Answer:
Explanation:
We are given that 25 mL of 0.10 M 
is titrated with 0.10 M NaOH(aq).
We have to find the pH of solution
Volume of 
Volume of NaoH=0.01 L
Volume of solution =25 +10=35 mL=
Because 1 L=1000 mL
Molarity of NaOH=Concentration OH-=0.10M
Concentration of H+= Molarity of 
=0.10 M
Number of moles of H+=Molarity multiply by volume of given acid
Number of moles of H+=
=0.0025 moles
Number of moles of 
=0.001mole
Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles
Concentration of H+=
pH=-log [H+]=-log [4.28
]=-log4.28+2 log 10=-0.631+2
Answer:
<u>84.00 kPa = 630.084 torr</u>
Formula for kPa to torr: For an approximate result, multiply the pressure value by 7.501. <em><u>84* 7.501</u></em>
<u>84.00 kPa = 0.831683168 atm</u>
Formula for kPa to atm: for an approximate result, divide the pressure value by 101. <em><u>84/101</u></em>
Answer: a . 152g/mol b. 102g/mol c. 183g/mol
Explanation:
By stating the atomic masses of each element in the questions, we have;
Fe= 56, S= 32, O= 16, Al = 27, C = 12, H =1 , N = 14, therefore
(a). FeSO4 = 56 + 32 + (16 x 4) = 152g/mol
(b). Al2O3 = (27 x 2) + (16 x 3) = 102g/mol
(c). C7H5NO3S ( Saccharin, an artificial Sweetner) =
(12 x 7) + (1 x 5) + 14 + (16 x 3) + 32 = 183g/mol
Answer:
the reaction is spontaneous at T > 2900 K
Explanation:
∴ ΔH = +120 KJ
∴ ΔS = (-42 J/K)*(KJ/1000 J) = -0.042 KJ/K
∴ ΔG < 0 ⇒ the reaction is spontaneous
⇒ at T = 2900 K:
⇒ ΔG = 120 - (2900)(-0.042) = 120 - 121.8 = - 1.8 KJ < 0
