The smallest particle of an element that still retains the chemical properties of it is an atom.
Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C°
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n = 1.45/mass weight of argon
= 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm
Neutral charges and are found in the center? I'm not sure for the last answer
Answer:
A) Sample B has more calcium carbonate molecules
Explanation:
M = Molar mass of calcium carbonate = 100.0869 g/mol
= Avogadro's number = ![6.022\times 10^{23}\ \text{mol}^{-1}](https://tex.z-dn.net/?f=6.022%5Ctimes%2010%5E%7B23%7D%5C%20%5Ctext%7Bmol%7D%5E%7B-1%7D)
For the 4.12 g sample
Moles of a substance is given by
![n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7Bm%7D%7BM%7D%5C%5C%5CRightarrow%20n%3D%5Cdfrac%7B4.12%7D%7B100.0869%7D%5C%5C%5CRightarrow%20n%3D0.0411%5C%20%5Ctext%7Bmol%7D)
Number of molecules is given by
![nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}](https://tex.z-dn.net/?f=nN_A%3D0.0411%5Ctimes%206.022%5Ctimes%2010%5E%7B23%7D%3D2.48%5Ctimes%2010%5E%7B22%7D%5C%20%5Ctext%7Bmolecules%7D)
For the 19.37 g sample
![n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B19.37%7D%7B100.0869%7D%5C%5C%5CRightarrow%20n%3D0.193%5C%20%5Ctext%7Bmol%7D)
Number of molecules is given by
![nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}](https://tex.z-dn.net/?f=nN_A%3D0.193%5Ctimes%206.022%5Ctimes%2010%5E%7B23%7D%3D1.16%5Ctimes%2010%5E%7B23%7D%5C%20%5Ctext%7Bmolecules%7D)
![1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}](https://tex.z-dn.net/?f=1.16%5Ctimes%2010%5E%7B23%7D%5C%20%5Ctext%7Bmolecules%7D%3E2.48%5Ctimes%2010%5E%7B22%7D%5C%20%5Ctext%7Bmolecules%7D)
So, sample B has more calcium carbonate molecules.
The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.
For the answer to the question above asking, h<span>ow many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
</span><span>moles glucose = 1.5 x 4.5 = 6.8 </span>