We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to 
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is 
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Answer:
2
Explanation:
Half-life
Concentration
![{[A]_0}_A=1.2\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_A%3D1.2%5C%20%5Ctext%7BM%7D)
![{[A]_0}_B=0.6\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_B%3D0.6%5C%20%5Ctext%7BM%7D)
We have the relation
![t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%5Cpropto%20%5Cdfrac%7B1%7D%7B%5BA%5D_0%5E%7Bn-1%7D%7D)
So
![\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%7Bt_%7B1%2F2%7D%7D_A%7D%7B%7Bt_%7B1%2F2%7D%7D_B%7D%3D%5Cleft%28%5Cdfrac%7B%7B%5BA%5D_0%7D_B%7D%7B%7B%5BA%5D_0%7D_A%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B2%7D%7B4%7D%3D%5Cleft%28%5Cdfrac%7B0.6%7D%7B1.2%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7Bn-1%7D)
Comparing the exponents we get
The order of the reaction is 2.
![t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7Bt_%7B1%2F2%7D%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7B2%5Ctimes%201.2%5E%7B2-1%7D%7D%5C%5C%5CRightarrow%20k%3D0.4167%5C%20%5Ctext%7BM%7D%5E%7B-1%7D%5Ctext%7Bmin%7D%5E%7B-1%7D)
The rate constant is
They or on the southern hysteric watch it on youtube (latitude and longitude song 1 direction remix by the history teachers)
Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt
= 11.75 g - 9.25 g = 2.50 g
number of water moles = 2.50 g / 18 g/mol = 0.139 mol
number of cobalt (II) chloride moles = 9.25 g / 130 g/mol = 0.0712 mol
ratio of water moles to CoCl₂ moles - 0.139 mol / 0.0712 mol = 1.95
rounded off 2 moles of water for every 1 mol of CoCl₂
formula - CoCl₂.2H₂O
name - Cobalt(II) chloride dihydrate
Answer:
20.9%
Explanation:
I took the test i hope this helps:)
