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Blizzard [7]
2 years ago
13

How many grams of Helium can be found in 3.4 moles of it.

Chemistry
1 answer:
MaRussiya [10]2 years ago
5 0

one mole

If we look this number up for helium, we find that helium has a molar mass of 4.0 grams per mole. So one mole of helium has a mass of 4.0 grams. We can use the molar mass as a conversion factor. Two moles of helium times four grams of helium per mole of helium is equal to eight grams of helium.

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B density

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A sample of the alloy “electrum” is 62% silver and 38% gold by mass; mass of part/
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c

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2 years ago
Calculate the concentration of hydronium and hydroxide ions in a 0.050 M solution of nitric acid.
Mazyrski [523]

Answer:

[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

  • HNO₃ is completely ionized in water as:

<em>HNO₃ + H₂O → H₃O⁺ + NO₃⁻.</em>

  • The concentration of hydronium ion is equal to the concentration of HNO₃:

[H₃O⁺] = 0.05 M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/0.05 = <em>2.0 x 10⁻¹³.</em>

3 0
2 years ago
Explain how energy causes changes to earths surface
dlinn [17]

Answer:

Explanation:

The absorbed sunlight drives photosynthesis, fuels evaporation, melts snow and ice, and warms the Earth system. Solar power drives Earth's climate. Energy from the Sun heats the surface, warms the atmosphere, and powers the ocean currents.

7 0
2 years ago
Balance the equation :
Sati [7]

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

<u>Explanation:</u>

  • Aluminum has a typical oxidation condition of  3+  , and that of iodine is  1-  .   Along these lines, three iodides can bond with one aluminum. You get  AlI3.  For comparable reasons, aluminum chloride is  AlCl3.  
  • Chlorine and iodine both exist normally as diatomic components, so they are  Cl2(  g  )  also,  I2(  g  ), individually. In spite of the fact that I would anticipate that iodine should be a strong.  

Balancing the equation, we get:  

            2AlI 3(  aq  )  +  3Cl2 (  g  )  →  2AlCl3 (  aq  )   +  3 I 2  (  g  )  

  • Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the  AlCl  3  on the right.  
  • Normally, presently we have two  Al  on the right, so I multiplied the  AlI  3  on the left. Hence, I have 6  I  on the left, and I needed to significantly increase  I 2  on the right.  
  • We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
3 0
3 years ago
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