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4 years ago

Ernest Rutherford discovered that atoms were mostly:

Chemistry

2 answers:

Bezzdna [24]4 years ago

7 0

The answer is empty space!!

Elden [556K]4 years ago

5 0

Answer:

The answer is positively charged.

Explanation:

The Rutherford model supplanted the atomic model of English physicist Sir J.J. Thomson, in where the electrons were set in a positively charged atom.

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The sharing of electrons between the oxygen atom and two hydrogen atoms form a

Sholpan [36]

Answer:

The answer is covalent bond

Explanation:

when oxygen atom and two hydrogen atoms are combined, water molecule is formed according to the equation;

2H2 + O2 ==> 2H2O

Water is a covalent compound.

A covalent bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

For many molecules, the sharing of electrons allows each atom to attain the equivalent of a full outer shell, corresponding to a stable electronic configuration.

8 0

4 years ago

When copper crystallizes, it forms face-centered cubic cells. The unit cell edge length is 361.5 pm. Calculate the density of co

Olegator [25]

Density of copper can be calculated using the following formula:

$d=\frac{m}{V}$

Here, m is mass and V is volume, thus, to calculate density first calculate mass and volume of copper.

In FCC, number of atoms in a unit cell are 4, atomic mass of Cu is 63.546 g/mol and number of atoms in 1 mole is $6.023\times 10^{23}$ .

Mass of Cu will be:

$m=4 atoms\times \frac{1 mol}{6.023\times 10^{23}}\times \frac{63.546 g}{mol}=4.22\times 10^{-22}g$

Now, volume can be calculate as:

$V=a^{3}$

Here, a is edge length.

First convert edge length from pm to cm

$1pm=10^{-10}cm$

Thus,

$361.5 pm=3.615\times 10^{-8}cm$

Putting the value,

$V=(3.615\times 10^{-8} cm)^{3}=4.72\times 10^{-23} cm^{3}$

Now, from mass and volume density can be calculated as follows:

$d=\frac{m}{V}=\frac{4.22\times 10^{-22}g}{4.72\times 10^{-23} cm^{3}}=8.932 g/cm^{3}$

Therefore, density of copper will be $8.932 g/cm^{3}$ .

8 0

4 years ago

A carbohydrate sample weighing .235 g was found to have a fuel value of 3.8 4 KJ what is the fuel value of 1 g of this carbohydr

coldgirl [10]

Answer:

The fuel value of 1 gram of this carbohydrate in a nutritional calorie is 4.0 kcal/G.

8 0

3 years ago

What information does the fossil record provide?

lukranit [14]

By studying the fossil record we can tell how long life has existed on Earth, and how different plants and animals are related to each other. Often we can work out how and where they lived, and use this information to find out about ancient environments. Fossils can tell us a lot about the past.

6 0

3 years ago

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) â†’ 2Al2O3 (s) In

jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

$\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of $Al_2O_3$

So, 0.092 moles of aluminium will produce = $\frac{2}{4}\times 0.092=0.046moles$ of $Al_2O_3$

Now, calculating the mass of aluminium oxide:

$\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

$\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%$

Hence, the percent yield of the reaction is 74 %

6 0

3 years ago