Answer:
The answer is "3".
Explanation:
acetic acid Dissociation:
Dissociation constant of the Ka:
using the ICE table:
x is negligible compared to Ka.
From the ICE table,
E= m× C2
... than:
m=E/c 2
m=1,366•10^3/3•10 A 8 2
m=1,518•10^14 kg
I think so that hypothesis is for isotopes and atomic mass.
Answer:
262 ppm of Na₃PO₄
Explanation:
In a dilution, the concentration of the initial solution is decreased. When you take 5.00mL of the solution that is diluted to 25.0mL The solution is diluted 25/5 = <em>5 times</em>
If you make another two serial dilutions the final solution wil decrease its concentration 5*5*5 = 125 times
As original solution containing 0.200 M of Na3PO4, the final solution will have a concentration of:
0.200M / 125 = <em>1.6x10⁻³M</em>
Molarity is defined as the ratio between moles and liters. 1.6x10⁻³ moles of Na3PO4 in 1L are:
1.6x10⁻³mol ₓ (164g/mol) = 0.262g Na₃PO₄ / L
Assuming density of Na3PO4 as 1g/mL the concentration of the solution is:
0.262mL Na₃PO₄ / L
As 1mL = 1000μL:
262μL Na₃PO₄ / L
μL of solute per L of solution is equal to ppm, that means the solution has:
<h3>262 ppm of Na₃PO₄</h3>
Answer:
4.952 × 10^28 atoms
Explanation:
1. Atoms in one molecule of glucose
The formula for glucose is C6H12O6.
In one molecule of glucose there are
6 + 12 + 6 = 24 atoms
2. Molecules of glucose
Molecules = 6853 mol × (6.022 × 10^23 molecules/1 mol)
= 4.1269 × 10^27 molecules
3. Atoms
Atoms = 4.1269 × 10^27 molecules × (12 atoms/1 molecule)
= 4.952 × 10^28 atoms
There are 4.952 × 10^28 atoms in 6853 mol of glucose