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2 years ago

What did Thomson’s and Rutherford’s experiments have in common?

Chemistry

1 answer:

Alchen [17]2 years ago

8 0

Answer:

The answer is A.

Explanation:

They both used charged particles in their experiments.

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Which contains elements with similar properties in the periodic table? A.) a period B.) a column C.) a row

zysi [14]

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3 years ago

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WILL GIVE 50 POINTS AND BRAINLIEST Describe the properties of alkaline earth metals. Based on their electronic arrangement, expl

Arisa [49]

Alkaline earth metals are metals of group two. They are divalent metals and they have a highly negative reduction potential hence the metals are mostly extracted by electrolysis.

They are highly reactive metals. They react with water but do so less readily than alkali earth metals.

Owing to their high reactivity, they are seldom found free in nature. They always occur in combined state with other highly reactive nonmetals.

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2 years ago

What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?

Vanyuwa [196]

Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

Explanation:

22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g. 1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

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3 years ago

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Draw structural formulas for the alkoxide ion and the alkyl(aryl)bromide that may be used in a williamson synthesis of the ether

Juliette [100K]

Williamson synthesis is the most common way for obtaining ethers, called after its developer Alexander Williamson. It is an organic reaction of forming ethers from an organohalide and an alkoxide. The reaction is carried out according to the SN2 mechanism.

On the attached picture it is shown required alkoxide ion, <span>alkyl(aryl)bromide and the ether that forms from the reactants. </span>

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3 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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3 years ago