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Otrada [13]
3 years ago
9

An unknown solid weighs 31.6 n. when submerged in water, its apparent weight is 19.8 n. what is the specific gravity of the unkn

own sample?
Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

specific gravity of the unknown sample = 2.68

Explanation:

True weight of solid=31.6 N

Apparent weight=19.8 N

loss in weight= 31.6-19.8=11.8 N

Loss in weight= weight of the liquid displaced

so weight of water displaced= 11.8 N

now specific gravity= weight of object/ weight of liquid displaced

specific gravity= 31.6/11.8

specific gravity= 2.68

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Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

5 0
3 years ago
Covalent bonds tend to make atoms more stable by helping them
Lelechka [254]
Achieve a full outer shell
6 0
3 years ago
Narysuj wykres zależności szybkości od czasu i drogi od czasu jeśli ciało porusza się ruchem jednostajnym z szybkością 45 m/s.
murzikaleks [220]
Lett me come back imma translate this... and then ill come to help
7 0
3 years ago
Does anyone know this one? Thanks
Inessa [10]

Answer:

3.844\,*\,10^5

So a=3.844 and b=5

Explanation:

Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

384,400=3.844 \,*\,100,000= 3.844 \,*\,10^5

with a=3.844, and "5" as the exponent of ten (so b=5)

6 0
4 years ago
Need help ASAP
VARVARA [1.3K]

Answer:

the correct one is D,  

Ultraviolet, x-ray, gamma ray

Explanation:

Electromagnetism radiation are waves of energy that is expressed by the Planck relationship

          E = h f

where h is the plank constant and f the frequency of the radiation.

Also the speed of light is

          c = λ f

         

we substitute

          E = h c /λ

therefore to damage the cells of the body radiation of appreciable energy is needed

microwave radiation has an energy of 10⁻⁵ eV

infrared radiation                E = 10⁻² eV

visible radiation                   E = 1 to 3 eV

radiation Uv                         E = 3 to 6 eV

X-ray                                    E = 10 eV

 

gamma rays                         E = 10 5 eV

therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.

When checking the answers, the correct one is D

6 0
3 years ago
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