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3 years ago

Each carbon atom of benzene is involved in two ________

Chemistry

1 answer:

coldgirl [10]3 years ago

7 0

Answer:

Benzene is a combination of carbon and hydrogen atoms. The hybridization is sp2 type. During the hybridization of benzene, each carbon atom forms different bonds with two other similar carbon atoms instead of just one.

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Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. when 2.59 g of magnesium ribbon bu

stepan [7]

Burning Mg in the air and reacting with O2 forming a white powder of MnO

So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)

to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
Δ
2Mg(s) + O2(g)⇒ 2MgO(s)

4 0

3 years ago

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

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

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

3 years ago

Calculate the speed of a marble that rolls 9 cm in 4 seconds

cupoosta [38]

Answer:

2.25

Explanation:

9/4

6 0

3 years ago

2 Na + Cl2 --> 2 NaCl What is the mole ratio of sodium to sodium chloride?

Irina18 [472]

Answer: The mole ratio of sodium to sodium chloride 2:2.

Explanation:

As the given reaction equation is as follows.

$2Na + Cl_{2} \rightarrow 2NaCl$

Here, 2 moles of sodium reacts with 1 mole of $Cl_{2}$ and leads to the formation of 2 moles of NaCl.

This means that 2 moles of sodium gives 2 moles of NaCl on reaction with chlorine.

Hence, the ratio of moles of sodium to sodium chloride is 2:2.

Thus, we can conclude that the mole ratio of sodium to sodium chloride 2:2.

6 0

3 years ago

34. 3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases

Digiron [165]

Answer:

ΔH = 2.68kJ/mol

Explanation:

The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:

q = m*S*ΔT

Where q is heat of reaction in J,

m is the mass of the solution in g,

S is specific heat of the solution = 4.184J/g°C

ΔT is change in temperature = 11.21°C

The mass of the solution is obtained from the volume and the density as follows:

150.0mL * (1.20g/mL) = 180.0g

Replacing:

q = 180.0g*4.184J/g°C*11.21°C

q = 8442J

q = 8.44kJ when 3.15 moles of the solid react.

The ΔH of the reaction is:

8.44kJ/3.15 mol

= 2.68kJ/mol

5 0

3 years ago