It states the fact, which we now know, that electrons are responsible for the chemical bonding. According to this theory, valency is the number of electrons present in the outermost energy shell of the atom. This energy shell is called valency shell.
I think sand is just glass and other stuff
Answer:
There is 1.6 L of NO produced.
Explanation:
I assume you have an excess of NH3 so that O2 is the limiting reagent.
<u>Step 1:</u> Data given
2.0 liters of oxygen reacts with ammonia
<u>Step 2:</u> The balanced equation
4NH3 + 5O2 → 4NO + 6H2O
For 4 moles of NH3, we need 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O
Consider all gases are kept under the same conditions for pressure and temperature, we can express this mole ratio in terms of the volumes occupied by each gas.
This means: when the reaction consumes 4 liters of ammonia (and 5 liters of oxygen) it produces 4 liters of nitrogen monoxide
Now, when there is 2.0 liters of oxygen consumed, there is 4/2.5 = 1.6 L of nitrogen monoxide produced.
There is 1.6 L of NO produced.
Answer:
1. Converging and Diverging plates
2. In the sunlit zone
3. a.Close to coastlines and upwelling zones b. Nutrients are brought to the surface ocean from depth by the upwelling process
4. Dead organic matter that come from phytoplankton and other plants
Answer:
pH = 12.43
Explanation:
<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
<em>Moles KOH:</em>
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
<em>Moles HN3:</em>
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
