The Earth is tilted, so it will be lit by different parts of the Sun as it orbits.
Hi hi hello hi hey hey do khan know if I have any pictures I have a picture of you
There are many safety precautions and rules you MUST follow during labs.
for this incident here is what you should do:
1)Notify your Instructor and partner
2) if the liquid is toxic (like not water or vinegar) then let your Instructor handle it properly, or follow your instructors orders (like if they say to put a towel over it or something like that)
3) Broken glassware, minus mercury thermometer, must be immediately cleaned up, do not use your bare hand, always wear gloves.
4)dispose of the broken glass properly and clean the liquid up (unless it is harmful, then let your instructor do it)
Answer:
pH = 4.164
Explanation:
The first process is to find the initial moles for the base (B) & the acid (HA)
i.e.
The acid with base reaction is expressed as;
HA + B → A⁻ + HB⁺
to 1.493 × 10⁻³ 2.047 × 10⁻³ - -
- 1.493 × 10⁻³ 1.493 × 10⁻³ 1.493 × 10⁻³ 1.493 × 10⁻³
0 5.54 × 10⁻⁴ 1.493 × 10⁻³ 1.493 × 10⁻³
From observation; both the acid & base weak
Given that:
The pKa for base = 4.594
The pKa for acid = 3.235
Recall that;
pKa = -log Ka
So; Ka =
By applying this:
For Base; Ka = = 2.5468 × 10⁻⁵
For Acid: Ka = = 5.821 × 10⁻⁴
After the reaction; we have the base with its conjugate acid & conjugate base of acid; Thus, since the conjugate acid of the base possesses a higher value of K, it is likely it would be the one to define the pH of the solution.
By analyzing the system, we have:
HB⁺ + H₂O ↔ B + H₃O⁺
to 0.01493 M 0.00554 M
- x x x
0.01493 - x 0.00554 - x x
Thus;
Using the common ion effect;
0.00554 - x 0.00554 &
0.01493 - x 0.01493
∴
x = [H₃O⁺] = 6.8635 × 10⁻⁵
∴
pH = -log(6.8635 × 10⁻⁵)
pH = 4.164
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.