6.02*10^23 is Avagadro's number representing the number of molecules per mole of substance.
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=
The answer for the question above is A. the gravitational pull of the moon on the water near the coast. The sun and and the moon are responsible for the rising and falling of the ocean tides. The gravitational pull of the moon and the sun makes the water in the oceans bulge, causing a continuous change between high and low tide.
Answer:
Explanation:
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In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:
Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:
Then, we obtain:
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