Answer:
9.9 ml of 0.200M NH₄OH(aq)
Explanation:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
?ml of 0.200M NH₄OH(aq) reacts completely with 12ml of 0.550M FeCl₃(aq)
1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution
1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)
=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters
The bohr model is arranged so there is a nucleus in the middle that holds the neutrons and protons. Around the nucleus, there is shells or energy levels that hold electrons in a formation of 2,8,8.
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Answer:
3.6 × 10²⁴ molecules
Explanation:
Step 1: Given data
Moles of methane (n): 6.0 moles
Step 2: Calculate the number of molecules of methane in 6.0 moles of methane
In order to convert moles to molecules, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ molecules of methane in 1 mole of molecules of methane.
6.0 mol × 6.02 × 10²³ molecules/1 mol = 3.6 × 10²⁴ molecules
Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate