I'm not sure, but maybe burning point...
Answer:125.84g
Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.
Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.
From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.
So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g
<h3>
Answer: b) 0.250 mol</h3>
============================================
Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
-------------------
Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
Answer:
15.04 mL
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 21 L
V₂ = ?
P₁ = 9 atm
P₂ = 15 atm
T₁ = 253 K
T₂ = 302 K
Using above equation as:
Solving for V₂ , we get:
<u>V₂ = 15.04 mL</u>