It's the person that you in love with.<span />
To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:



Then you add all of them together:

Therefore, the molar weight of glucose is 180.15 grams.
Answer:
0.36 m/s
Explanation:
im in 8th grade in honors science and i think, since u have distance and time you have to divide distance by time
Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;
Moles = n = 1.20 mol
Volume = V = 4 L
Temperature = T = 30 + 273 = 303 K
Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹
Putting Values in Eq.1,
P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L
P = 7.45 atm