅)₂NH₂Br
Explanation:
It seems your question lacks the values required to solve the problem. However, an internet search tells me these are the values (if your values are different, keep those in mind when solving the problem, but the methodology remains the same):
" A chemistry graduate student is given 125 mL of a 0.90 M diethylamine ((C₂H₅)₂NH) solution. Diethylamine is a weak base with Kb=1.3×10⁻³. What mass of (C₂H₅)₂NH₂Br should the student dissolve in the (C₂H₅)₂NH solution to turn it into a buffer with pH=10.53? You may assume that the volume of the solution doesn't change the (C2H5)2NH2Br is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. "
To solve this problem we use the <em>Henderson-Hasselbach equation</em>:
- pH = pKa + log
We <u>calculate pKa from Kb</u>:
- Ka = Kw/Kb ⇒ Ka = 1x10⁻¹⁴/1.3×10⁻³ = 7.69x10⁻¹²
Now we possess all the required data to calculate the concentration of the bromide salt (C₂H₅)₂NH₂Br :
- pH = pKa + log
- 10.53 = 11.11 + log
- -0.58 = log
- 10⁻⁰°⁵⁸=
Now we use the total volume to<u> calculate the moles of (C₂H₅)₂NH₂Br required to have that molar concentration</u>:
- 3.42 M * 0.125 L = 0.4275 mol (C₂H₅)₂NH₂Br
Finally we calculate <u>the mass of (C₂H₅)₂NH₂Br</u>:
- 0.4275 mol (C₂H₅)₂NH₂Br * 154.05 g/mol = 65.86 g
Which rounding to 2 significant figures is 65 g (C₂H₅)₂NH₂Br
Answer:
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Explanation:
ans: gas
The same number of particles in a gas spread further apart than in the liquid or solid states. The same mass takes up a bigger volume . This means the gas is less dense. Density also depends on the material.
Because seasons are caused by the changing angles that sunlight strikes the Earth (due to it's tilted axis), a decrease in tilt would mean less extreme seasons. Spring and fall are not affected by tilt because they are at the point in orbit where sunlight is hitting the earth evenly. However, less tilt would mean less of a change in angle around the orbit and therefore cooler summers and warmer winters.