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joja [24]
3 years ago
5

A 0.100 g sample of a carbon–hydrogen–oxygen compound is combusted in a stream of pure oxygen and produces 0.220 g co2 and 0.239

g h2o. what is the mass percent of hydrogen in the sample?
a.13.3
Chemistry
1 answer:
beks73 [17]3 years ago
8 0

First we assume that the compound containing only C,H,and O is combusted completely in the presence of excess oxygen, so that the only things that can be produced are water and carbon dioxide.

 

From there we should back calculate the amount of Hydrogen that is in the original sample by taking all of the hydrogen in the 0.239g to came from the organic compound.

 

And since we know that the original mass of the sample was .100g, we can also easily get a mass % H by taking the mass Hydrogen calculated over the total original mass (.100 g)

 

So that:

 

0.239g H2O / (18.01 g/mol) = .01327 moles H20

 

.01327 Moles H20 * 2.02g H (per every mole H2O) = .0268g H initially present in the sample

 

.0268g H / .100g sample = 26.8% H by mass

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andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

q=c\times \Delta T

where,

c = heat capacity of calorimeter = 1620 J/K

\Delta T = change in temperature = 0.150^oC=0.150K   (Change remains same)

Putting values in above equation, we get:

q=1620J/K\times 0.15K=243J

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}

We are given:

Moles of monoclinic sulfur = 1 mole

  • To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,  

q = amount of heat released = -243 J

n = number of moles = 1 mole

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol

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8 0
3 years ago
8. (02.04 MC)
elena55 [62]

Answer:

3p^2

Explanation:

after filling 3s^2 only two electrons  left out of 14 so the next sub shell is 3p therefore ,X represents 3p^2

3 0
2 years ago
3. Hydrogen reacts with nitrogen to produce ammonia according to the equation:
lys-0071 [83]

Mass of ammonia produced : 121.38 g

<h3>Further explanation</h3>

Given

Reaction

3H₂(g) + N₂(g) ⇒ 2NH₃(g)

100g of N₂

Required

Ammonia produced

Solution

mol of N₂ :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{100}{28}\\\\mol=3.57

From the equation, mol ratio of N₂ and NH₃ = 1 : 2, so mol NH₃ :

\tt \dfrac{2}{1}\times 3.57=7.14~moles

mass of NH₃(MW=17 g/mol) :

\tt mass=mol\times MW\\\\mass=7.14\times 17\\\\mass=121.38~g

8 0
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