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nlexa [21]
3 years ago
12

(a) if a sample containing 2.00 ml of nitroglycerin is detonated, how many total moles of gas are produced? (b) if each mole of

gas occupies 55 l under the conditions of the explosion, how many liters of gas are produced? (c) how many grams of n2 are produced in the detonation?
Chemistry
1 answer:
KATRIN_1 [288]3 years ago
3 0
<span>(2.09 mL) x (1.592 g/mL) / (227.0871 g C3H5O9N3/mol) = 0.014652 mole C3H5O9N 4 moles C3H5O9N produce 12 + 6 + 1 + 10 = 29 moles of gases, so: (0.014652 mole C3H5O9N) x (29/4) = 0.106 mole of gases (b) (0.106 mol) x (46 L/mol) = 4.88 L gases (c) (0.014652 mole C3H5O9N) x (6/4) x (28.0134 g/mol) = 0.616 g N2</span>
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Which of the following reactions are redox reactions? 4Li(s)+O2(g)→2Li2O(s) Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) Sr(NO3)2(aq)+Na2SO4(aq
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Answer:  

As2O3 → H3AsO4

As2O3 → 2H3AsO4             balance As

5H2O + As2O3 → 2H3AsO4       balance O by adding H2O to one side

5H2O + As2O3 → 2H3AsO4 + 4H+    balance H by adding H+ to one side

5H2O + As2O3 → 2H3AsO4 + 4 H+ + 4e-     balance charge by adding electrons to one side

 

Now do the same for the other part of the reaction

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3e- + 4H+ + NO3- → NO + 2H2O

 

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4 0
3 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
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