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>
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>
<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol
</span>
Answer: 207.2
Explanation:
In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 x 10 -27 kilogram (kg), or 1.67377 x 10 -24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus.
Answer:
A
Explanation:
Nitrogen is an atom made up of 7 electrons.
To draw the orbital energy level diagram, let us write the orbital notation of the atom;
7 electrons of Nitrogen:
1s² 2s² 2p³
So,
The orbital notation diagram is :
1s² 2s² 2p³
↑↓ ↑↓ ↑↑↑
<em>Answer:</em>
- The concentration of new solution will be 1×10∧-7 M.
<em>Solution:</em>
<em>Data Given </em>
given mass of fluoxymesterone =16.8mg = 0.0168 g
molar mass of fluoxymesterone = 336g/mol
vol. of fluoxymesterone = 500.0 ml = 0.500 L
Stock Molarity of fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M
So applying dilution formula
Stock Solution : New Solution
M1.V1 = M2.V2
( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L
[( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2
1 × 10∧-7 = M2
<em>Result:</em>
- The concentration of new solution M2 will be 1 × 10∧-7
