Answer:
V = 5.17L
Explanation:
Mass of gas = 8.7g
T = 23°C = (23 + 273.15)K = 296.15K
P = 1.15 atm
V = ?
R = 0.082atm.L / mol.K
From ideal gas equation
PV = nRT
P = pressure of the gas
V = volume of the gas
n = no. Of moles
R = ideal gas constant
T = temperature of the gas
no of moles = mass / molar mass
Molar mass of Chlorine = 35.5g / mol
No. Of moles = 8.7 / 35.5
No. Of moles = 0.245 moles
PV = nRT
V = nRT / P
V = (0.245 * 0.082 * 296.15) / 1.15
V = 5.9496 / 1.15
V = 5.17L
The volume of the gas is 5.17L
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
-google
its probably 40
