A chemical element that has an atomic number less than 58 and an atomic mass greater than 135.6m is barium (atomic no. 56 and atomic mass137.13 ) and lanthanum (atomic no. 57 and atomic mass 135.6).
<h3>Give a brief introduction about Barium and Lanthanum.</h3>
Barium is an element with the symbol Ba and atomic number 56. It is an alkaline earth metal that is soft and silvery, and it is the fifth element in group 2. Barium is never found in nature as a free element due to its extreme chemical reactivity. Oil well drilling fluid uses barium sulfate as an insoluble ingredient. It is employed as an X-ray radiocontrast agent in a purer form to image the human gastrointestinal tract. Barium compounds that dissolve in water have been employed as rodenticides despite being hazardous.
Chemical element lanthanum has the atomic number 57 and the symbol La. It is a silvery-white, ductile, soft metal that slowly tarnishes when exposed to air. It serves as the eponym for the group of 15 related elements in the periodic table between lanthanum and lutetium, of which lanthanum is the first and prototype. The rare earth elements traditionally include lanthanum.
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Answer:
6 Cl0 + 6 e- → 6 Cl-I (reduction)
2 Fe0 - 6 e- → 2 FeIII (oxidation)
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
Thus, the mass of sodium bromide added should be, 18.3 grams.
Answer:
16.6 mg
Explanation:
Step 1: Calculate the rate constant (k) for Iodine-131 decay
We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.
k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹
Step 2: Calculate the mass of iodine after 8.52 days
Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.
ln I = ln I₀ - k × t
ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day
I = 16.6 mg
Answer:
0.00369 moles of HCl react with carbonate.
Explanation:
Number of moles of HCl present initially = 
moles = 0.00600 moles
Neutralization reaction (back titration): 
According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.
So, excess number of moles of HCl present = number of NaOH added for back titration = 
moles = 0.00231 moles
So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles
Hence, 0.00369 moles of HCl react with carbonate.
