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3 years ago

Carbon can react with oxygen to form carbon dioxide. Which of the following statements about this chemical change is true?

1 answer:

6 0

Carbon can react with oxygen to form carbon dioxide. Which of the following statements about this chemical change is true? ... Carbon and oxygen atoms are destroyed as new atoms are formed. Carbon and oxygen atoms have the same properties as molecules of carbon dioxide.

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The atmospheric pressure on the surface of Venus is 6.84X10^4. Calculate the atmospheric pressure in atm and torr. Round each of

quester [9]

Answer:

0.675 atm

513 Torr

Explanation:

Given is that, the atmospheric pressure on the surface of Venus is

6.84 X 10⁴ Pa.

1 atm (atmospheric pressure) is equal to 101325 pascal (Pa).

To convert divide the pressure value by 101325.

Pressure in atm = $\frac{6.84 \times 10^{4} }{101325}$

= 0.675055 atm

Rounding it off to 3 significant digits: 0.675 atm

Now, one Torr is 133.322 Pa. For conversion, divide the pressure value by 133.322.

Pressure in Torr = $\frac{6.84 \times 10^{4} }{133.322}$

=513.04219 Torr

Rounding it off to 3 significant digits: 513 Torr

5 0

3 years ago

In the molecule of SO2 there are two pie bonds these are formed Due to

Aleks [24]

These are formed when, after formation of the first atom, the pi-orbitals of Sulfur and Oxygen align and share another pair of electrons. This occurs between sulfur and both of the oxygen atoms, with each oxygen atom forming one pi-bond.

4 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

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

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

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

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

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

What is the area of a medium triangle?

galben [10]

Answer: Area of a Triangle Equals Base x Height / 3

Explanation: Hope this works

6 0

2 years ago

What type of energy is present in wound-up rubber band in a toy airplane?

andriy [413]

It is elastic potential energy

3 0

3 years ago