Concept 1.2 Energy for Launch

The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o

belka [17]

Answer:

Explanation:

We can calculate the volume of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

volume=33.40×10⁻³⁰m³

Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

k=R/Na

R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

p is pressure of gas

v is volume of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

p=12.39×10⁷Pascal

8 0

3 years ago

Two brothers are skateboarding down a hill. Boy 1 has a mass of 100 kg and is moving 10 meters per second. Boy 2 has a mass of 2

sukhopar [10]

The second brother Brother because he is much heavier and therefore has more energy to be released hope this helps

6 0

3 years ago

Read 2 more answers

If a solution has a concentration of 12 M when the volume is 150 mL. What is the new concentration if you increase the volume to

swat32

Answer:

The idea with diluting a solution is that the number of moles of solute will remain constant after the initial solution is diluted. The only ...

Explanation:

hope it helps you

6 0

1 year ago

The photograph below shows the East African Rift Valley in Africa. Which tectonic movement

GalinKa [24]

Answer:

divergence of continental crust

Explanation:

8 0

2 years ago

How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of

xeze [42]

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

$moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}$

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

$moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}$

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

The excess reactant that will be left after the reaction is 3.45 moles.

If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

$moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}$

moles of H₂O= 4.3 moles

4.3 moles of water can be produced.

8 0

2 years ago