Chemical reaction: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄<span>(s).
</span>Ksp(Ag₂CrO₄) = 1,1·10⁻<span>¹².
</span>[Ag⁺] = c(AgNO₃) = 8.0·10⁻⁴ M.
[CrO₄²⁻] = c(K₂CrO₄) = 1.8·10⁻⁴ M.
Q(Ag₂CrO₄) = [Ag⁺]²·[CrO₄²⁻].
Q = (8.0·10⁻⁴ M)² · 1.8·10⁻⁴ M.
Q = 1.152·10⁻¹⁰.
Q<span> > Ksp, </span><span>a precipitate of silver chromate will form.
</span>Ksp is the solubility
product constant for a solid substance dissolving in an aqueous solution.
<span>
[Ag</span>⁺<span>] is equilibrium concentration of silver cations.
[CrO</span>₄²⁻] is equilibrium
concentration of chromate anions.
Answer:
1. Data:
a) Half-life: 5730 years
b) Final radioactivity: 68%
2. Solution:
a) <u>Determine the number of half-lives undergone</u>
- Since, <em>the radioactivity has decreased to 68%, means that the carbon-14 contanined is has been reduced in </em> 32%: 100% - 68% = 32%.
- 32 = 2⁵, meaning that five half-lives have passed since the <em>plant material</em> that formed the<em> parchment fragment died</em>.
b) <u>Compute the time of five half-lives</u>:
- 5 × half-life time = 5 × 5730 years = 28,650 years.
c) <u>Round to the nearest hundred:</u>
- 28,650 years ≈ 28,700 years
And that is <em>the age of the parchment</em>.
Answer:
White being the color and coming in small grains.
Explanation:
Physical properties are something you can clearly see about the object.
Answer:
Striped should be the answer
Formation of ammonia by nitrogen and hydrogen is habers process wher 28g N2 results in formation of 34g NH3
so 35g N2 will form 34*35/28=42.5g NH3 where it given that reaction takes place in excess of H2
N2+3H2 gives 2NH3
