Your gonna half to space this out in order for me to answer if you don’t mind. I’ll answer in the comments
Answer:
Mass of glucose = 515.34 g
Explanation:
We are given;
Mass; m = 60 kg
Elevation; h = 1550 m
Acceleration due to gravity; 9.8 m/s²
Now, work performed to lift 60kg by 1550m is given by the formula;
W = mgh
W = 60 × 9.8 × 1550
W = 911400 J
We are told the actual work is 4 times the one above.
Thus;
Actual work = 4W = 4 × 911400 = 3,645,600 J
Now,
Molar mass of Glucose(C6H12O6) = 180 g/mol
We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300
Moles of glucose = 3645600/1273300 = 2.863mol
Mass of glucose = 2.863 mol × 180 g/mol
Mass of glucose = 515.34 g
Answer:
5.22*1022 molecules of glucose (C6H12O6) is equal to 15.08 grams
Explanation:
Number of molecule of glucose in one mole 
Now mass of one mole of glucose 
grams
Number of moles in 
molecules of glucose 
Number of moles in 
molecules of glucose = 
Number of moles in 
molecules of glucose = 
= 
moles
Weight of moles
grams
It's answer A!
Reactants on left (see on the arrow), products on right.
Answer:
M = No.of moles of solute / Volume of solution in litres
HBr: H = 1 x 1 = 1 Br = 1 x 80
Molar mass : 81 g / mol
a. n = 27.7% / 100 =0.277 g / (81 g / mol) = 0.0034 mol
density = 1.24 g/ml ; D= m / v
b. 65.32 ml x 1 L / 1000 mL = 0.065 L
c. M= n / V
= 0.0034 mol / 0.065 L
M= 0.052 mol / L
The molarity of the concentrated HBr solution is 0.052 mol/L
