Why did Rutherford choose alpha particles in his experiment?

1 answer:

5 0

Rutherford used gold for his scattering experiment because gold is the most malleable metal and he wanted the thinnest layer as possible. The goldsheet used was around 1000 atoms thick. Therefore, Rutherford selected a Gold foil in his alpha scatttering experiment.

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Which element would mostly have chemical properties similar to that of magnesium (Mg)?

DanielleElmas [232]

The elements of same group are placed in same group as they have similar properties. So Calcium is the element which resembles Mangesium.

Calcium and Magnesium both are alkaline earth metals.

They have similar properties as they have similar outer or valence shell electronic configuration

The general electronic configuration of alkaline earht metal is ns2

The electronic configuration of Mg [atomic number 12] and Ca [atomic number 20] are

$[Ne]3s^{2}$ and $[Ar]4s^{2}$ respectively.

7 0

3 years ago

Which would most likely be joined by an ionic bond?

KATRIN_1 [288]

An ionic bond is composed of a cation and anion, which are charged ions. You cannot have atoms with ionic bonds, because there is no opposite attraction between charges due to atoms having a net nuclear charge of zero. Therefore, a cation and anion would most likely be joined by an ionic bond.

3 0

3 years ago

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$