PV=nRT
n=2.4 moles
T=273.15+50=323.15K
P=2*101325=202650 Pa
R=8.31
Solve for V:
V=nRT/P=2.4*8.31*323.15/202650=.032m^3
Answer:
A) Mass = 32 g of KCl
Explanation:
Given data:
Mass of potassium chloride produced = ?
Mass of potassium chlorate = 52 g
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Number of moles of KClO₃:
Number of moles = mass/molar mass
Number of moles = 52 g/ 122.55 g/mol
Number of moles = 0.424 mol
Now we will compare the moles of KClO₃ and KCl
KClO₃ : KCl
2 : 2
0.424 : 0.424
Mass of KCl:
Mass = number of moles × molar mass
Mass = 0.424 mol × 74.55 g/mol
Mass = 32 g
Answer:
I would describe the jumping pattern of the green frog bellow as triangular or random it depend on you P.O.V.
Answer: The concentration of KOH for the final solution is 0.275 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
n = moles of solute
= volume of solution in ml = 150 ml
moles of solute =
Now put all the given values in the formula of molality, we get
According to the dilution law,
where,
= molarity of stock solution = 1.19 M
= volume of stock solution = 15.0 ml
= molarity of diluted solution = ?
= volume of diluted solution = 65.0 ml
Putting in the values we get:
Therefore, the concentration of KOH for the final solution is 0.275 M
mole=10 x 10⁻³ : 46 g/mol = 2.17 x 10⁻⁴
