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2 years ago

Where are most organisms that live in the ocean found ? why ?

1 answer:

3 0

Most ocean life is found in coastal habitats. Coastal habitats have more light than the deep ocean resulting in a higher concentration of organisms. Plant species are especially abundant in coastal habitats.

You might be interested in

How might a molecule with two strong bond dipoles have no molecular dipole at all?

Reil [10]

A molecule with two strong bond dipoles can have no molecular dipole if the bond dipoles cancel each other out by pointing in exactly opposite directions. For example, in carbon dioxide (a linear molecule), the carbon-oxygen bonds have a <span>large dipole moment. However, because one dipole points to the left and the other to the right the dipole is cancelled.</span>

5 0

2 years ago

How are acids and ionic compounds similar?

11Alexandr11 [23.1K]

Answer: Ionic compounds are held together by the virtue of their opposing charges. Na+Cl- for example. If we consider Hg+(2Cl-)2, a mercuric chloride, the solubility is much less. Ba++(SO)4 Barium Sulphate, is highly insoluble; all differ by the relative attractiveness by Differing opposing charge(s).

Acids are very similar, consider Formic Acid, HCOOH, the simplest of the Carboxylic Acids. It dissociates more than say Benzoic Acid, C6H5-COOH. But neither disassociate as fully as Nitric Acid HNO3.

So the relative disassociation of the H+ (proton), or H3O+, (Hydronium ion), from any of these in water vary for a number of reasons we need not consider now.

Here is a “Tricky One!” (And very nasty). Take HF liquid or gas. This is one of the strongest acids on Earth - AS A LIQUID compound OR GAS. It will dissociate essentially near completion! Eat the floor, and is very dangerous.

NOW - HF (aqueous). The HF is in water. Very like HCl? NO! Why you may ask...The Electrophilic nature of Fluorine, “bathed in water, with an H+ all its own”, doesn’t let it go as easily!

HF is HIGHLY ordered in water, you can almost imagine a sort of “Hydrated matrix”, little HFs in endless rows...

BUT BE WARNED - even the aqueous HF is so reactive it will dissolve bone!

(I was told it was extremely painful; and did not appear to heal for weeks!)

Explanation: so, both types of compounds have a similarity, held together by the strength of their opposing charges or the degree of dissociation, (using water for simplicity).

That should do it.

8 0

2 years ago

Given the balanced equation representing a reaction at 101.3 kPa and 298 K:

ser-zykov [4K]

Answer (4)

Endothermic processes: are those in which occurs the absorption of heat.
the change of enthalpy (ΔH) is greater <span>than zero.
</span>
hope this helps!

8 0

3 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

2 years ago

Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles of O₂ to t

djverab [1.8K]

Answer : The correct option is (c) $\frac{2\text{ mole of }Fe_2O_3}{3\text{ mole of }O_2}$

Explanation :

The given balanced chemical reaction is,

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

From the balanced chemical reaction, we conclude that

As, 3 moles of $O_2$ react to give 2 mole of $Fe_2O_3$

So, 1 mole of $O_2$ react to give $\frac{2\text{ mole of }Fe_2O_3}{3\text{ mole of }O_2}$ moles of $Fe_2O_3$

Thus, the conversion factor needed to convert the number of moles of $O_2$ to the number of moles of $Fe_2O_3$ produced is $\frac{2\text{ mole of }Fe_2O_3}{3\text{ mole of }O_2}$

Hence, the correct option is (c) $\frac{2\text{ mole of }Fe_2O_3}{3\text{ mole of }O_2}$

7 0

2 years ago

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