Hello! I have a green pigment called the chlorophyll. I also absorb solar

ohaa [14]

What you can do is organize them by color, what matter they are in room temperature, their molecular structure, or what kind of conductor in electricity and heat it is. I'm not sure what the format is supposed to look like but first just organize them all in different categories.

5 0

3 years ago

Read 2 more answers

Solid silicon and solid magnesium chloride form when silicon tetrachloride gas reacts with magnesium metal. Write a word equa- t

antiseptic1488 [7]

SiCl4 + 2Mg ---> 2Si + 2MgCl2

8 0

3 years ago

This fine-grained dark colored igneous rock is most likely to

Serggg [28]

Most likely basalt....

6 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

3 years ago

A compound has the empirical formula . a 256-ml flask, at 373 k and 750. torr, contains 0.527 g of the gaseous compound. give th

CaHeK987 [17]

<span>Answer: From the ideal gas law, MM=mRTPV; where MM = molecular mass; m = mass; P = pressure in atmospheres; V= volume in litres; R = gas constant with appropriate units. So, 0.800â‹…gĂ—0.0821â‹…Lâ‹…atmâ‹…Kâ’1â‹…molâ’1Ă—373â‹…K0.256â‹…LĂ—0.987â‹…atm = 97.0 gâ‹…molâ’1. nĂ—(12.01+1.01+2Ă—35.45)â‹…gâ‹…molâ’1 = 97.0â‹…gâ‹…molâ’1. Clearly, n = 1. And molecular formula = C2H2Cl2. I seem to recall (but can't be bothered to look up) that vinylidene chloride, H2C=C(Cl)2 is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.</span>

4 0

3 years ago