Answer:
0.544 M
Explanation:
First find the moles in the final solution
0.8 mols/L *1.7L
1.36 mols
so there is 1.36 mols in 2.5L
concentration will be 1.36/2.5
0.544 M
<span>If you do not wash and dry the thermometer after every time you use it in this scenario, the temperature could be affected by the NaOH residue. This could make it so that your findings were inaccurate, so in order to be efficient, this precaution needs to be taken.</span>
Answer:
1859.4 g of ZnCrO₄ in 10.25 moles
Explanation:
First of all, we determine the molecular formula of the compound:
Zinc → Zn²⁺ (cation)
Chromate → CrO₄⁻² (anion)
Zinc chromate → ZnCrO₄
Molar mass for the compound is:
Molar mass of Zn + Molar mass of Cr + (Molar mass of O) . 4 = 181.41 g/mol
65.41 g/mol + 52 g/mol + 16 g/mol . 4 = 181.41 g/mol
Let's apply this conversion factor: 10.25 mol . 181.41 g/mol = 1859.4 g
This is false because catalytic converters in an automobile don't act upon carbon dioxide. They burn carbon monoxide and other carbon gasses to create carbon monoxide and water.
Answer:
![[I_2]=[Br]=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D0.31M)
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:
Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:
![K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%5C%5C%5C%5C1.2x10%5E2%3D%5Cfrac%7B%282x%29%5E2%7D%7B%282.0-x%29%5E2%7D)
Thus, we solve for x as show below:
Therefore, the concentrations of both bromine and iodine are:
![[I_2]=[Br]=2.0M-1.69M=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D2.0M-1.69M%3D0.31M)
Regards!
