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nataly862011 [7]
3 years ago
8

Which answer below correctly identifies the type of change and the explanation for the boiling of water?

Chemistry
1 answer:
Trava [24]3 years ago
6 0

Answer:

The second choice: physical change because even though gas formation was observed, the water was undergoing a state change, which means that its original properties are preserved

(WAS ALREADY ANSWERED PREVIOUSLY)

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SHOW WORK
stealth61 [152]
PV = nRT —> n = PV/RT

P = 2.90 atm
V = 4.80 L
R = 0.08206 L atm / mol K
T = 62.0 + 273 = 335 K (make sure you convert from celsius to kelvin)

n = (2.90 • 4.80) / (0.08206 • 335) = 0.506 moles of gas
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Which group would Ns2np2 go into
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The group 15 elements: the pnicogens

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The group 15 elements, nitrogen, phosphorus, arsenic, antimony and bismuth, all have the general valence shell electronic configuration ns2np3. They can all exist in the +3 or +5 oxidation state, with the +3 state increasing in stability as we move vertically down the group.

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For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
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