Answer:
0.187 mol
Explanation:
A solution is created by dissolving 10.0 grams of ammonium chloride in enough water to make 315 mL of solution.
The formula of ammonium chloride is NH₄Cl and its molar mass is 53.49 g/mol. We can use this information to calculate the moles of ammonium chloride associated with 10.0 grams.
10.0 g × (1 mol/53.49 g) = 0.187 mol
Answer: a) Por la estequiometría de la reacción, vemos que: 2. 4. 2. 4. 3. 2. 4. 98 g de H ... a) ¿Qué volumen de ácido sulfúrico concentrado de densidad 1'84 g/mL y 96 ... a) ¿Cuántos moles de átomos de carbono hay en 1'5 moles se sacarosa 12 ... b) El volumen de la misma que debe tomarse para preparar 1 L de disolución de.
Answer:
Explanation:
Hello there!
Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:
In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
Best regards!
Answer:
34g of potassium chlorate.
Explanation:
A saturated solution is a solution that, under a temperature, has the maximum amount of solute possible. The maximum amount that a solvent can dissolve of a solute is called <em>solubility.</em>
<em> </em>
The solubility of potassium chlorate in water at 70°C is 34g/ 100g of water.
That means, to saturate 100g of water at 70°C you need yo add:
<h3>34g of potassium chlorate.</h3>
<span>The answer is
0.124639157 pounds.
I converted it. </span><span />
