Answer:
Since molarity is defined as moles of solute per liter of solution, we need to find the number of moles of nitric acid, and the volume of solution.
molar mass of nitric acid (HNO3) = 1 + 14 + (3x16) = 15 + 48 = 63 g/mole
1.50 g/ml x 1000 ml = 1500 g/liter
1500 g/liter x 0.90 = 1350 g/liter of pure HNO3 (the 0.9 is to correct for the fact that it is 90% pure)
1350 g/liter x 1 mole/63 g = 21.43 moles/liter = 21 Molar HNO3
= 21 Molar of HNO3
Answer:
The answer to your question is:
Explanation:
Data
carbon 7.3% = 7.3g
hydrogen 4.5% = 4.5g
oxygen 36.4% = 36.4 g
nitrogen 31.8% = 31.8 g
Now
For carbon
12 g --------------------1 mol
7.3 g ------------- x
x = 7.3/12 = 0.608 mol
For hydrogen
1 g -------------------- 1 mol
4.5 g ------------------ x
x = 4.5 mol
For oxygen
16 g ------------------- 1 mol
36.4 g ---------------- x
x = 2.28 mol
For nitrogen
14 g ---------------- 1 mol
31.8 g --------------- x
x = 2.27 mol
Now divide by the lowest result, the is 0.608 from carbon
carbon 0.608/0.608 = 1
hydrogen 4.5/ 0.608 = 7.4
oxygen 2.28/0.608 = 3.75
nitrogen 2.27/0.608 = 3.73
Empirical formula = CH₇O₄N₄
Well, we need to find the ratio of Al to the other reactant.
Al:HCl = 1:3
--> this means that for every 1 Al used, you have to use 3 HCl.
6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.
The ratio of HCl:AlCl = 3:1
13/3 = 4.3333...
The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.
Hope this helps!!! :)
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
Answer:
(a): 2,300 kilograms
(b): 0.005 kilograms
(c): 2.3 × 10^-5 kilograms
(d): 155 kilograms
Explanation:
Formulas:
(a); divide the mass value by 1000
(b); divide the mass value by 1e+6
(c); divide the mass value by 1e+9
(d); multiply the mass value by 1000
