Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
Answer:
Mole fraction of C₄H₄S = 0.55
Explanation:
Mole fraction is moles of solute / Total moles
Total moles are the sum of moles of solute + moles of solvent.
Let's find out the moles of our solute and our solvent.
Mass of solute: 55g
Mass of solvent: 65g
Mol = Mass / molar mass
55 g / 84.06 g/mol = 0.654 moles of C₄H₄S
65 g /123 g/mol = 0.529 moles of C₂H₃BrO
Total moles = 0.654 + 0.529 = 1.183 moles
Mole fraction of thiophene = Moles of tiophene / Total moles
0.654 / 1.183 = 0.55
Answer:
V₂ = 5.97 L
Explanation:
Given data:
Initial temperature = 9°C (9+273 = 282 K)
Initial volume of gas = 6.17 L
Final volume of gas = ?
Final temperature = standard = 273 K
Solution:
Formula:
The Charles Law will be apply to solve the given problem.
According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 6.17 L × 273K / 282 k
V₂ = 1684.41 L.K / 282 K
V₂ = 5.97 L
Answer:
350 g dye
0.705 mol
2.9 × 10⁴ L
Explanation:
The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:
70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye
The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:
350 g × (1 mol / 496.42 g) = 0.705 mol
The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:
3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L
Answer:
The equation is balanced
Explanation:
NaCl (aq) + AgNO3(aq) ––> AgCl (s) + NaNO3 (aq)
NaCl (aq) + AgNO3 (aq)
Na = 1 , Cl=1 , Ag = 1 , No3= 1
AgCl (s) + NaNO3 (aq)
Ag = 1 , Cl=1 , Na = 1 , No3= 1
