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2 years ago

The reaction

Chemistry

1 answer:

Aliun [14]2 years ago

7 0

From he calculations, we can see that the total pressure at equilibrium is 21 atm.

<h3>What is equilibrium constant?</h3>

The term equilibrium constant commonly describes the constant that that shows the extent of conversion of reactants to products.

We have to find the pressure of each gas as follows;

For H2

P = nRT/V = 4.553 /2 × 0.082 × 1000/8.89 L = 21 atm

Using the ICE table;

C(s) + 2H2(g) ⇌ CH4(g)

I 21 atm 0

C -x +x

E 21 - x x

0.263= x/(21 - x )^2

0.263(21 - x )^2 = x

38 - 11x - 0.263x^2 = x

0.263x^2 + 12x - 38 = 0

x=2.97 atm

At equilibrium, we have;

(21 - 2.97) + 2.97 = 21 atm

Learn more about equilibrium constant: brainly.com/question/17960050

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ikadub [295]

Answer:

a. Heterogeneous
b. Homogeneous
c. Homogeneous
d. Heterogeneous
e. Heterogeneous

Explanation:

A heterogeneous mixture is a mixture in which you can see multiple different ingredients in, for example vegetable soup, tea with ice and lemon slices, or fruit salad.

A homogeneous mixture is a mixture in which you can only see one thing, for example tea, seawater, or milk.

8 0

3 years ago

A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is

aliina [53]

Answer:

$\large \boxed{\text{67 000 g}}$

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$