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monitta
2 years ago
12

Suppose that 10.0 L of Carbon Dioxide gas are produced by this reaction, 4C3H5N3O9 -> 12 CO2 + 10H2O + 6N2 +O2, at a temperat

ure of -5 degrees C, and a pressure of exactly 1 atm. Calculate the mass of nitroglycerin that must have reacted in grams.
Chemistry
1 answer:
siniylev [52]2 years ago
4 0

The mass of nitroglycerin : 34.52 g

<h3>Further explanation</h3>

Reaction

4C₃H₅N₃O₉ ⇒ 12 CO₂ + 10H₂O + 6N₂ +O₂

Volume = 10 L

Temperature = -5°C=268 °K

Pressure = 1 atm

mol of CO₂ (ideal gas) :

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 10}{0.082\times 268}\\\\n=0.455

mol ratio C₃H₅N₃O₉ : mol CO₂= 4 : 12, so mol C₃H₅N₃O₉ :

\tt \dfrac{4}{12}\times 0.455=0.152

mass C₃H₅N₃O₉ (MW=227,0865 g/mol):

\tt 0.152\times 227.0865=\boxed{\bold{34.52~g}}

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3 years ago
The pressure in car tires is often measured in pounds per square inch ( lb/in.2 ), with the recommended pressure being in the ra
Goshia [24]

when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm

<h3>Conversion scale</h3>

14.6959 lb/in² = 1 atm

<h3>Data obtained from the question</h3>
  • Pressure (in lb/in²) = 32.5 lb/in²
  • Pressure (in ATM) =?

<h3>How to convert 32.5 lb/in² to atm</h3>

14.6959 lb/in² = 1 atm

Therefore

32.5 lb/in² = 32.5 / 14.6959

32.5 lb/in² = 2.21 atm

Thus, 32.5 lb/in² is equivalent to 2.21 atm

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1 year ago
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3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc
jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2HBr(g)\rightarrow H_2(g)+Br_2(g

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}

Given: -\frac{d[HBr]}{dt}]=0.301

Putting in the values we get:

\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}

+\frac{1d[Br_2]}{dt}=0.151

Thus the rate of appearance of Br_2 is 0.151

8 0
3 years ago
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