First, we need to find the atomic mass of

.
According to the periodic table:
The atomic mass of Carbon = C = 12.01
The atomic mass of Hydrogen = H = 1.008
The atomic mass of Oxygen = O = 16
As there are 6 Carbons, 12 Hydrogens and 6 Oxygens, therefore:
The
molar mass of

= 6 * 12.01 + 12 * 1.008 + 6 * 16
The
molar mass of

= 180.156
grams/moleNow that we have the molar mass of

, we can find the grams of glucose by using:
mass(of glucose in grams) = moles(of glucose given in moles) * molar mass(in grams/mole)
Therefore,
mass(of glucose in grams) = 2.47 * 180.156
mass(of glucose in grams = 444.99 grams
Ans: Mass of glucose in grams in 2.47 moles =
444.99 grams
-i
<h2>Energy is conserved. It never appears or disappears. It just changes from one kind of type to another. The Universe contains the same amount of energy it started with. This amount can never change.</h2><h2>
</h2>
You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.
<h3>What is the equilibrium constant?</h3>
In a reaction, we can judge using the value of the equilibrium constant weather or not the reaction moves on to completion. If the reaction moves up to completion, it the follows that the value of the equilibrium constant ought to be large.
On the other hand, when we have a case that the equilibrium constant is small and is not so large, then the reaction does not proceed essentially to completion.
Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.
Learn more about equilibrium constant:brainly.com/question/10038290
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