Answer:
40.5 g of P₄O₁₀ are produced
Explanation:
We state the reaction:
P₄ + 5O₂ → P₄O₁₀
We do not have data from P₄ so we assume, it's the excess reactant.
We need to determine mass of oxygen and we only have volumne so we need to apply density.
Density = mass / volume, so Mass = density . volume
Denstiy of oxygen at STP is: 1.429 g/L
1.429 g/L . 16.2L = 23.15 g
We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles
5 moles of O₂ can produce 1 mol of P₄O₁₀
Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles
We determine the mass of product:
0.138 mol . 292.88 g/mol = 40.5 g
Using accurate measurements, using pure chemicals and performing the reaction under the most ideal conditions is important to get a valuable percent yield.
<h3>How we calculate the percent yield?</h3>
Percent yield of any chemical reaction is define as the ratios of the actual yield to the theoretical yield of the product and multiply by the 100.
To get the high percent yield or actual yield of any reaction, we have to perform the reaction under ideal condition because if we not use the standard condition then we get the low rate of reaction. Reactants should be present in the pure form as impurity make unwanted products and reduce the productivity of main product and accurate amount of reactants also important for the spontaneous reaction.
Hence, options (a), (b) & (c) are correct.
To know more about percent yield, visit the below link:
brainly.com/question/8638404
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
Answer: The value of acid ionization constant 
for acetic acid is 
Explanation:
cM 0 0
So dissociation constant will be:
Give c= 0.85 M and 
= 2.4
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Putting in the values we get:
Thus the value of acid ionization constant for acetic acid is
