<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
Answer:
Explanation:
Data:
Mass of NaCl = 4.6 g
Mass of water = 250 g
Calculations:
Mass of solution = mass of NaCl + mass of water = 4.6 g + 250 g = 254.6 g.
Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
The decay mode of cesium-137 is beta decay. This means that the cesium-137 decays into a beta particle and a nuclide with the same mass number, but with a charge number that is 1 more than that of cesium.
Therefore, this means Cs-137 decays into an electron and Barium-137, meaning the answer is choice 1.
Answer:
[Kr] 4d10 5s2 5p4
Explanation:
The Symbol I represents Iodine. It has atomic number of 53. The full electronic configuration is given as;
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p5
However the question requested for the configuration of I+.
I+ is a cation and it simply refers to an iodine atom that has lost a single electron. The electronic configuration of I+ is given as;
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4
Using Noble gas shorthand representation, we have;
[Kr] 4d10 5s2 5p4
