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3 years ago

According to the Kinetic Molecular Theory, the absolute temperature of a gas is directly related to average molecular kinetic

Chemistry

2 answers:

astra-53 [7]3 years ago

7 0

Answer;

-energy

According to the Kinetic Molecular Theory, the absolute temperature of a gas is directly related to average molecular kinetic energy.

Explanation;

Kinetic molecular theory is a theory that explains that the particles of a gas are in constant random motion, exhibiting perfectly elastic collisions.
This theory may be explained using Boyle's and Charles's law. Such that the average kinetic energy of a given particles of a gas is directly proportion to the absolute temperature.

Alekssandra [29.7K]3 years ago

6 0

Answer: The correct answer is Option D.

Explanation:

Average kinetic energy is defined as the average of kinetic energies of all the particles in a system. It is a measure of temperature. The chemical equation used to calculate average kinetic energy is given as:

$K=\frac{3RT}{2N_A}$

where,

K = average kinetic energy

R = Gas constant

$N_A$ = Avogadro's number

T = temperature of the system

Average kinetic energy of the system is directly proportional to the temperature of the system. Thus, if temperature increases, the average kinetic energy also increases and vice-versa.

From the above information, it is concluded that the correct answer is Option D.

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MgCl2 + AgNO3 → AgCl + Mg(NO3)2

Law Incorporation [45]

Answer:

MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2

Explanation:

I'm assuming you want to balance it so...

The first thing I see is that there are two chlorines on the reactant side and one on the product side

Adding a coefficient of 2 would get 2AgCl2

Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.

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6 0

3 years ago

Read 2 more answers

What is the change in enthalpy when 4.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas? 250 2(g) + O2(g) ®

lidiya [134]

Answer: A. -396 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

$2SO_2(g)+O_2 (g)\rightarrow 2SO_3(g)$ $\Delta H^0=198kJ$

Reversing the reaction, changes the sign of $\Delta H$

$2SO_3(g)\rightarrow 2SO_2(g)+O_2 (g)$ $\Delta H^0=-198kJ$

On multiplying the reaction by 2, enthalpy gets multiplied by 2:

$4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)$ $\Delta H=2\times -198kJ=-396kJ$

Thus the enthalpy change for the reaction $4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)$ is -396 kJ.

8 0

3 years ago

16.25 g of water at 54 C relaeases 402.7 J. What will be its final temp?

leonid [27]

Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?

Solving:

Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
$\Delta\:T = \frac{-402.7}{67.99}$
$\boxed{\Delta\:T \approx -5.92\:^0C}$

If: ΔT (T final - T initial) = ?
$-5.92^0 = T_{final} - 54^0$
$T_{final} = 54^0 - 5.92^0$
$\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark$

8 0

3 years ago

The rate of disappearance of HCl was measured for the following reaction:

AURORKA [14]

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min =0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) after 54mins, t = 54*60s = 3240s

average reaction rate = (1.58 - 1.85)M / (3240 * 0.0)s

= -0.27M/3240

= 0.000083M/s

after 107mins, t = 107*60s = 6420s

average reaction rate = (1.36 - 1.58)M/ (6420 - 3240)s

= -0.22M/3180s

= 0.000069M/s

after 215mins, t = 215*60s = 12900s

average reaction rate = (1.02 - 1.36)M/ (12900 - 6420)s

= -0.34M/6480s

= 0.000052M/s

after 430mins,t = 430*60 = 25800s

average reaction rate = (0.580 - 1.02)M / (25800 - 12900)s

= -0.44M/12900s

= 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min

= (0.580 - 1.85)M/ (25800 - 0.0)s

= -1.27M/25800s

=0.000049M/s

= (1.02 - 1.58)M / (12900 - 3240)s

= -0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36)M / (25800 - 6420)s

= 0.78M /19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

6 0

3 years ago

Rutherfordium-261 has a half-life of 1.08 min. How long will it take for a sample of rutherfordium to lose one-third of its nucl

Ira Lisetskai [31]

Answer:

$t=1.712min$

Explanation:

Hello!

In this case, since the radioactive decay equation is:

$\frac{A}{A_0}=2^{-\frac{t}{t_{1/2} }$

Whereas A stands for the remaining amount of this sample and A0 the initial one. In such a way, since the sample of rutherfordium is reduced to one-third of its nuclei, the following relationship is used:

$A=\frac{1}{3} A_0$

And we plug it in to get:

$\frac{\frac{1}{3} A_0}{A_0}=2^{-\frac{t}{t_{1/2}} } \\\\\frac{1}{3}=2^{-\frac{t}{t_{1/2}} }$

Now, as we know its half-life, we can compute the elapsed time for such loss:

$log(\frac{1}{3})=log(2^{-\frac{t}{t_{1/2}} })\\\\log(\frac{1}{3})=-\frac{t}{t_{1/2}} }*log(2)$

$t=-\frac{log(\frac{1}{3})t_{1/2}}{log(2)} \\\\t=1.71min$

Best regards!

8 0

2 years ago