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The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.
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B) a molecule
A molecule is formed when two atoms join together with a covalent bond.
Answer:
After complete reaction, 0.280 moles of ammonia are produced
Explanation:
Step 1: Data given
Number of moles N2 = 0.140 moles
Number of moles H2 = 0.434 moles
Step 2: The balanced equation
N2(g) + 3H2 (g) ⟶ 2NH3 (g)
Step 3: Calculate the limiting reactant
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
N2 is the limiting reactant. It will completely be consumed (0.140 moles).
H2 is in excess. There will react 3*0.140 = 0.420 moles
There will remain 0.434 - 0.420 = 0.014 moles
Step 4: Calculate moles NH3
For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3
After complete reaction, 0.280 moles of ammonia are produced
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
Answer:
7,94 minutes
Explanation:
If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>
For the zero-order reactions, concentration-time equation can be written as follows:
[A] = - Kt + [Ao]
where:
- [A]: concentration of the reactant A at the <em>t </em>time,
- [A]o: initial concentration of the reactant A,
- K: rate constant,
- t: elapsed time of the reaction
<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>
Data:
K = 4.2 ×10−3atm/s,
[A]o=[HBr]o= 2 atm,
[A]=[HBr]=0 atm (all HBr(g) is gone)
<em>We clear the incognita :</em>
[A] = - Kt + [Ao]............. Kt = [Ao] - [A]
t = ([Ao] - [A])/K
<em>We replace the numerical values:</em>
t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes
So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).
