To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.
A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2
*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*
B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles.
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH
Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6
Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 * (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6
C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2
I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!
Answer:
No, it is not appropriate to mix water and DMSO
Explanation:
We have to realize that DMSO is a highly polar solvent and water is a highly polar solvent. The question explicitly says that our target is to produce a solvent of<u><em> intermediate polarity</em></u>.
We can only do this by mixing a polar and a nonpolar solvent. We have been given the example of the mixture of acetone/hexane which is quite a perfect mixture.
Thus, it is inappropriate to mix DMSO and water.
This is a true statement if it is density you are looking for... Density problem.....
Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits
The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.
Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.
The accepted value for the density of tin is 7.31 g/cm³
<em><u>Unit conversion is very important because the rest of the world other than three contries uses the metric system. So, converting unitsis important in science because it uses the metric system. The metric system aremeasurements like cm, m, l, mL, etc.</u></em>
