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3 years ago

6

Acetylene, C2H2, burns according to the following reaction: C2H2 5O2 --> 4CO2 2H2O. Suppose 1.20 g of C2H2 is mixed with 3.50

g of O2 in a closed, steel container, and the mixture is ignited. What substances will be found in the mixture left when the burning is complete

1 answer:

OLEGan [10]3 years ago

8 0

WWWWWWAAAAAAPPPPPP!!!!!

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N2O5 decomposes to form NO2 and O2 with first-order kinetics. The initial concentration of N2O5 is 3.0 M and the reaction runs f

kow [346]

Answer : The final concentration of $N_2O_5$ is, 2.9 M

Explanation :

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $5.89\times 10^{-3}\text{ min}^{-1}$

t = time passed by the sample = 3.5 min

a = initial concentration of the reactant = 3.0 M

a - x = concentration left after decay process = ?

Now put all the given values in above equation, we get

$3.5=\frac{2.303}{5.89\times 10^{-3}}\log\frac{3.0}{a-x}$

$a-x=2.9M$

Thus, the final concentration of $N_2O_5$ is, 2.9 M

3 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Match the correct definition with the correct term from questions 10-13: A. Internal energy B. Latent heat C. Chemical (bond) en

Marizza181 [45]

Answer: A. Internal energy : May be viewed as the sum of the kinetic and potential energies of the molecules

B. Latent heat: The internal energy associated with the phase of a system.

C. Chemical (bond) energy : The internal energy associated with the atomic bonds in a molecule

D. Nuclear energy : The internal energy associated with the bonds within the nucleus of the atom itself

Explanation:

Internal energy is defined as the total energy of a closed system. Internal energy is the sum of potential energy of the system and the kinetic energy of the system. It is represented by symbol U.

Latent heat is the thermal energy released or absorbed by a thermodynamic system when the temperature of the system does not change. It is thus also called as hidden heat.

Chemical energy is the energy stored in the bonds of molecules.

Nuclear energy is the energy which is stored in the nucleus of an atom called as binding energy within protons and neutrons.

5 0

3 years ago

21. _____ Which of the following entities has the greatest mass?

IrinaK [193]

I believe it’s D because phosphorus has the mass of 30.9.

8 0

3 years ago

A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.

Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L$

Therefore, the new volume of the air is, 6.83 L

6 0

4 years ago