Answer:
8.7g/mL
Explanation:
I used the density formula to solve. d=m/v= 20g/2.3mL
Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry) 
= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
7.4x10^23 = molecules of silver nitrate sample
6.022x10^23 number of molecules per mole (Avogadro's number)
Divide molecules of AgNO3 by # of molecules per mol
7.4/6.022 = 1.229 mols AgNO3 (Sig Figs would put this at 1.3)
(I leave off the x10^23 because they both will divide out)
Use your periodic table to find the molar weight of silver nitrate.
107.868(Ag) + 14(N) + 3(16[O]) = 169.868g/mol AgNO3
Now multiply your moles of AgNO3 with your molar weight of AgNO3
1.229mol x 169.868g/mol = 208.767g AgNO3
Answer: all I know it’s not -31.5 for ppl taking the k12 test
Explanation: I took the test
Answer:
Mark has a speed of 6 MPS (Miles Per Hour)
Explanation:
It took mark 2 hours to ride his bike to his grandma's house, which was 12 miles away. I divided 2 by 12 and got 6. Remember this: mph = miles away ÷ time.
