Answer is: mass of silver chloride is 34,82 g.
Chemical
reaction: AgNO₃ +
NaCl → AgCl + NaNO₃.
<span>m(AgNO</span>₃) = 53,42 g..<span>
m(NaCl) = 14,19 g.
n(AgNO</span>₃) = m(AgNO₃) ÷ M(AgNO₃).<span>
n(AgNO</span>₃) = 53,42 g ÷ 169,87 g/mol.
n(AgNO₃) = 0,314 mol.
n(NaCl) = 14,19 g ÷ 58,4 g/mol.
n(NaCl) = 0,242 mol; limiting reactant.
From chemical reaction: n(NaCl) : n(AgCl) = 1 : 1.<span>
n</span>(AgCl)<span> = 0,242 mol.
m</span>(AgCl) = 0,252 mol · 143,32 g/mol.
m(AgCl) = 34,82 g.
The H⁺ ion concentration can be calculated from pH values using the following equation:
![pH=-log[H⁺]](https://tex.z-dn.net/?f=pH%3D-log%5BH%E2%81%BA%5D)
1.) Given pH = 2
Using the above equation, 2 = - log [H⁺]
Therefore, [H⁺] = 10⁻² mol/L
2.) Given pH = 6
Using the same equation, we have 6 = - log [H⁺]
Hence, [H⁺] = 10⁻⁶ mol/L
3.) Taking the ratio of [H⁺] for pH = 2 and pH = 6, we have
= 10⁴
So, there are 10,000 times more H⁺ ions in a solution of pH = 2 than that of pH = 6.
The weight of aluminum are required to produce 8.70 moles of aluminum chloride is 234.9 g
<h3>
What is the use of aluminium chloride ?</h3>
Aluminum chloride is useful for the treatment of palmar, plantar, and axillary hyperhidrosis.
Aluminum chloride has also been reported to be useful in facial and scalp hyperhidrosis
The balanced chemical equation represents the mole ratio in which the chemicals combine.
In this case, illustrates that 2 mol Al produces 2 mol Al Cl₃, hence these 2 chemicals are in a 1:1 ratio.
Thus, to produce 8.70 mol aluminium chloride, it will require 8.70 mol aluminium.
But this quantity of Al has a mass in grams of
m = n × Mr
= 8.70 mol × 27g/mol
= 234.9 g
Hence, The weight of aluminum are required to produce 8.70 moles of aluminum chloride is 234.9 g
Learn more about mole concept here ;
https://brainly.in/question/12599804
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Answer:
Place 20.0 g NaOH(s) in a flask and dilute to 1.00 L with water.
Explanation:
carbon won't have same number of valence electrons as others
