1 mole of any substance contains 6.022 × 1023 particles.
⚛ 6.022 × 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA
N = n × NA
· N = number of particles in the substance
· n = amount of substance in moles (mol)
· NA = Avogardro Number = 6.022 × 10^23 particles mol-1
For H2O we have:
2 H at 1.0 each = 2.0 amu
1 O at 16.0 each = 16.0 amu
Total for H2O = 18.0 amu, or grams/mole
It takes 18 grams of H2O to obtain 1 mole, or 6.02 x 1023 molecules of water. Think about that before we answer the question. We have 25.0 grams of water, so we have more than one mole of water molecules. To find the exact number, divide the available mass (25.0g) by the molar mass (18.0g/mole). Watch how the units work out. The grams cancel and moles moves to the top, leaving moles of water. [g/(g/mole) = moles].
Here we have 25.0 g/(18.0g/mole) = 1.39 moles water (3 sig figs).
Multiply 1.39 moles times the definition of a mole to arrive at the actual number of water molecules:
1.39 (moles water) * 6.02 x 1023 molecules water/(mole water) = 8.36 x 1023 molecules water.
That's slightly above Avogadro's number, which is what we expected. Keeping the units in the calculations is annoying, I know, but it helps guide the operations and if you wind up with the unit desired, there is a good chance you've done the problem correctly.
N = n × (6.022 × 10^23)
1 grams H2O is equal to 0.055508435061792 mol.
Then 23 g of H2O is 1.2767 mol
To calculate the number of particles, N, in a substance:
N = n × NA
N = 1.2767 × (6.022 × 10^23)
N= 176.26
N=
Answer:
When you breathe in, or inhale, your diaphragm contracts and moves downward. This increases the space in your chest cavity, and your lungs expand into it. The muscles between your ribs also help enlarge the chest cavity. They contract to pull your rib cage both upward and outward when you inhale.
Explanation:
^^^
Given: C3H8(g) + O2(g) ----> CO2 (g) + H2O (g)
Step : Put a 3 in front of CO2 (g) to balance C
=> C3H8(g) + O2(g) ----> 3CO2 + H2O to balance H
Step 2: Put a 4 in front of H2O
=> C3H8 (g) + O2(g) -----> 3CO2 (g) + 4H2O (g)
Step 3: Given that there are 3*2 + 4 = 10 O to the right side, put a 5 in front of O2 to balance O:
=> C3H8(g) + 5O2(g) -----> 3CO2(g) + 4H2O(g)
You can verify that the equation is balanced.
So, the answer is that the coefficient in front of O2 is 5.
Answer:
The answer you have selected in the screenshot is correct.
Its tendency to react with oxygen is correct.
Hope this helps.
