A 3kg ball is kicked at 4m/s squared .How much force is used to move the ball?

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

Answer: The average atomic mass of element Zinc is 65.40 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

For $_{30}^{67}\textrm{Zn}$ isotope:

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

For $_{30}^{68}\textrm{Zn}$ isotope:

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

For $_{30}^{70}\textrm{Zn}$ isotope:

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0

3 years ago

Calculate the mass of water produced when 2.06 g of butane reacts with excess oxygen.

dem82 [27]

2C4H10 + 13O2 = 8CO2 + 10H2O

1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10

2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O

3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O

8 0

3 years ago

How is science used on a daily basis give 3 reasons how

goldenfox [79]

1-Autonomy: Our desire to direct our own lives. In short: “You probably want to do something interesting, let me get out of your way!”
2-Mastery: Our urge to get better at stuff.
3-Purpose: The feeling and intention that we can make a difference in the world.

7 0

4 years ago

What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?

tensa zangetsu [6.8K]

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

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3 0

3 years ago

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