Answer:
They will die.
Explanation:
Polar Bears were made to live in the frigid temperatures of the tundras. If there is a temperature change it can affect their habitat. It can destroy it and they would have to move out to another place. They can die due to global warming.
Answer:
P=atm
![b=\frac{L}{mol}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BL%7D%7Bmol%7D)
Explanation:
The problem give you the Van Der Waals equation:
![(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT](https://tex.z-dn.net/?f=%28P%2B%5Cfrac%7Bn%5E%7B2%7Da%7D%7BV%5E%7B2%7D%7D%29%28V-nb%29%3DnRT)
First we are going to solve for P:
![(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}](https://tex.z-dn.net/?f=%28P%2B%5Cfrac%7Bn%5E%7B2%7Da%7D%7BV%5E%7B2%7D%7D%29%3D%5Cfrac%7BnRT%7D%7B%28V-nb%29%7D)
![P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BnRT%7D%7B%28V-nb%29%7D-%5Cfrac%7Bn%5E%7B2%7Da%7D%7Bv%5E%7B2%7D%7D)
Then you should know all the units of each term of the equation, that is:
![P=atm](https://tex.z-dn.net/?f=P%3Datm)
![n=mol](https://tex.z-dn.net/?f=n%3Dmol)
![R=\frac{L.atm}{mol.K}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BL.atm%7D%7Bmol.K%7D)
![a=atm\frac{L^{2}}{mol^{2}}](https://tex.z-dn.net/?f=a%3Datm%5Cfrac%7BL%5E%7B2%7D%7D%7Bmol%5E%7B2%7D%7D)
![b=\frac{L}{mol}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BL%7D%7Bmol%7D)
![T=K](https://tex.z-dn.net/?f=T%3DK)
![V=L](https://tex.z-dn.net/?f=V%3DL)
where atm=atmosphere, L=litters, K=kelvin
Now, you should replace the units in the equation for each value:
![P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B%28mol%29%28%5Cfrac%7BL.atm%7D%7Bmol.K%7D%29%28K%29%7D%7BL-%28mol%29%28%5Cfrac%7BL%7D%7Bmol%7D%29%7D-%5Cfrac%7B%28mol%5E%7B2%7D%29%28%5Cfrac%7Batm.L%5E%7B2%7D%7D%7Bmol%5E%7B2%7D%7D%29%7D%7BL%5E%7B2%7D%7D)
Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:
![P=\frac{L.atm}{L-L}-atm](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BL.atm%7D%7BL-L%7D-atm)
Then operate the fraction subtraction:
P=![P=\frac{L.atm-L.atm}{L}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BL.atm-L.atm%7D%7BL%7D)
![P=\frac{L.atm}{L}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BL.atm%7D%7BL%7D)
And finally you can find the answer:
P=atm
Now solving for b:
![(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT](https://tex.z-dn.net/?f=%28P%2B%5Cfrac%7Bn%5E%7B2%7Da%7D%7BV%5E%7B2%7D%7D%29%28V-nb%29%3DnRT)
![(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}](https://tex.z-dn.net/?f=%28V-nb%29%3D%5Cfrac%7BnRT%7D%7B%28P%2B%5Cfrac%7Bn%5E%7B2%7Da%7D%7BV%5E%7B2%7D%7D%29%7D)
![nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}](https://tex.z-dn.net/?f=nb%3DV-%5Cfrac%7BnRT%7D%7B%28P%2B%5Cfrac%7Bn%5E%7B2%7Da%7D%7BV%5E%7B2%7D%7D%29%7D)
![b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BV-%5Cfrac%7BnRT%7D%7B%28P%2B%5Cfrac%7Bn%5E%7B2%7Da%7D%7BV%5E%7B2%7D%7D%29%7D%7D%7Bn%7D)
Replacing units:
![b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BL-%5Cfrac%7B%28mol%29.%28%5Cfrac%7BL.atm%7D%7Bmol.K%7D%29.K%7D%7B%28atm%2B%5Cfrac%7Bmol%5E%7B2%7D.%5Cfrac%7Batm.L%5E%7B2%7D%7D%7Bmol%5E%7B2%7D%7D%7D%7BL%5E%7B2%7D%7D%29%7D%7D%7Bmol%7D)
Multiplying and dividing units,(please see the second photo below), we have:
![b=\frac{L-\frac{L.atm}{atm}}{mol}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BL-%5Cfrac%7BL.atm%7D%7Batm%7D%7D%7Bmol%7D)
![b=\frac{L-L}{mol}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BL-L%7D%7Bmol%7D)
![b=\frac{L}{mol}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7BL%7D%7Bmol%7D)
Answer:
A solution in which no more solute can be dissolved in is referred to as SATURATED. In such a solution, the concentration of solute is called SOLUBILITY . When that concentration is reported in moles per liter, it is more specifically called MOLAR SOLUBILITY. A special equilibrium constant called the SOLUBILITY PRODUCT constant is calculated from the molar concentrations of the aqueous components of the dissolution equation.
Explanation:
The solubility of a solute in a solvent is the maximum amount of solute in moles that will be dissolved in 1dm3 of the solvent at a specified temperature. Once the maximum number or concentration has been reached, the solvent can no longer take in solutes and this point in the reaction, the solution is said to be saturated. That is the composition of the saturated solution is not affected by the presence of excess solute. An unsaturated solution has a lower concentration of solute and can dissolve more solutes if added until it becomes saturated.
Solubility when reported in moles per liter is called molar solubility of the solution and it gives a more accurate measurement of yh solubility of a solution. The solubility product constant is calculated from the molar concentrations of the aqueous components of the dissolution equation. This solubility product constant explains the balance between dissolved ions from the salt and undissolved salt in a dissolution equation.
1.) Mass
2.) Can occupy space (Volume)
Answer:
392g sulfuric acid are produced
Explanation:
Based on the balanced equation:
2HCl + Na2SO4 → 2NaCl + H2SO4
<em>2 moles of HCl produce 1 mole of sulfuric acid</em>
<em />
To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -<em>Molar mass H2SO4 = 98g/mol- </em>we can find the mass of sulfuric acid produced:
<em>Moles sulfuric acid:</em>
8mol HCl * (1mol H2SO4 / 2mol HCl) = 4 mol H2SO4
<em>Mass sulfuric acid:</em>
4mol H2SO4 * (98g / mol) =
392g sulfuric acid are produced