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:
A. weak acid and its conjugate base
Explanation:
A buffer solution can be made with a weak acid and conjugate base or a weak base and conjugate acid.
This may help you:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers/Introduction_to_Buffers
Answer:
Polythene has the lowest melting/boiling point from all the other covalent structures mentioned in this question.
Answer:
97 000 g Na
Explanation:
The absortion (or liberation) of energy in form of heat is expressed by:
q=m*Cp*ΔT
The information we have:
q=1.30MJ= 1.30*10^6 J
ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)
Cp=30.8 J/(K mol Na)
If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.
To do so, we use the molar mass of Na= 22.99g/mol
Now, we are able to solve for m:
=97 000 g Na
Answer:
4. A concentration of a trace mineral was found to be 2.53 ppb in an aqueous solution. A. What mass of solution contains 2.53 g of the mineral? (1 point - Knowledge, 2 points - Inquiry, 1 point - Communication) m=? C = 2.53 ppm or 2.53X106 n = Con ܗ B. What volume of solution contains 2.53 g of the mineral? (1 point - Knowledge, 2 points - Inquiry, 1 point - Communication) VE? C: 2,53 ppm ns Cen
