0.001 would be the smallest.
Good Luck! :)
Answer:
.0924 moles of NaCl
Explanation:
So you know you have 5.4 g of NaCl and you need to know how many moles there are in this amount of NaCl
- You'll need to find the atomic mass of the compound NaCl to help you solve for moles
- Sodium (Na) on the periodic table has a mass of 22.99
- Chlorine (Cl) on the periodic table has a mass of 35.45
Add these two together----> 22.99 + 35.45 = 58.44
Now you can calculate for moles
<u>Written-out method:</u>
<u>5.4 grams of NaCl | 1 mole of NaCl </u>
| 58.44 grams NaCl = .0924 moles of NaCl
<u>Plug into calculator method:</u>
(5.4 g of NaCl/ 58.44g NaCl= .0925 moles)
Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67
Answer:
Throughout the explanation section, the reason behind the given statement is described.
Explanation:
- The chemicals thus produced were indeed opposite or separate from the reaction mixture, this same reaction wouldn’t change when it's more balanced than some of the reactants.
- Another reason is that the development of advanced organisms, as well as chemical alterations, is irreversible throughout nature cant undo the chemical modifications.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of 
= 
= 0.0458 M
Concentration of 
= 
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72
pH ≅ 4.80
