Answer:
Robert
Explanation:
There is not more than one colour
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.
<h2>
Answer:</h2>
In <u>Combination reaction</u>, two or more elements combined to form one compound of different properties.
- C(s) + O2(g) ⇢ CO2(g).
- H2(g) + O2(g) ⇢ H20(l).
In <u>Displacement reation</u>, the high reactive element displaces the low reactive element and formed compound of different properties.
- Fe(s) + CuSo4(aq) ⇢ FeSo4(aq) + Cu(s).
- AgNO3(aq) + Cu(s) ⇢ CuNO3(aq) + Ag(s).
Answer:- The natural abundance of is 0.478 or 47.8% and is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for and as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of as n then the abundance of would be 1-n .
Let's plug in the values in the formula:
151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:
negative sign is on both sides so it is canceled:
The abundance of is 0.478 which is 47.8%.
The abundance of is =
= 0.522 which is 52.2%
Hence, the natural abundance of is 0.478 or 47.8% and is 0.522 or 52.2% .