Answer:
6.88 mg
Explanation:
Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄
The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.
175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P
Step 2: Calculate the rate constant for the decay of ³²P
The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.
k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹
Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days
For first-order kinetics, we will use the following expression.
ln P = ln P₀ - k × t
ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d
P = 6.88 mg
0.300 M IKI represents the
concentration which is in molarity of a potassium iodide solution. This means
that for every liter of solution there are 0.300 moles of potassium iodide. Knowing
that molarity is a ratio of solute to solution.
By using a conversion factor:
100 ml x (1L / 1000 mL) x (0.300
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g
Therefore, in the first
conversion by simply converting the unit of volume to liter, Molarity is in L
where the volume is in liters. The next step is converted in moles from volume
by using molarity as a conversion factor which is similar to how density can be
used to convert between volume and mass. After converting to moles it is simply
used as molar mass of Kl which is obtained from periodic table to convert from
mole to grams.
In order to get the grams of IKI
to create a 100 mL solution of 0.600 M IKI, use the same formula as above:
100 ml x (1L / 1000 mL) x (0.600
mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g
Molarity is calculated by using following formula,
Molarity = Moles / Volume
Data Given:
Moles = 23 moles
Volume = 100 ml ÷ 1000 = 0.1 L
Putting values in eq. 1,
Molarity = 23 mol / 0.1 L
Molarity = 230 mol/dm³
Result:
When 23 mol of solute is dissolved in a solvent to make a solution of 100 ml, then it will have a Molarity of 230 mol/dm³.
Answer:
Please find the complete question in the attached file.
Explanation:
Balanced equation:
Ionic equation:
Net Ionic equation:
