The chemical equation needs to be balanced so that it follows the law of conservation of mass.
<span>The best choice is hypochlorous acid nitrous acid (HNO2) because it has the nearest value of pK to the desired pH.
pKa of </span>nitrous acid<span> is 3.34
If we know pKa and pH values, we can calculate the required ratio of conjugate base (NO2⁻) to acid (HNO2) from the following equation:
pH=pKa + log(conc. of base)/( conc. of acid)
</span><span>3.19=3.34 + log c(NO2⁻)/c(HNO2)
</span><span>3.19 - 3.34 = log c(NO2⁻)/c(HNO2)
-0.15 = log c(NO2⁻)/c(HNO2)
c(NO2⁻)/c(HNO2) = 10⁰¹⁵ = 1.41
</span>
Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Answer: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^4
Explanation:
I suggest looking at the electron configuration chart, it has really helped me a lot :)
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
You can learn more about solutions here: brainly.com/question/2412491
