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stellarik [79]
3 years ago
7

The kinetic energy of an object can be determined using the equation , where is the kinetic energy, is the mass, and is the spee

d of the object. What is the mass of an object moving at a speed of 12 that has 36 J of kinetic energy?
Chemistry
1 answer:
Cloud [144]3 years ago
8 0

Answer:

0.5kg

Explanation:

Given parameters:

Speed of the object  = 12m/s

Kinetic energy  = 36J

Unknown:

Mass of the object  = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically expressed as;

     K.E = \frac{1}{2} m v²  

 Insert the parameters and solve;

    36  =  \frac{1}{2} x m x 12²

     36  = 72m

       m  = 0.5kg

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A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of
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<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100

Mass percent means that the mass of a substance is present in 100 grams of mixture

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

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3 years ago
Which part of the crust is the thickest
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2 years ago
Which of these is not a layer of skin?
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Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
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