The electron configuration of an element is shown below.

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc

pantera1 [17]

Answer:

$5.35 *10^{-4}$ M

Explanation:

Equation for the reaction is as follows:

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

By Applying the ICE Table; we have

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

Initial x 0.0025 M 0.0010 M

Change 0 0 0

Equilibrium x 0.0025 M 0.0010 M

$K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}$

Given that $K_c = 1.4*10^2$ ; Then:

$1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}$

$1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}$

$3.5 =( \frac{(0.001)}{(x)})^2$

$\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}$

$1.87=\frac{(0.001)}{(x)}$

$(x)= \frac{(0.001)}{1.87 }$

$x = 5.35 *10^{-4}$ M

∴ The equilibrium concentration of CO = $x = 5.35 *10^{-4}$ M

7 0

2 years ago

I need help with part B . please

myrzilka [38]

So it's good to map out what you know you have and work from there:
We have two liter measurements and one mole measurement, and we need to find the moles.

For this problem, think of it this way: 46 liters of gas = 1.4 moles.
If one side changes, the other has to as well (if the liters decrease, the moles decrease. if the liters increase, so do the moles.) What you can do is put this into a fraction:

 1.4 moles
 46 L 

if we know that each liter of gas is equal to x amount of moles, we know that 11.5 liters equals some amount of moles. You can put this into a fraction too, and make it equal to the other fraction:

1.4 moles = x moles
46 L 11.5 L

Then get your calculator out and do some algebra.

11.5 * (1.4/46) = x

The answer should come out to be: 0.35 moles

8 0

2 years ago

Which conditions suggest that a bond will be covalent?

Neporo4naja [7]

Explanation:

c is the answer because it's right now I lost my acc

5 0

2 years ago

How many atoms are in 2.5 moles of beryllium atoms?

Irina-Kira [14]

As we know that there are avogadro no. of atoms in 9 g of beryllium.
1 mole of beryllium = 6.02 * 10^23 atoms
so 2.5 mole= 6.02*10^23*2.5 i.e = 15.055 × 10^23 atoms

8 0

2 years ago

Matt did an experiment to study the solubility of two substances. He poured 100 mL of water at 20 °C into each of two beakers la

Black_prince [1.1K]

Part 1 : Answer is only B substance is soluble in water.

In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.

Part 2 : Substance A is not soluble in water and substance B is soluble in water.

According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.

But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.

4 0

2 years ago

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