So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.
using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water
27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g
therefore,
you need 310.34g of water is in the jar.
Answer:
Alkali metals are soft and have low melting points.
Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:
Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
Answer:
The molarity of the solution is 0,31 M
Explanation:
We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)
Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g
58, 5 g-----1 mol NaCl
13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl
727 ml solution------ 0, 224 mol NaCl
1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution
x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>
