100ml volume of 0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.
Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,
C1 V1 =C2 V2
where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1) HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1) Caoh2 solution.
While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give in the expression we get the volume of the final concentration.
V1= C2 V2/ C1
= 0.0100m . 150ml /0.0150M
= 100ml
The volume of the hcl solution is 100ml.
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<span>Answer:
2 C8H18 + 25 O2 => 16 CO2 + 18 H2O
1.0 kg = 1000 g C8H18 = 1000 g / 114.2293 g/mole = 8.75 moles C8H18
8.75 moles C8H18 produce (16/2) (8.75) = 70 moles CO2
70 moles CO2 = (70 moles) (44.0096 g/mole) = 3081 g CO2 = 3.1 kg CO2</span>
D.) "Vaporization" <span>is not a form of heat transfer.
It is just conversion of the state of matter from Liquid to Vapor
Hope this helps!</span>
Answer:
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Answer:
Explanation:
Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)
Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V
Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V
Anode reaction :
Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V
Cathode reaction :
Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V
Cell potential = Ecathode - Eanode
Ecell = .80 - ( - 2.90 )
Ecell = 3.7 V .
equilibrium constant ( K ) :
Ecell = .059 log K / n
n = 2
3.7 = .059 log K / 2
log K = 125.42
K = 2.63 x 10¹²⁵ .
Free energy change :
ΔG = - n F Ecell
= - 2 x 96500 x 3.7
= 714100 J
= 7.141 x 10⁵ J .
