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zmey [24]
2 years ago
15

g 8. A hydrogen atom is initially a rest and in the ground state. A proton with kinetic energy 1000 eV scatters off the atom. In

the process the hydrogen atom is ionized. The kinetic energy of the ionized electron is measured to be 15.2 eV and the kinetic energy of the proton produced from the ionization of hydrogen is 4.3 eV. What would be the final kinetic energy of the projectile proton after the collision
Chemistry
1 answer:
aleksley [76]2 years ago
8 0

Answer:

Explanation:

Ionization energy of hydrogen atom is 13.6 eV . This energy will be provided by energetic proton , the kinetic energy of which is 1000 eV.  The kinetic energy of ionized electron is 15.2 eV . Kinetic energy of proton produced from from the ionization of hydrogen or the nucleus of the hydrogen atom is 4.3 eV . All these energy must have come from kinetic energy of initial proton.

So kinetic energy of projectile proton after collision

= 1000 - ( 13.6 + 15.2 + 4.3 ) eV.

= 966.9 eV .

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13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
6 0
2 years ago
Use this equation for the following problems: 2NaN3 --&gt; 2Na+3N2
olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

<em>2NaN₃ → 2Na + 3N₂,</em>

  • It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

<em>Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?​ </em>

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

  • Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

  • We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

  • We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

<em><u>using cross multiplication:</u></em>

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

  • Finally, we can get the grams of NaN₃ needed:

<em>mass = no. of moles x molar mass</em> = (0.517 mol)(65.0 g/mol) =<em> 33.6 g.</em>

<em />

<em>Q2: How many mL of N₂ result if 8.3 g Na are also produced?</em>

  • We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

  • We can get the no. of moles of N₂ produced with 0.36 mol of Na:

<em><u>using cross multiplication:</u></em>

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

  • We can get the mass of 0.54 mol of N₂:

mass = no. of moles  x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

  • Now, we can get the mL of 15.12 g of N₂:

<em><u>using cross multiplication:</u></em>

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

<em>∴ The volume of N₂ result </em>= (1.0 L)(15.12 g)/(0.92 g) = <em>16.434 L = 16434 mL.</em>

4 0
2 years ago
Energy in a chemical reaction can be identified as which of the following?
daser333 [38]
I believe it’s A..but I’m not quite sure.
6 0
2 years ago
Thea is doing a chemistry experiment. The instructions say she needs to use pure water. Hamza offer her a bottle labelled “100%
Diano4ka-milaya [45]
How is this going to help us in life wat the fawk
5 0
2 years ago
Chemistry help!!
Scrat [10]

The student originally has 252 grams of water in this experiment.

LAW OF CONSERVATION OF MASS:

  • The law of conservation of mass explains that matter (mass) can neither be created nor destroyed but can be changed from one form to another.

  • This means that in a chemical reaction, the sum of the masses of the reactants must equate to the total mass of product(s).

  • According to this question, a student conducts an experiment to separate water into hydrogen and oxygen. The student collects 28.0 g of hydrogen and 224.0 g of oxygen.

  • Since hydrogen and oxygen are the constituent elements of water, the sum of their masses must equate the mass of water.

  • Therefore, 224g of oxygen + 28g of hydrogen = 252g of water.

Learn more at: brainly.com/question/24996173?referrer=searchResults

8 0
2 years ago
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