All categories

2 years ago

If an equilibrium mixture of the three gases at 600K contains 2.9210^-2 M COCH(g) and 1.7610^2 M CO, what is the equilibrium

Chemistry

1 answer:

Vlad [161]2 years ago

7 0

Answer:

Explanation:

You might be interested in

Can someone help me with this, what even is this? I didn't even learn this

Lunna [17]

It’s hard which class are u

3 0

2 years ago

Why Formaldehyde is more reactive than Acetaldehyde?

tatyana61 [14]

Aldehydes are usually more reactive<span> toward nucleophilic substitutions </span>than<span> ketones because of both steric and electronic effects. In aldehydes, the relatively small hydrogen atom is attached to one side of the carbonyl group, while a larger R group is affixed to the other side.</span>

7 0

2 years ago

Which of the following will have the slowest rate of diffusion at a given temperature?

Orlov [11]

So diffusion is inversely proportional to mass !
so as mass of the particle increases, diffusion decreases !

4 0

3 years ago

14. What is the actual temperature 3000 km below the surface of the Earth?

Alexus [3.1K]

Answer:

temperature 3000 km below the surface of earth is varied, it's harder to know the actual temperature, it can be found around 3,000 - 3,500 degrees Celsius

7 0

2 years ago

CaO + H2O -> Ca(OH)2

yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075$

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

$\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g$

% yield :

$\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%$

8 0

2 years ago

If an equilibrium mixture of the three gases at 600K contains 2.92*10^-2 M COCH(g) and 1.76*10^2 M CO, what is the equilibrium

If an equilibrium mixture of the three gases at 600K contains 2.9210^-2 M COCH(g) and 1.7610^2 M CO, what is the equilibrium