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AysviL [449]
2 years ago
7

Exposure to what type of radiant energy is sensed by human skin as warmth?

Chemistry
2 answers:
larisa86 [58]2 years ago
3 0
The answer is C.gamma rays
Hatshy [7]2 years ago
3 0
I think it is gamma rays as well so id go with c.

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It’s probably probably 01/16 because of the other number don’t make sense except that one
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WILL GIVE BRAINLIEST
Sav [38]

Answer:

the answer is A

Explanation:

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3 years ago
2. An element is all the way through,​
elena55 [62]
?? Is that the whole question?
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2 years ago
A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
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.
3 0
3 years ago
A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 295 mL of solution. How many moles o
Masja [62]

Answer:

(a) Moles of ammonium chloride = 0.243 moles

(b) Molarity_{ammonium\ chloride}=0.824\ M

(c) 60.68 mL

Explanation:

(a) Mass of ammonium chloride = 13.0 g

Molar mass of ammonium chloride = 53.491 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{13.0\ g}{53.491\ g/mol}

<u>Moles of ammonium chloride = 0.243 moles</u>

(b) Moles of ammonium chloride = 0.243 moles

Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{ammonium\ chloride}=\frac{0.243}{0.295}

Molarity_{ammonium\ chloride}=0.824\ M

(c) Moles of ammonium chloride = 0.0500 moles

Volume = ?

Molarity = 0.824 M

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

0.824\ M=\frac{0.0500}{Volume}

<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>

6 0
3 years ago
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