The question is incomplete, the complete question is:
What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.
<u>Answer:</u> 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
<u>Explanation:</u>
We first calculate the number of moles of soft drink in a volume of 10 mL
The formula used to calculate molarity:
.....(1)
Taking the concentration of soft drink from the example be = 0.375 M
Volume of solution = 10 mL
Putting values in equation 1, we get:
<u>Calculating volume of sweetened tea:</u>
Moles of sugar = 0.00375 mol
Molarity of sweetened tea = 0.05 M
Putting values in equation 1, we get:
Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Answer:
<em>Hi Todoroki here!!! </em>
Explanation:
Chlorine has the electron configuration [Ne]3s 2 3p 5, with the seven electrons in the third and outermost shell acting as its valence electrons. Like all halogens, it is thus one electron short of a full octet, and is hence a strong oxidising agent, reacting with many elements in order to complete its outer shell.
<em>Your welcome!!</em>
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
<span>C7H8
First, lookup the atomic weight of all involved elements
Atomic weight of carbon = 12.0107
Atomic weight of hydrogen = 1.00794
Atomic weight of oxygen = 15.999
Then calculate the molar masses of CO2 and H2O
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol
Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol
Now calculate the number of moles of each product obtained
Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK.
moles CO2 = 3.52 / 44.0087 = 0.079984
moles H2O = 0.822 / 18.01488 = 0.045629
Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen.
moles C = 0.079984
moles H = 0.045629 * 2 = 0.091258
So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting
1: 1.140953
The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like)
1 ( 1 : 1.140953) = 1 : 1.140953
2 ( 1 : 1.140953) = 2 : 2.281906
3 ( 1 : 1.140953) = 3 : 3.422859
4 ( 1 : 1.140953) = 4 : 4.563812
5 ( 1 : 1.140953) = 5 : 5.704765
6 ( 1 : 1.140953) = 6 : 6.845718
7 ( 1 : 1.140953) = 7 : 7.986671
8 ( 1 : 1.140953) = 8 : 9.127624
That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
