Answer:
The precipitated are option a and d.
Explanation:
2 LiI(aq) +Hg2(NO3)2(aq) → Hg2I2(s) ↓ + 2 LiNO3(aq)
Cation Hg2+ 2 in the presence of iodide, a precipitated is formed.
Zn(s) + 2AgNO3(aq) → 2 Ag(s) ↓ +Zn(NO3)2(aq)
Zinc starts to get rid, and some white particles also stick to it. Afterwards the solution becomes cloudy and a precipitate appears, which is the solid silver
B. 11,540
<h3>Further explanation
</h3>
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually radioactive elements have an unstable atomic nucleus.
General formulas used in decay:
T = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Nt=25 g
No=100 g
t1/2=5770 years
Gle's cache of http://www.middleschoolchemistry.com/lessonplans/chapter5/lesson4<span>. It is a snapshot of the page as it appeared on 21 Oct 2017 07:24:57 GMT.</span>
Answer: Come Onshore
Explanation:
Sea waves are harmless when present at Sea. However, it becomes DESTRUCTIVE as it travels to the land surface.
Its effects include the sweeping off (erosion) of features, BENEFICIAL soil organisms and the transport of sand and sediment along coastal areas which might include farms, residential houses etc.
Thus, the effects of Sea waves ONSHORE affects the following: agricultural activities
Domestic activities
Environmental balance etc.
Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
