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2 years ago

4. Chemical energy is the energy stored in a

Chemistry

1 answer:

Sergeu [11.5K]2 years ago

8 0

Answer:

Explanation:

Potential energy. Potential energy is energy that is stored.

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The equilibrium constant, Kc, for the following reaction is 1.55 at 667 K.

scoray [572]

Answer: The concentration of ammonia in the equilibrium mixture is 0.022 M

Explanation:

We are given:

Equilibrium concentration of hydrogen gas = 0.324 M

For the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

Initial: a

At eqllm: a-2x x 3x

Evaluating the value of 'x':

$\Rightarrow 3x=0.324\\\\\Rightarrow x=\frac{0.324}{3}=0.108$

So, equilibrium concentration of nitrogen gas = x = 0.108 M

Equilibrium concentration of ammonia gas = (a - 2x) = [a - 2(0.108)] = (a - 0.216) M

The expression of $K_c$ for above equation follows:

$K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}$

We are given:

$K_c=1.55$

Putting values in above expression, we get:

$1.55=\frac{0.108\times 0.324}{(a-0.216)}\\\\a=0.238$

Equilibrium concentration of ammonia gas = (a - 0.216) = [0.238 - 0.216] = 0.022 M

Hence, the concentration of ammonia in the equilibrium mixture is 0.022 M

3 0

3 years ago

If a ball rolling down a hill is half way between the top and bottom, how much potential energy does the ball have compared to k

madreJ [45]

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength $g$ is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If $m$ denote the mass of this ball, the gravitational potential energy ( $\rm GPE$ ) of this ball at height $h$ would be ${\rm GPE} = (m \cdot g) \cdot h$ , which is proportional to $h\!$ .

The value of $g$ near the surface of the earth is indeed approximately constant (typically $g \approx 9.8\; \rm m \cdot s^{-2}$ .)

At halfway between the top and bottom of this hill, the height of this ball would be $(1/2)$ of its initial value (the value when the ball was at the top of the hill.) Because the $\rm GPE$ of this ball is proportional to its height, at halfway down the hill, the $\rm GPE\!$ of this ball would also be $(1/2)\!$ its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that $(1/2)$ of the initial $\rm GPE\!$ of this ball was not lost. Rather, these $(1/2)\!$ of the initial $\rm GPE$ would have been converted to the kinetic energy ( $\rm KE$ ) of this ball.

Hence, when the ball is halfway down the hill:

$\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}$ .

$\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}$ .

Therefore:

$\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}$ .

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

3 0

2 years ago

A pump contains 1.5 L of air at 175 kPa. You draw back on the piston of the pump, expanding the volume until the pressure reads

Tresset [83]

Answer:

$\large \boxed{\text{86.8 L}}$

Explanation:

The temperature and amount of gas are constant, so we can use Boyle’s Law.

$p_{1}V_{1} = p_{2}V_{2}$

Data:

$\begin{array}{rclrcl}p_{1}& =& \text{0.579 atm}\qquad & V_{1} &= & \text{150 L} \\p_{2}& =& \text{1.00 atm}\qquad & V_{2} &= & ?\\\end{array}$

Calculations:

$\begin{array}{rcl}0.579 \times 150 & =& 1.00V_{2}\\86.85 & = & 1.00V_{2}\\V_{2} & = &\dfrac{86.85}{1.00}\\\\& = &\textbf{86.8}\\\end{array}\\\text{The new volume of the gas is } \large \boxed{\textbf{86.8 L}}$

6 0

3 years ago

The free energy change for the following reaction at 25 °C, when [Cd2+] = 1.20 M and [Fe2+] = 1.40×10-3 M, is -23.9 kJ: Cd2+(1.2

vladimir2022 [97]

Answer:

It's spontaneous in the reverse direction

Explanation:

A negative voltage indicate s that the reverse reaction is spontaneous (i.e. oxidation at the cathode, and reduction at the anode; by convention you would need to swap the labels on the electrodes)

8 0

3 years ago

If ammonia were formed from dalton's simplest formula of one atom of each element, what would he have concluded about the relati

baherus [9]

He would have concluded that the order of operations would be different. This therefore would affect the periodic table's order.

8 0

3 years ago