Answer:
The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
Explanation:
As HCl is a strong acid and hence a strong electrolyte, it will dissociate as
HCl ⟶ H⁺ + Cl⁻
So, The concentration of H⁺ will be 2.5 M (same as HCl)
Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
<u>-TheUnknownScientist</u><u> 72</u>
Answer:
The volume will be "2.95 L".
Explanation:
Given:
n = 0.104
p = 0.91 atm
T = 314 K
Now,
The Volume (V) will be:
= 
By putting the values, we get
= 
= 
= 
Answer:
= -356KJ
<em>therefore, the reaction where heat is released is exothermic reaction since theΔH is negative</em>
Explanation:
given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ
then,
change in enthalpy (ΔH) = -162( released energy)
work(w) = -194KJ
change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.
At constant pressure the change in enthalpy is given as
ΔH = ΔU + PΔV
ΔU = change in energy
ΔV = change in volume
P = pressure
w = -pΔV
therefore,
ΔH = ΔU -W
to evaluate energy change we have,
ΔU =ΔH + W
ΔU = -162+ (-194KJ)
= -356KJ
<em>therefore, the reaction where heat is released is exothermic reaction since theΔH is negative</em>
Answer:
30.0 mol CO₂
Explanation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
To answer this problem we need to convert moles of C₃H₈ into moles of CO₂: We'll do that by using the <u>stoichiometric coefficients</u>, using a conversion factor that has C₃H₈ moles in the denominator and CO₂ moles in the numerator:
10.0 mol C₃H₈ * 
= 30.0 mol CO₂
Answer:
4L
Explanation:
To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.
From the question given above,
Mass of O2 = 5.72g
Molar Mass of O2 = 32g/mol
Number of mole =Mass/Molar Mass
Number of mole of O2 = 5.72/32
Number of mole of O2 = 0.179 mole
Now, we can calculate the volume of O2 at stp as follow:
1 mole of a gas occupy 22.4L at stp.
Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L
Therefore, the volume occupied by the sample of O2 is 4L
