CHEMISTRY HIGH SCHOOL

A student adds 0.40g of sodium hydroxide to a clean, dry 250 mL beaker and leaves the beaker and leaves the beaker on the lab table. The mass of the empty beaker is 112.58g. After 24 hours, the student observes that the beaker contains a dry, solid white residue. The total mass of the beaker and its contents is 113.09g. Following are three possible chemical reactions that could have occured: Sodium Hydroxide (solid) + Carbon Dioxide (gas) rightarrow Sodium Bicarbonate (solid) Sodium Hydroxide (solid) + Carbon Dioxide (gas) rightarrow Sodium Carbonate (solid) + Water (gas) Sodium Oxide (solid) rightarrow Sodium Oxide (solid) + Water (gas) Use your knowledge of stoichiometry to determine which chemical reaction occurred. Show your work, and explain your reasoning whenever needed.

Answers

Answer 1
Answer:

2 NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

Explanation:

The most probably chemical reaction is:

2 NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

Sodium Hydroxide (solid) + Carbon Dioxide (gas) → Sodium Carbonate (solid) + Water (gas)

In this reaction the water you obtain have to be in liquid form because the reaction takes place at room temperature. However because the student left the beaker for 24 hours, the water have a plenty of time to evaporate, so in the end only sodium carbonate (Na₂CO₃) remained in the beaker.

number of moles = mass / molar weight

number of moles of NaOH = 0.4  / 40 = 0.01 moles

Taking in account the chemical reaction, we devise the following reasoning:

if          2 moles of NaOH produce 1 mole of Na₂CO₃

then    0.01 moles of NaOH produce X mole of Na₂CO₃

X = (0.01 × 1) / 2 = 0.005 moles of Na₂CO₃

mass = number of moles × molar weight

mass of Na₂CO₃ = 0.005 × 106 = 0.53 g

mass of Na₂CO₃ = mass of beaker and its contents after 24h - mass of empty beaker

mass of Na₂CO₃ = 113.09 - 112.58 = 0.51 g which is close the theoretical value of 0.53 g.

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number of moles

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Answers

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Answers

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Answers

Answer:

1)conduction

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4) radiation

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When a nucleus of ^235 U undergoes fission, it breaks into two smaller, more tightly bound fragments.1) Calculate the binding energy per nucleon for ^235U? (Express with appropriate units)2) Calculate the binding energy per nucleon for the fission product ^137 Cs? (Express in appropriate units)

Answers

1) The binding energy per nucleon is 7.59 MeV/nucleon and 2) The binding energy per nucleon is  8.39 MeV/nucleon.

1) The binding energy per nucleon is as follows:

m_proton + m_neutron + m_electron - m_U

= 92*1.007276466812 + 92*5.4857990943e-4 + 143*1.008664916 - (235.0439299)

= 1.9151 u

= 931.494061 * 1.9151 MeV

= 1784 MeV

= 1784/235 = 7.59 MeV/nucleon

2) The binding energy per nucleon is as follows:

m_proton + m_neutron + m_electron - m_Cs

= 55*1.007276466812 + 55*5.4857990943e-4 + 82*1.008664916 - (136.9070895)

= 1.2338 u

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Answers

The correct option for above statement is:

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Explanation:

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The scientist who provided a foundation for Dalton's work on the atomic structure is Ernest Rutherford

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Answers

Of course , B. 4,000 g
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