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Allushta [10]
3 years ago
11

What are examples of actinides

Chemistry
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

Neptunium, Protactinium, plutonium and more

Explanation:

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Which statement explains why meiosis haploid cells rather than diploid cells
Julli [10]
In meiosis there are haploid cells because two cells are being combined to make one, while in mitosis there are diploid cells because they are splitting apart.
4 0
4 years ago
Asteroids in the main asteroid belt at certain distances from the Sun are in an orbital resonance with Jupiter, creating gaps in
tatyana61 [14]

Asteroids in the main asteroid belt at certain distances from the Sun are in an orbital resonance with Jupiter, the period of a belt asteroid in an 11:5 orbital resonance with Jupiter is mathematically given as

TA=5.4090years

<h3>What is the period of a belt asteroid in an 11:5 orbital resonance with Jupiter?</h3>

Generally, the equation for the ratio is mathematically given as

asteroid in an 11:5 orbital resonance= 11/5

Therefore, with Jupiter at

TJ=11.9years

Hence

11/4=TJ/TA

TA=5/11)*11.9

TA=5.4090years

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5 0
2 years ago
ANSWER: b <br> i hope i helped
Sergio [31]

Answer:

thanks

Explanation:

7 0
3 years ago
Read 2 more answers
How many molecules of hydrogen will be required to react completely with 0.600 mol of o2 in order to form water?
olga nikolaevna [1]
The  number  of  molecules  of hydrogen  that will be required   to react  completely  with  0.600 mol  of  O2  is  calculated  as  follows
equation  for  reaction

2H2+O2  = 2H2O
by  use  of  mole  ration of H2  to O2  (2:1)  the  moles  of H2  = 0.600 x2  = 1.2  moles

by  use of Avogadro  law   that is  1  mole =  6.02  x10^23  molecules  what about  1.2  moles

=  1.2 moles  x (6.02  x10^23)/  1mole= 7.224  x10^23 molecules
8 0
3 years ago
Tungsten and molybdenum both have the BCC crystal structure, and Mo forms a substitutional solid solution for all concentrations
goldenfox [79]

Answer:

the unit cell edge length is 3.1585 × 10⁻⁸ cm

Explanation:

Given the data in the question;    

density = 100 / [ (84/19.3) + (16/10.22) = 100 / 5.917889 = 16.8979 g/cm³

so Aug density p_aug = 16.8979 g/cm³

Aug Atomic weight = 100 / [ (84/183.85 ) + (16/95.94) = 100 / 0.623665

= 160.3424 g/mol

now

volume = ( no. of atoms in BCC × Aug Atomic weight) / ( Aug density × Avogadro's number)

we substitute

volume = ( 2 × 160.3424 ) / ( 16.8979 × 6.023 × 10²³ )

volume =  320.6848 / 1.017760517 × 10²⁵

volume = 3.15088 × 10⁻²³ cm³

Now, unit cell edge length will be;

= Volume^{1/3}

= (3.15088 * 10^{-23})^{1/3}

= 3.1585 ×  10⁻⁸ cm  

Therefore, the unit cell edge length is 3.1585 × 10⁻⁸ cm  

8 0
3 years ago
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