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3 years ago

A solution with a higher osmotic pressure than body fluids is said to be

Chemistry

1 answer:

Vsevolod [243]3 years ago

7 0

A solution that has a higher osmotic pressure than body fluids is said to be a <span>hypertonic solution.</span>

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How many moles are in 8.90 X 10^24 atoms of Na?

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1.48 x 10^47 mol of Na

Explanation:

8.90 x 10^24 atoms of Na (1 mol of Na/6.022 x 10^23 atoms of Na)=

1.48 x 10^47 mol of Na

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3 years ago

The element Y has 13 electrons. What is the electronic configuration of

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0,2,8,3

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3 years ago

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2 years ago

He nitrogen cycle is one of the most important (and complex!) of the biogeochemical cycles. Why is nitrogen important?

gayaneshka [121]

Nitrogen is an essential part of proteins and nucleic acids (DNA).

<h3>Why is nitrogen important?</h3>

All living things require nitrogen. It is a part of proteins, ATP, chlorophyll, DNA, and RNA. Given that it makes up a significant portion of the structures of biomolecules in living organisms, its synthesis and consumption in the biogeochemical cycle are of highest importance. Proteins are made up of amino acids (NH2 group).

The structural development of living species' genetic material (DNA, RNA) also involves nitrogen. It is a part of the nitrogenous bases, which together with other nucleotides create RNA and DNA.

Therefore, a nitrogen deficit can cause major structural issues for such an organism.

<h3>What happens if nitrogen cycle is disrupted?</h3>

A number of undesirable consequences can result from the nitrogen cycle being disrupted. For instance, an abundance of nitrogen in aquatic systems leads to eutrophication. Acid rain can be caused by a rise in atmospheric nitrogen.

The majority of ecosystems can tolerate low nitrogen levels since the majority of nitrogen is not biologically accessible in nature. Ecosystems become out of balance when humans significantly increase the quantity of nitrogen that is accessible through fertilizers, fertilizer use, and other means.

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2 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

3 years ago