Answer:
option c is correct
Explanation:
the addition of catalyst does not effect the position or equilibrium constant and increase the forward and backward reaction in equal rates so no effect would be observed
Answer:
5 moles of electrons
Explanation:
The balance equation is as follow,
<span> 5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O →<span> 5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺
Reduction of Ag:
Ag⁺ + 1 e⁻ → Ag
Or,
5 Ag⁺ + 5 e⁻ → 5 Ag
Oxidation of Mn:
Mn⁺² → MnO₄⁻ + 5 e⁻
Result:
Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
When an electron transit from lower level to higher level it emits energy
When an electron transit from higher level from a lower level it absorbs energy
n=3 to n=5, : absorption
n=1 to n=3, : absorption
n=3 to n=2, : emission
n=2 to n=1 : emission
the highest energy means highest wave number
wave number = Rh (1/ n1^2 - 1/n2^2)
Rh = 1.1 X 10^7 m^-1
1) n=3 to n=5
wavenumber = Rh ( 1/ 9 - 1/ 25) = 0.071 Rh
2) n=1 to n=3
wavenumber = Rh ( 1- 1/9 ) = 0.89 Rh
3) n=3 to n=2
wavenumber = Rh ( 1/4 - 1/9) = 0.139 Rh
4) n=2 to n=1
wavenumber = Rh ( 1- 1/4) = 0.75 Rh
Thus for 2) n=1 to n=3 energy will be maximum
