Answer:
Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:
Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:
By plugging in the moles and molarity, we obtain:
Which in mL is:
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Answer:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.
55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g
To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:
Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5
The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.
HA ⇄ H⁺ + A⁻
so:
![\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D%201.5%20x%2010%5E%7B-5%7D%20%20)
and now:
= 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83
