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
Answer: The reactivity of acetic acid with various chemicals (A)
Explanation: engenuity 2021
Answer:
506.912 L
Explanation:
From the question given above, the following data were obtained:
Number of mole of O₂ = 22.63 moles
Volume of O₂ =?
Recall:
1 mole of a gas occupy 22.4 L at STP.
With the above information, we obtained the volume occupied by 22.63 moles of O₂ as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 22.63 moles of O₂ will occupy = 22.63 × 22.4 = 506.912 L at STP.
Thus, 22.63 moles of O₂ is equivalent to 506.912 L.
Answer:
A. Coal
Explanation:
= water displaced by rock
Water displaced by rock = volume of water after rock is dropped into the cylinder - volume of water before the rock was dropped into the water
Water displaced by the rock = 180 ml - 150 ml = 30 ml
Density of rock:
40 grams => 30 ml
x grams => 1 ml
Cross multiply
1*40 = 30*x
40 = 30x
40/30 = 30x/30
1.3 = x
Density of rock = 1.3 g per 1 ml
Recall: 1 ml = 1 cm³
Therefore,
Density of the rock = 1.3 g/cm³
1.3 g/cm³ falls within the range of 1.1 - 1.4 g/cm³
Therefore, the rock is identified as Coal.
