To counter the removal of A the equilibrium change by <u>s</u><em>hifting toward the left</em>
<em> </em><u><em>explanation</em></u>
<u><em> </em></u>If the reaction is at equilibrium and we alter the condition a new equilibrium state is created
<u><em> </em></u>The removal of A led to the shift of equilibrium toward the left since it led to less molecules in reactant side which favor the backward reaction.( equilibrium shift to the left)
From the question you will find that:
one capsule of tamiflu is obtained from 2.6 g of star anise.
1 capsule = 2.6 g tamiflu
? capsules = 155 g tamiflu
by cross multiplication =
= 59 capsules
Answer:
Removal of Third Electron
Explanation:
a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.
Ionization energy increases from bottom to top within a group, and increases from left to right within a period.
Answer:
Mass of liquid B = 271.2 gram
Explanation:
Given:
Density of liquid A = 1000 kg/m³
Density of liquid B = 600 kg/m³
Density of mixture = 850 kg/m³
Mass of mixture = 1 kg
Assume:
Volume of liquid A = Va
Volume of liquid B = Vb
So,
Volume of mixture = Va + Vb
Mass of liquid A = 1000(Va)
Mass of liquid B = 600(Vb)
Mass of mixture = Mass of liquid A + Mass of liquid B
1 = 1000(Va) + 600(Vb)
Volume of mixture = 1 / 850
So,
(1/850) = Va + Vb
Vb = (1/850) - Va
1 = 1000(Va) + 600[(1/850) - Va]
Va = 7.25 × 10⁻⁴
Vb = (1/850) - Va
Vb = (1/850) - [7.25 × 10⁻⁴]
Vb = 4.25 × 10⁻⁴
Mass of liquid B = 600(Vb)
Mass of liquid B = 600(4.25 × 10⁻⁴)
Mass of liquid B = 271.2 gram
Answer:
0.0222 mole of NaOH is needed to react with NH4F
Explanation:
NH4F + NaOH --> NaF + NH3 + H2O
Data given
Mass of NH4F =0.821g, Concentration of NaOH= 1M, volume of NaoH =25ml
But mole = (CV)/1000
given mole of NaoH = (1 * 25)/1000 = 0.025moles of NaOH used
Molar mass of NH4F = 37g/mol
mole of NH4F used 0.821 / 37 = 0.0222 mole NH4F
Determine the excess and limiting reactant,
NaOH is in excess
0.025 - 0.0222 = 0.0028 mole NaOH excess
0.0222 mole of NaOH is required to react with NH4F
