Answer:
2
Explanation:
C⁴H¹⁰ + 2O² —> 4CH²O + H²
C) It determines the concentration of an unknown substance in neutralization reactions.
It's where chemical elements are organized based on atomic weight. Also, like/similar physical and chemical properties in each interval of 7 elements.
It was a silly reason it was rejected. It was rejected because it's octaves were too similar to those in music.
I hope this helps!
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
Answer:
The correct answer is option B.
Explanation:
As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.
After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.