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Setler79 [48]
1 year ago
12

A certain radioactive isotope has a half life of one hour. I you start with 1.0 g of the material at 1:00 P. M., how many grams

will be present at 2:00 P. M. on the same day
Chemistry
1 answer:
geniusboy [140]1 year ago
6 0

A certain radioactive isotope has a half life of one hour. I you start with 1.0 g of the material at 1:00 P. M. 0.5 grams will be present at 2:00 P.M. on the same day.

<h3>What is half Life ?</h3>

Half life is the amount of time required to reduce to one half of its initial value. The symbol of half life is t_{1/2}.

<h3 /><h3>How to calculate the quantity remaining when half life given ?</h3>

It is expressed as:

N_{t} = N_{0} (\frac{1}{2})^{n}

where,

N(t) = quantity remaining

N₀ = initial quantity

n = half life that have passed

Now put the values in above expression we get

N_{t} = 1.0\ g \times (\frac{1}{2})^{1}

     = \frac{1}{2} g

      = 0.5 gram

Thus from the above conclusion we can say that A certain radioactive isotope has a half life of one hour. I you start with 1.0 g of the material at 1:00 P. M. 0.5 grams will be present at 2:00 P.M. on the same day.

Learn more about the Half life here: brainly.com/question/25750315

#SPJ4

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omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
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3 years ago
Members of which group easily lose an electron to form a +1 cation?
vaieri [72.5K]
The alkali metals, which occupy group 1 of the periodic table. This is because the valence shells of these elements have only 1 electron, so easily form an ionic bond with a non-metal compound by donating this. A cation is formed by this donation, since there is one fewer electron orbiting the nucleus than there is in the atomic form - conversely an anion is formed when an atom gains an extra electron to become negatively charged.
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How to answer this ?
tamaranim1 [39]

Answer:

as follow

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2

a acceptor

b Donor electron

3

A. MgBr2 B. Na2O

C. HCl D. AlCl3

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DIMENSIONAL ANALYSIS PLSSSS<br>convert 19 inches to feet(show work)
Anna [14]

19 in = 1.6 ft

You know that

12 in = 1 ft

If you divide both sides by 12 in, you get the <em>conversion factor</em>: 1 = 1 ft/12 in.

If you divide both sides by 1 ft, you get the <em>conversion factor</em>: 12 in/1 ft = 1.

Both are conversion factors because they <em>both equal one</em> and <em>multiplying a measurement by one does not change its value</em>.

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Thus, to convert 19 in to ft, you use the conversion factor with “ft” on the top.

∴ Length = 19 in × (1 ft/12 in) = 1.6 ft

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