Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
fireproofing spray used in firefighting can be termed as one of the possible discovery that would be attributed to an inorganic molecule.
Explanation: • There a number of advancement been made each day to improve the different process either domestic, industrial, or related to research work. As a number of professionals are always in search to get the right set of information or data to get to a more proper conclusion.
Answer:
Explanation:
There are no molecules in NaCl, because it consists only of ions.
However, we can calculate the number of formula units (FU) of NaCl.
Step 1. Calculate the moles of NaCl
Step 2. Convert moles to formula units
There are 
in 3.6 g of NaCl.
Answer:
Explanation:
The density formula is:
Let's rearrange the formula for 
. the volume. Multiply both sides by , then divide by 
.
The volume can be found by dividing the mass by the density. The mass of the object is 30.07 grams and the density is 1.48 grams per milliliter.
Divide. Note, when dividing, the grams, or 
will cancel out.
The volume of the object is 20.317567567568 milliliters.
