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:
please select my answer as brainlist answers
Explanation:
plastic is harmful for the environment
Mass %abundance # of atoms (pick 1000 sample)
68.9257 60.4% 604
70.9249 39.6% 396
*39.6* ( i got this by subtracting 100 as most percentage is out of 100)
*# of atoms* ( i just moved it one decimal place, then i am going to make the sample out of 1000, to get the decimal place)
Formula= total mass/ total #atoms
x= (68.9257u* 604) + (70. 9249*396)/ 1000
x- 69717. 3832/ 1000
x= 69.717amu or 69.72 amu
if you look on the periodic table the closet element with that mass is gallium ( amu is 69.72)
Answer:
10.88 g
Explanation:
We have:
[CH₃COOH] = 0.10 M
pH = 5.25
Ka = 1.80x10⁻⁵
V = 250.0 mL = 0.250 L
The pH of the buffer solution is:
(1)
By solving equation (1) for [CH₃COONa*3H₂O] we have:
![[CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%20%3D%2010%5E%7B-0.495%7D%20%3D%200.32%20M)
Hence, the mass of the sodium acetate tri-hydrate is:
![m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g](https://tex.z-dn.net/?f=m%20%3D%20moles%2AM%20%3D%20%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%2AV%2AM%20%3D%200.32%20mol%2FL%2A0.250%20L%2A136%20g%2Fmol%20%3D%2010.88%20g)
Therefore, the number of grams of CH₃COONa*3H₂O needed to make an acetic acid/sodium acetate tri-hydrate buffer solution is 10.88 g.
I hope it helps you!