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Delvig [45]
3 years ago
14

What is the acceleration of an object if it goes from a velocity of +25m/s to rest in 5.0s?

Chemistry
1 answer:
kari74 [83]3 years ago
5 0

Vf = Vi + at

0 m/s = 25 m/s + a(5sec)

-5 m/s2 = acceleration

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A swimmer is moving at a speed of 2.0 meters/second. How long will it take for the swimmer to go 100 meters?
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Two atoms that share one electron each between have a ?
evablogger [386]

Answer:

Covalent bond

Explanation:

A covalent bond is defined as the sharing of election pairs between atoms

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Natural gas is a mixture of many substances, primarily CH₄, C₂H₆, C3H8, and C4₄H₁₀. assuming that the total pressure of the gase
olchik [2.2K]

Answer:

A. The partial pressure for CH4 = 0.0925atm

B. The partial pressure for C2H6 = 0.925atm

C. The partial pressure for C3H8 = 0.346atm

D. The partial pressure for C4H10 = 0.115atm

Explanation:

Total pressure = 1.48atm

Total mole = 0.4+4+1.5+0.5=6.4

A. Mole fraction of CH4 = 0.4/6.4 = 0.0625

The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm

B. Mole fraction of C2H6 = 4/6.4 = 0.625

The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm

C. Mole fraction of C3H8 = 1.5/6.4 = 0.234

The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm

D. Mole fraction of C4H10 = 0.5/6.4 = 0.078

The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm

7 0
3 years ago
The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie
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<u>Answer:</u> Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

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

  • <u>For ozone:</u>

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol

  • <u>For nitric oxide:</u>

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol

For the given chemical equation:

O_3+NO\rightarrow O_2+NO_2

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitric oxide

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

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

  • Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

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