Explanation:
34.2g of C12H22O11 is dissolved in 180g of H20.
Molar mass of sucrose = 342g/mol
Moles of sucrose = 342 / 34.2 = 10 mol.
Molar mass of water = 18g/mol
Moles of water = 180 / 18 = 10 mol.
For hydrogen atoms, there are 22 * 10 in sucrose and 2 * 10 in water, which gives a total of 240.
For oxygen atoms, there are 11 * 10 in sucrose and 1 * 10 in water, which gives a total of 120.
Answer:
pH of resulting solution = 7.98
Explanation:
The balanced equation
HA + NaOH - Na+ + A- + H2O
Number of moles of A = Number of moles of HA = Number of moles of NaOH
= 35.8/1000 * 0.020 = 0.000716 mol
Initial concentration of A = 0.000716/0.0608 = 0.01178 M
pKb = 14 – pKa = 14 -3.9 = 10.1
Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11
Kb = [HA][OH-]/[A-]
Kb = a^2/(0.01178 -a) = 7.943 * 10^-11
a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0
a = 9.673 * 10^-7
OH- = a = 9.673 * 10^-7 M
pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02
pH = 14-6.02 = 7.98
Answer:
Heat of solution= - 63KJ/mole
Heat evolved = - 18.82KJ
Explanation:Heat of solution = Heat of solute + Heat of hydration
◇Hsoln = - ( - 7.3 × 10 ^ 2) + ( - 793 ) KJ / mol
◇H soln= - 63 KJ
Use enthalpy of solution equation to determine the enthalpy of solution for LiI
LiI ---> Li^ + + I^-
Molecular mass of LiI=133.85g
The chemical equation for the dissociation will be:
40gLiI (1molLiI)/133.85g LiI - 63KJ/1 mol LiI
18.82KJ
The heat was evolved because the value of enthalpy was negative
Answer:
Velocity = 4.41×10⁵ m/s
Explanation:
The force of attraction between the nucleus and the electron is equal to the centrifugal force acting on the outermost electron due to circular rotation. It can be mathematically written as:

Where,
K is the Coulomb's constant having value 9×10⁹ N. m²/C²
q is the charge on the electron or the proton which is 1.6×10⁻¹⁹ C
r is the radius of the atom
m is the mass of the electron having value 9.1×10⁻³¹ Kg
v is the velocity of the electron.
The equation can be written to calculate the velocity as:

Given that:
r = 1.3×10⁻⁹ m
So,

<u>Velocity = 4.41×10⁵ m/s</u>