All categories

3 years ago

If a sample containing 2.50 ml of nitroglycerin (density=1.592g/ml) is detonated, how many total moles of gas are produced?

1 answer:

5 0

Density = mass / volume
mass of nitroglycerin = density x volume = 1.592 x 2.5 = 3.98 grams
molar mass of nitroglycerin = 3(12) + 5(1) + 3(14) + 9(16) = 227 grams
number of moles reacting = mass / molar mass = 3.98/227 = 0.017 moles

The balanced chemical equation for this reaction is as follows:
4C3H5N3O9 ..............> 12CO2(g) + 6N2(g) + O2(g) + 10H2O(g)

From the balanced equation:
4 moles of nitroglycerin produces a total of 12 + 6 + 1 + 10 = 29 moles of gas.
Therefore:
0.017 moles of nitroglycerin produces a total of:
(0.017 x 29) / 4 = 0.12325 moles of gases

You might be interested in

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g)⟶2NH3(g) Ass

yaroslaw [1]

Answer:

moles of ammonia produced = 0.28 moles

Explanation:

The reaction is

$N_{2}(g)+3H_{2}(g) --> 2NH_{3}(g)$

As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia

So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen

= 0.42 moles of hydrogen molecule.

this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia

the moles of ammonia produced = 0.28 moles

Here the nitrogen is limiting reagent.

4 0

3 years ago

A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0

3 years ago

In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide

Black_prince [1.1K]

Answer:

Explanation:

The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:

Σ $P_g_a_s$ $= P_1+P_2+P_3+...+P_n$

The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:

$P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr$