Answer:
150 g/mol
Explanation:
Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.
H₂X + 2 NaOH → Na₂X + 2 H₂O
40.0 mL of 0.200 M NaOH. were required to reach the endpoint. The reacting moles of NaOH are:
0.0400 L × 0.200 mol/L = 8.00 × 10⁻³ mol
The molar ratio of H₂X to NaOH is 1:2. The reacting moles of H₂X are 1/2 × 8.00 × 10⁻³ mol = 4.00 × 10⁻³ mol.
4.00 × 10⁻³ moles of H₂X have a mass of 0.600 g. The molar mass of H₂X is:
0.600 g/4.00 × 10⁻³ mol = 150 g/mol
The answer to this is a solution
Since sodium chloride contains both a metal AND a nonmetal, the combination of those would result in an ionic bond.
For the reaction 2 K + F2 --> 2 KF,
consider K atomic wt. = 39
23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)
Now, following the ideal gas equation, PV = nRT
P = 0.98 atm
V = unknown
n = 0.3015 moles
R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)
T = 298 K
Solving for V,
V = (nRT)/P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K)/(0.98 atm)
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solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed. </span>
There are 337.23 × 10²³ atoms in 4 moles of aluminum sulfite Al₂(SO₃)₃.
Explanation:
The questions ask how many atoms are in 4 moles of aluminum sulfite Al₂(SO₃)₃?
To answer this we use the Avogadro's number to devise the following reasoning:
if in 1 mole of Al₂(SO₃)₃ there are 14 × 6.022 × 10²³ atoms
then in 4 moles of Al₂(SO₃)₃ there are X atoms
X = (4 × 14 × 6.022 × 10²³) / 1 = 337.23 × 10²³ atoms
Learn more about:
Avogadro's number
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