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galben [10]
3 years ago
6

Shown below is a table with plants and animals with their adaptations.

Chemistry
1 answer:
zhenek [66]3 years ago
8 0

Answer:

Desert

Explanation:

The adaptation shown by the given plants and animals shows that they will adapted to the desert biome.

It is so because, due to high temperature of desert some desert animals like camel have the storage of fat in humps or tails; some animals have large ears such as Jackrabbits, it helps to release body heat and adapt in high temperature; plants have thick water holding tissues to reduce water loss in heat and waxy coating that keeps  the plants cooler and reduce moisture loss.

Hence, the correct answer is "Desert".

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For a fixed amount of gas at a fixed volume, what will happen if the absolute temperature is doubled?
Firlakuza [10]
The answer to this question is: The pressure will be increased two times

The gas law includes an interaction of pressure(p), volume(V) and temperature(T). In this case, the gas has same amount and volume but then the temperature is doubled. The calculation would be:

p1 * V1 / T1= p2*V2 /T2 
p1 * 1 / 1 = p2 * 1 / 2
p1 / (1/2)=p2
p2= 2* p1
8 0
3 years ago
What does Ca stand for in chem?
abruzzese [7]

Answer:

Calcium

Explanation:

Periodic table

7 0
3 years ago
Read 2 more answers
25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
Kamila [148]

Answer:

Concentration of the original \rm KOH solution: approximately 1.05\; \rm mol \cdot dm^{-3}.

Explanation:

Notice that the concentration of the \rm HCl solution is in the unit \rm mol\cdot dm^{-3}. However, the unit of the two volumes is \rm cm^{3}. Convert the unit of the two volumes to \rm dm^{3} to match the unit of concentration.

\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}.

\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 0.0350\; \rm dm^{3} of 0.75\; \rm mol \cdot dm^{3} \rm HCl\! solution:

\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}.

\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

7 0
3 years ago
A substance that contains more hydroxide ions than hydronium ions in solution could have which pH value?
Hunter-Best [27]
Hydronium ions are basically H+ ions.
So if [OH-]>[H+]
This means that the substance is more basic.
 and basic substances a pH>7.
Therefore a possible pH value would be 10.
6 0
3 years ago
if the density of copper is 8.9g/cm^3 and the Volume of the sample is 23.4 mL what would be the mass? ​
Black_prince [1.1K]

Answer:

The mass of copper = 208.26 grams.

Explanation:

Density = mass / volume

8.9 = mass / 23.4

mass = 8.9 * 23.4

= 208.26 grams.

8 0
3 years ago
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