Answer:
10.07 L is the volume for the produced N₂
Explanation:
This is a question for the Ideal Gases Equation:
P . V = n . R . T
In STP conditions, the pressure is 1 atm and the T° is 273K
Let's determine the moles of produced N₂ by the decomposition of TNT
The reaction is:
2 C₇H₅(NO₂)₃ (s) → 7 C (s) + 7 CO (g) + 3 N₂ (g) + 5 H₂O (g)
We convert the mass to moles (mass / molar mass)
68.2 g . 1 mol / 227.13 g = 0.300 moles
Ratio is 2:3. 2 moles of TNT can produce 3 moles of N₂
Then, 0.300 moles will produce (0.300 . 3) / 2 = 0.450 moles
We replace data:
1 atm . V = 0.450 mol . 0.082 L.atm/mol.K . 273K
V = (0.450 mol . 0.082 L.atm/mol.K . 273K) / 1 atm
V = 10.07 L
B. Intake valve is the correct answer
I’m am most positive the answer is c. For cat
Answer:
0.2042 M is the original concentration of 
(aq) in the titrating solution.
Explanation:
Mass of 
= 0.217 g
Moles of 
1 mole of have 2 mole of As and 1 mole of 
have 1 mole of As.
So, from 1 mole of we will have 2 moles of
Then from 0.001096 mol of :
of
According to reaction, 1 mole of reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of
of cerium (IV) ions.
Volume of the acidic cerium{IV) sulfate = 21.47 ml =0.02147 L
1 mL = 0.001 L
![[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M](https://tex.z-dn.net/?f=%5BCe%5E%7B4%2B%7D%5D%3D%5Cfrac%7B0.004384%20mol%7D%7B0.02147%20L%7D%3D0.2042%20M)
0.2042 M is the original concentration of (aq) in the titrating solution.
Answer:
E. None of the above statements are true.
Explanation:
