Answer:
It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.
Explanation:
The question is not very much clear.
If you are asking for molecules then 1 mole water= 6.023 * 10^23
If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3
If you are asking for particles then,
So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.
I hope that was helpful!
H=1 proton,1 electron
O=8 protons,8 neutrons and 8 electrons
total particles in one H2O molecule-28
total no. of particles in 1 mole of water- 6.023 * 10^23 * 28
Answer:
V = 0.0327 L.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the liters of C3H6O by the definition of density. We can tell the density of this substance as that of acetone (0.784 g/mL) and therefore calculate the liters as shown below:
Regards!
Answer:
Option 4 with o-h in the most polar bond, since the two atoms in the bond have the greatest difference in electronegativity. This is assuming there are no other factors in other atoms bound to either of the elements in the bond.
Explanation:
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
A fact can be disapproved with further evidence different details. So basically evidence can change a fact!! Hope this helps
