Answer:
Is not possible to make a buffer near of 7.
Optimal pH for sulfate‑based buffers is 2.
Explanation:
The dissociations of H₂SO₄ are:
H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10
HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.
The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between <em>-11 and -9 and between 1 and 3</em>, having in mind that pH's<0 are not useful. For that reason, <em>is not possible to make a buffer near of 7.</em>
The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is <em>2.</em>
<em />
I hope it helps!
Answer:
Mass of solution=100g
mass of salt=20g
so; mass of solute=80g
percentage composition =(mass of salt/total
mass) ×100
= \frac{20}{100} \times 100 \\ = 20\%
glad to help you
hope it helps
<h2>
Hello!</h2>
The answer is:
The percent yield of the reaction is 32.45%
<h2>
Why?</h2>
To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.
We are given that:
Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.
So, calculating we have:
Hence, we have that the percent yield of the reaction is 32.45%.
Have a nice day!
Answer:
Mass of Ca in sample, Mass of Br in sample, Number of moles of Ca in sample, Number of moles of Br in sample, Mass or moles of element other than Ca or Br in sample
Explanation:
The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.
Good luck,
I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.
Answer:
The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Explanation:
Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg
The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)
where u = initial rms speed of the gas = 850 m/s
Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J
Then, 3600 J is added to the gas,
New kinetic energy of the gas = 4335 + 3600 = 7935 J
New kinetic energy of Helium atoms = (1/2)(m)(v²)
where v = final rms speed of the gas = ?
7935 = (1/2)(0.012)(v²)
v² = (7935×2)/0.012
v² = 1,322,500
v = 1150 m/s
Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.
Hope this Helps!!!
