Cadmium chloride is a highly soluble compound. The equation for its dissolution is:
CdCl₂(s) → Cd⁺²(aq) + 2Cl⁻(aq)
This dissociation in water allows for the cadmium and chlorine ions to take part in reactions. This is the reason that solutions of chemicals are prepared when a reaction needs to take place.
Taking into account the animals and the landscape being mentioned, we can confirm that the combination of these forms<u><em> part of an </em></u><u><em>ecosystem</em></u><u><em>.</em></u>
An ecosystem is described as an area consisting of:
- Plants
- Animals
- Bacteria
- Weather
- Landscape
and many other factors, which all work together to form a small pocket of life.
Ecosystems are described as containing both living and non-living elements, meaning biotic and non-biotic parts. A community or population, on the other hand, <u>are used to consider only the </u><u>biotic </u><u>or </u><u>living factors </u><u>of an </u><u>ecosystem</u><u>. </u>Therefore, since the question includes the rock bottom (a<em><u> non-biotic factor</u></em>) as a part of the makeup, the only correct grouping is that of an ecosystem.
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Answer:
All of the given statements are true.
Explanation:
All the elements are heavier than Bismuth (Bi) are radioactive.
The time for half of the original sample to spontaneous decay is called half life (
)

All radioactive elements are spontaneously decaying towards formation of a stable element.
Radioactive elements undergo decay in order to attain stability.
Radioactivity is a natural part of our environment. The earth also contains several primordial long-lived radioisotopes that have survived to the present in significant amounts.
Hence, all the given statements are true.
Filtration
distillation
crystallization
chromatography
Answer:
C₄H₂N₂
Explanation:
First we<u> calculate the moles of the gas</u>, using PV=nRT:
P = 2670 torr ⇒ 2670/760 = 3.51 atm
V = 300 mL ⇒ 300/1000 = 0.3 L
T = 228 °C ⇒ 228 + 273.16 = 501.16 K
- 3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K
Now we<u> calculate the molar mass of the compound</u>:
- 2.00 g / 0.0256 mol = 78 g/mol
Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:
- C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4
- H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2
- N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2
So the empirical formula is C₄H₂N₂