Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
9.6 %
Explanation:
<u>Step 1: How to define percent error ? </u>
⇒ % error is the difference between a measured value and the known or accepted value
⇒Percent error is calculated using the following formula:
⇒%error = | Experimental value-theoretical value/theoretical value | x100%
⇔ this can be written as well as : error = (| Experimental value/ theoretical value | - | Theoretical value / Theoretical value | ) x100%
<u>Step 2: Calculate % error</u>
In this case, this means :
%error = ( |(4.45 cm - 4.06cm ) / 4.06cm | ) x100%
%error = 0.096 x100%
%error =9.6 %
<u>Answer:</u> The solubility of
in water is 
<u>Explanation:</u>
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

3s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D%5E2)
We are given:

Putting values in above equation, we get:

Hence, the solubility of
in water is 