The answer should be <span>enteropeptidase
</span>
Answer: pH=12.69
Explanation:
Initial 0.12 0 0
Eqm 0.12-x x x
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
(neglecting small value of x in comparison to 0.12)
Moles of 
0.06 moles of NaOH will give 0.06 moles of ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
Now 
moles of 
will be neutralized by moles of 
and 
moles of will be left.
Molarity of 
![pOH=-\log[OH^-]=-\log[0.049]=1.31](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B0.049%5D%3D1.31)
pH = 14 - pOH= 14 - 1.31 = 12.69
B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.
mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)
mass nh3 = 63.58 g nh3 is produced</span>
<h3>Answer:</h3>
Mass = 96.47 g
<h3>
Solution:</h3>
Data Given:
M.Mass = 28.97 g.mol⁻¹
Moles = 3.33 mol
Mass = ??
Formula Used:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 3.33 mol × 28.97 g.mol⁻¹
Mass = 96.4701 g
Rounding to four significant numbers,
Mass = 96.47 g
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.
