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dezoksy [38]
3 years ago
12

What properties prompted Rutherford to use alpha particles and gold? (What set of experimental conditions would make it easiest

to detect differences in atomic structure?)
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0
<span>Gold is  a stable heavy metal that is easily formed into uniform and thin sheets of measurable property. Alpha particles are relatively massive (compared to electrons) and so deflection is less dramatic and more easily measured. This is the property that prompted Rutherford to use alpha particles and gold. </span>
The particles that are positively charged were fired at the gold foil and when it reached that goldatoms’ nuclei, which are also positively charged, the particles were repelled back
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How do you determine the chemical reactivity for metals by using the periodic table?
vesna_86 [32]

Answer:

Chemical reactivity increases down a group and decreases from left to right of a period.

Explanation:

The higher the ionization energy is, the lower the reactivity is. Since the ionization energy is highest in the top right corner of the periodic table, we can assume that the most reactive elements are in the opposite bottom left corner. This is because the electrons that react are farther away from the nucleus thus experience less attraction to the nucleus (called nuclear shielding). Therefore their electrons are more easily removed than elements that don't ecperience nuclear shielding.

4 0
3 years ago
The chemical formula for diiodine nonaflouride
stira [4]

Answer:

The chemical formula is I2F9

Explanation:

3 0
3 years ago
1) You are asked to make 10 miles of iron (Fe) from iron oxide (Fe2O3) and excess carbon monoxide (CO). Fe2O3(s) + 3CO(g)—&gt; 2
Inessa05 [86]

For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.

1) D 5 mols

2) A 0.55 mols

8 0
3 years ago
if you react 28 grams of butene with excess hydrogen how many grams of butane would you expect to get
liberstina [14]

Answer:

The correct answer is 29 grams.

Explanation:

Based on the given question, the reaction will be,

CH3CH=CHCH3 + H2 ⇒ CH3CH2CH2CH3

The molecular weight of butene is 56 grams per mole, and the molecular weight of butane is 58 grams per mole.

Thus, it can be said that 56 grams of butene reacts with hydrogen gas and produces 58 grams of butane.

Therefore, 28 grams of butene when reacts with hydrogen gas to give,

= 58/56 * 28 = 29 grams of butane.

Hence, the mass of butane produced will be 29 grams.

8 0
2 years ago
Given 4.80g of ammonium carbonate, find:
V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

<em>1) Number of moles of the compound:</em>

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

<em>2) Number of moles of ammonium ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

<em>∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ </em>= (2.0)(0.05 mol) =  <em>0.1 mol.</em>

<em>3) Number of moles of carbonate ions :</em>

  • Ammonium carbonate is dissociated according to the balanced equation:

<em>(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.</em>

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

<em>4) Number of moles of hydrogen atoms:</em>

  • Every 1.0 mol of (NH₄)₂CO₃  contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

<em>∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO</em>₃ = (8.0)(0.05 mol) = <em>0.4 mol.</em>

<em>5) Number of hydrogen atoms:</em>

  • It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

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

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

<em>∴ The no. of atoms in  0.4 mol of H atoms</em> = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = <em>2.4 x 10²³ molecules.</em>

8 0
3 years ago
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