Answer:
2 atoms of nitrogen are present.
Answer:
74.4 ml
Explanation:
C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)
Given 15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate
From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.
Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.
The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution
=> Volume (Liters) = moles citric acid / Molarity of citric acid solution
=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml
Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;
Moles = n = 1.20 mol
Volume = V = 4 L
Temperature = T = 30 + 273 = 303 K
Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹
Putting Values in Eq.1,
P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L
P = 7.45 atm
Answer:
It is a eukaryotic organism
It belongs to kingdom protista
It does not have membrane bounded organelles
Explanation:
