part 1 : the final volume : 1.404 L
part 2 : the initial concentration : 4.06 M
<h3>Further explanation
</h3>
Dilution is the process of adding a solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
part 1 :
M₁=44.8%
V₁=0.73 L
M₂=23.3%
part 2 :
V₁=739 ml=0.739 L
V₂=1.5 L
M₂=2
Answer:
Explanation:
a. Oxidation : 2O + 4e^- ------> 2O^2-
b. Reduction: 2Sr - 4e- -------> Sr^2+
c. Balanced redox reaction
2Sr + O2 ------------> 2Sr O
Oxidation and reduction can be defined by various means, addition of oxygen, removal of hydrogen, removal of electrons. For this reaction, this definition is used, oxidation is the loss of electrons while reduction is the gaining of electrons.
In (a) oxidation half reaction, the valency of oxygen is zero and then moves into lossing two electrons resulting into -2 valency.
In (b) reduction half reaction, the valency of Sr is zero and gains electrons resulting into valency of 2.
In the overall redox reaction, Sr and O2 with valency of 0 each reacts together and form SrO with valency of 2 and -2 respectively, which gives 0 and then balances the equation.
Hello!
I believe the answer is A) Anaphase.
I hope it helps!
Answer:
The enthalpy of the solution is -35.9 kJ/mol
Explanation:
<u>Step 1:</u> Data given
Mass of lithiumchloride = 3.00 grams
Volume of water = 100 mL
Change in temperature = 6.09 °C
<u>Step 2:</u> Calculate mass of water
Mass of water = 1g/mL * 100 mL = 100 grams
<u>Step 3:</u> Calculate heat
q = m*c*ΔT
with m = the mass of water = 100 grams
with c = the heat capacity = 4.184 J/g°C
with ΔT = the chgange in temperature = 6.09 °C
q = 100 grams * 4.184 J/g°C * 6.09 °C
q =2548.1 J
<u>Step 4:</u> Calculate moles lithiumchloride
Moles LiCl = mass LiCl / Molar mass LiCl
Moles LiCl = 3 grams / 42.394 g/mol
Moles LiCl = 0.071 moles
<u>Step 5:</u> Calculate enthalpy of solution
ΔH = 2548.1 J /0.071 moles
ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)
The enthalpy of the solution is -35.9 kJ/mol
