What volume (in L) of a 1.25 M
1 answer:
Answer:
V1= 0.305L
Explanation:
To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2
Since the question wants the volume in litres, convert 455 mL to L
455/ 1000
= 0.455 L
Now make the substitution
1.25 × V1 = 0.838 × 0.455
Rearrange to make V1 the subject
V1=
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1. ₉¹⁹F
2. Mg(OH)₂
3. 2H₂ + O₂ ⇒ 2H₂O
<h3>Further explanation</h3>1. Fluorine, atomic number : 9 , mass number = 19
Symbol : ₉¹⁹F
protons=electrons=atomic number = 9
neutrons = mass number - atomic mass
Configuration : [He] 2s² 2p⁵
2. Magnesium hydroxide is an ionic compound and is a strong base consisting of 2 ions:
Positive ion: Magnesium: Mg²⁺
negative ion: Hydroxide: OH⁻
The charges of the two are crossed, so that the compound becomes:
Mg(OH)₂
3. Reaction :
H₂ + O₂ --------> H₂O
give coefficient :
aH₂ + bO₂ --------> H₂O
H, left = 2a, right 2⇒2a=2⇒a=1
O, left = 2b, right 1⇒2b=1⇒b=0.5
Reaction becomes :
H₂ + 0.5O₂ --------> H₂O x 2
2H₂ + O₂ --------> 2H₂O