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Lilit [14]
2 years ago
11

How many grams of nitroglycerin are needed to produce 25.0 grams of CO2

Chemistry
1 answer:
LekaFEV [45]2 years ago
8 0
N<em>itroglycerin  decomposes    according  to   the   following  equation

4 C3H5N309  --->  12 CO2 +  10 H20  +  6 N2 +O2
The  moles  of  CO2 =  25 /44(molar  mass  of  CO2)=  0.568  moles

the  mole  ratio  of  C3H5N3O9  to  CO2  IS  4:12   from  equation  above
therefore  the  moles  of  C3H5N3O9=( 0.568 x 4)/12= 0.189  moles
mass  of C3H5N3O9=moles  x  molar  mass  of  C3N5N3O9

That  is    0.189 mol x  227  g/mol=  42.903  grams</em>
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The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.
ASHA 777 [7]
We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:

PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas

Hope this answers the question. Have a nice day.
7 0
2 years ago
What is the mass of 1.82 moles of lithium carbonate?
Svetlanka [38]
You can view more details on each measurement unit: molecular weight of Lithium Carbonate or grams The molecular formula for Lithium Carbonate is Li2CO3. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Lithium Carbonate, or 73.8909 grams.
5 0
2 years ago
Read 2 more answers
Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu
Nitella [24]

Answer:

\Delta \:H_{vap}=40383.88\ J/mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)

Given:

P_1 = 13.95 torr

P_2 = 144.78 torr

T_1 = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K

T_1 = 298.15 K

T_2 = 75°C  = 348.15 K

So,

\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)

\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}

\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=40383.88\ J/mol

4 0
2 years ago
What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its
pishuonlain [190]

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Solution:

Q = m.c. ΔT

ΔT  = T2 - T1

ΔT  = 20°C - 25°C

ΔT  = -5°C

50208 j = 2.50 g . c. -5°C

50208 j = -12.5 g.°C .c

50208 j /-12.5 g.°C =  c

c = 4016.64 j/g.°C

7 0
2 years ago
The empirical formula of a gaseous fluorocarbon is CF2. At a certain temperature and pressure, a 1-L volume holds 8.93 g of this
dimaraw [331]

Answer:

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

molar mass of fluorine gas = 37.99681

let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

molecular formula of CF₂ = 4 (CF₂) = C₄F₈

3 0
2 years ago
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