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3 years ago

Consider the bonds in iron(III) oxide. Are the bonds more ionic or more covalent?

Chemistry

2 answers:

kvv77 [185]3 years ago

4 0

<span>the bonds in iron(III) oxide are more ionic</span>

kap26 [50]3 years ago

3 0

Answer: Option (1) is the correct answer.

Explanation:

A bond formed by transfer of electrons from one atom to another is known as an ionic bond.

Generally, an ionic bond is formed between a metal and a non-metal.

When a bond is formed by sharing of electrons then it is known as a coavlent bond.

A covalent bond is formed only within non-metals.

Chemical formula of iron(III) oxide is $Fe_{2}O_{3}$ . Here, iron is the metal and oxygen atom is the non-metal.

Hence, bond formed within both the elements will be ionic in nature due to transfer of electrons from iron to oxygen atom.

Thus, we can conclude that the bonds in iron(III) oxide are more ionic.

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In a chemical equation, what do the coefficients indicate?

Mekhanik [1.2K]

In a chemical equation coefficients indicate the number of molecules/atoms involved in the reaction.

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3 years ago

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Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$

$75 \times \frac{1}{2} =48.2$

$34.5 \times \frac{1}{2} =72.3$

$17.25 \times \frac{1}{2} =96.4$

$8.625\times \frac{1}{2} =120.5$

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

$A=A_{0} \cdot \frac{1}{2^{n}}$

Where

$A_0$ is the initial sample amount

n = the number of half-lives that pass in a given period of time.

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3 years ago

At 25°C and constant pressure, carbon monoxide gas combines with oxygen gas to give carbon dioxide gas with the evolution of 10.

stealth61 [152]

Answer : The value of $\Delta H$ for the reaction is, -565.6 kJ

Explanation :

First we have to calculate the molar mass of CO.

Molar mass CO = Atomic mass of C + Atomic mass of O = 12 + 16 = 28 g/mole

Now we have to calculate the moles of CO.

$\text{Moles of }CO=\frac{\text{Mass of }CO}{\text{Molar mass of }CO}=\frac{1g}{28g/mole}=\frac{1}{28}mole$

Now we have to calculate the value of $\Delta H$ for the reaction.

The balanced equation will be,

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

From the balanced chemical reaction we conclude that,

As, $\frac{1}{28}mole$ of CO release heat = 10.1 kJ

So, 2 mole of CO release heat = $2\times 28\times 10.1=565.6kJ$

Therefore, the value of $\Delta H$ for the reaction is, -565.6 kJ (The negative sign indicates the amount of energy is released)

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