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2 years ago

Which statement best explains the difference between polar and nonpolar covalent bonds?(1 point)

Chemistry

2 answers:

Mars2501 [29]2 years ago

8 0

Answer:

Let me help you out fr:

Explanation:

1. 0.5 to 1.7

2. fluorine (F)

3. KNO3

4. Polar covalent bonds share electrons unequally, while nonpolar covalent bonds share electrons equally.

5. Fe and S have an ionic bond, while S and O have covalent bonds.

miv72 [106K]2 years ago

6 0

Answer:

Answer is 4

Explanation:

Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared between two atoms. ... If the electronegativity of two atoms is basically the same, a nonpolar covalent bond will form, and if the electronegativity is slightly different, a polar covalent bond will form.

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A mixture of 5L of H2 and 3L O2 reacts to form H2O (g) at constant T and P .Find the

slamgirl [31]

The volume of H₂O = 5 L

<h3>Further explanation</h3>

Given

5L of H₂ and 3L O₂

Reaction

2H₂ (g) + O₂(g) ⇒2H₂O(g)

Required

The volume of H₂O

Solution

Avogadro's hypothesis:

<em>In the same T,P and V, the gas contains the same number of molecules </em>

So the ratio of gas volume will be equal to the ratio of gas moles

mol H₂ = 5, mol O₂ = 3

Find limiting reactants

From equation, mol ratio H₂ : O₂ = 2 : 1, so :

$\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)$

Find volume H₂O

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

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Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is

erik [133]

Answer : The percent yield of $MgO$ is, 64.13 %

Solution : Given,

Mass of Mg = 10 g

Mass of $O_2$ = 6 g

Molar mass of Mg = 24 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg and $O_2$ .

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{10g}{24g/mole}=0.4167moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{6g}{32g/mole}=0.1875moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $Mg$

So, 0.1875 moles of $O_2$ react with $0.1875\times 2=0.375$ moles of $Mg$

From this we conclude that, $Mg$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $MgO$

From the reaction, we conclude that

As, 1 mole of $O_2$ react to give 2 mole of $MgO$

So, 0.1875 moles of $O_2$ react to give $0.1875\times 2=0.375$ moles of $MgO$

Now we have to calculate the mass of $MgO$

$\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO$