The RDA for iodine in adults is 150 micrograms
To meet this, one must consume:
150 / 0.7645
= 196.2 micrograms of potassium iodide per day.
Answer:
-66.88KJ/mol
Explanation:
It is possible to obtain the heat involved in a reaction using a calorimeter. Formula is:
q = -C×m×ΔT
<em>Where q is heat of reaction, C is specific heat capacity (4.18J/°Cg), m is mass of solution (100.0g) and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)</em>
Replacing:
q = -4.18J/°Cg×100.0g×0.80°C
q = -334.4J
Now, in the reaction:
Ag⁺ + Cl⁻→ AgCl
<em>AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻</em>
Moles that react are:
0.050L× (0.100mol /L) = 0.0050moles
If 0.0050 moles produce -334.4J. Heat of reaction is:
-334.4J / 0.0050moles = -66880J/mol = <em>-66.88KJ/mol</em>
Answer:
The concentration of chloride ions in the final solution is 3 M.
Explanation:
The number of moles present in a solution can be calculated as follows:
number of moles = concentration in molarity * volume
In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻
For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:
number of moles of Cl⁻ = 2 * number of moles of CaCl₂
number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻
The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.
Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:
Concentration = number of moles of Cl⁻ / volume
Concentration = 150.2 mol / 50.1 l = 3.0 M
A, I believe it is 810liters of oxygen! Please correct me if im wrong
