Answer:
1.21 g of Tris
Explanation:
Our solution if made of a solute named Tris
Molecular weight of Tris is 121 g/mol
[Tris] = 100 mM
This is the concentration of solution:
(100 mmoles of Tris in 1 mL of solution) . 1000
Notice that mM = M . 1000 We convert from mM to M
100 mM . 1 M / 1000 mM = 0.1 M
M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris
0.1 M = mmoles of Tris / 100 mL
mmoles of Tris = 100 mL . 0.1 M → 10 mmoles
We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol
And now we determine the mass of solute, by molecular weight
0.010 mol . 121 g /mol = 1.21 g
So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
The answer is D, species!
Hope this helps!
Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
Answer:
In first shell 2 electrons are present and 7 electrons are present in last shell
Explanation:
