They usually get it from an old well they dug up and have a pump and strainer that go to their house.
Answer:
The molecular weight is ![Z = 111.2 \ g/mol](https://tex.z-dn.net/?f=Z%20%3D%20%20111.2%20%5C%20g%2Fmol)
Explanation:
From the question we are told that
The mass of the sample is ![m = 0.98 \ g](https://tex.z-dn.net/?f=m%20%3D%20%200.98%20%5C%20%20g)
The temperature is ![T = 348 K](https://tex.z-dn.net/?f=T%20%20%3D%20%20348%20K)
The volume which the gas occupied is ![V = 265 \ ml = 265 *10^{-3} L](https://tex.z-dn.net/?f=V%20%20%3D%20%20265%20%5C%20ml%20%20%3D%20265%20%2A10%5E%7B-3%7D%20L)
The pressure is ![P = 0.95 \ atm](https://tex.z-dn.net/?f=P%20%20%3D%20%200.95%20%5C%20%20atm)
Generally from the ideal gas equation we have that
![PV = n RT](https://tex.z-dn.net/?f=PV%20%20%3D%20%20n%20RT)
Here n is the number of moles of the gas while the R is the gas constant with value ![R = 0.0821 \ atm \cdot L \cdot mol^{-1} \cdot K^{-1}](https://tex.z-dn.net/?f=R%20%20%3D%20%200.0821%20%5C%20atm%20%5Ccdot%20L%20%20%5Ccdot%20mol%5E%7B-1%7D%20%5Ccdot%20K%5E%7B-1%7D)
![n = \frac{PV}{ RT}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BPV%7D%7B%20RT%7D)
=> ![n = \frac{ 0.95 * 265 *10^{-3} }{ 0.0821 * 348}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B%200.95%20%2A%20265%20%2A10%5E%7B-3%7D%20%7D%7B%20%20%200.0821%20%2A%20348%7D)
=> ![n = 0.00881 \ mol](https://tex.z-dn.net/?f=n%20%3D%200.00881%20%5C%20%20mol)
Generally the molecular weight is mathematically represented as
![Z = \frac{m}{n}](https://tex.z-dn.net/?f=Z%20%3D%20%20%5Cfrac%7Bm%7D%7Bn%7D)
=> ![Z = \frac{0.98 }{0.00881}](https://tex.z-dn.net/?f=Z%20%3D%20%20%5Cfrac%7B0.98%20%7D%7B0.00881%7D)
=> ![Z = 111.2 \ g/mol](https://tex.z-dn.net/?f=Z%20%3D%20%20111.2%20%5C%20g%2Fmol)
Answer:
![Kc=0.0851](https://tex.z-dn.net/?f=Kc%3D0.0851)
Explanation:
Hello there!
In this case, according to the given chemical reaction:
CaCO3(s)→CaO(s)+CO2(g)
In which the amounts are unfortunately given, we can however, assume the information of similar problems so you can further modify the numbers but follow the same work:
CaCO3 = 25.3 g
CaO = 14.9 g
CO2 = 33.7 g
Thus, since just CO2 is involved on the equilibrium expression because CaCO3 and CaO are solid, we can compute the moles of the CO2 at equilibrium and further compute the concentration in the 9.0-L vessel:
![n_{CO_2}=33.7g*\frac{1molCO_2}{44.01gCO_2}=0.766molCO_2\\\\](https://tex.z-dn.net/?f=n_%7BCO_2%7D%3D33.7g%2A%5Cfrac%7B1molCO_2%7D%7B44.01gCO_2%7D%3D0.766molCO_2%5C%5C%5C%5C)
![[CO_2]=\frac{0.766molCO_2}{9.0L}=0.0851M](https://tex.z-dn.net/?f=%5BCO_2%5D%3D%5Cfrac%7B0.766molCO_2%7D%7B9.0L%7D%3D0.0851M)
Thus, we proceed with the equilibrium expression to obtain:
![Kc=[CO_2]\\\\Kc=0.0851](https://tex.z-dn.net/?f=Kc%3D%5BCO_2%5D%5C%5C%5C%5CKc%3D0.0851)
Best regards!
Answer:
For every 2 moles of Na used, 2 moles of NaCl are produced.
Explanation: