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Evgen [1.6K]
4 years ago
14

Why does the band of stability curve upward at high atomic numbers? excess neutrons are required due to the repulsion between th

e protons. atoms with high atomic numbers have a large number of electrons orbiting the nucleus. this increased number of electrons requires a lot of extra mass in the nucleus to keep the electrons in their orbit?
Physics
1 answer:
ella [17]4 years ago
3 0
The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.

Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.


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A student was given a sample of crude acetanilide to recrystallize. The initial mass of the the crude acetanilide was 166 mg. Th
padilas [110]

Answer:

The percentage of mass recover from recrystallization = 24.69 %

Explanation:

Given that

Initial mass of crude acetanilide = 166 gm

The mass after recrystallization = 125 gm

The mass recover from recrystallization= 166 - 125 gm

The mass recover from recrystallization= 41 gm

The percentage of mass recover from recrystallization can be find as

percentage\ of\ mass\ recover\ from\ recrystallization\ =\dfrac{166-125}{166}

The percentage of mass recover from recrystallization = 24.69 %

4 0
3 years ago
Una cámara fotográfica analógica (no digital) tiene dos lentes intercambiables. Uno de foco 55mm y el otro de 200 mm. Toma una f
Sergeu [11.5K]

Answer:

f = 55mm,     h ’= -9.89 cm

f = 200 mm,  h ’= 42.5 cm

Explanation:

For this exercise let's start by finding the distance to the image, using the equation of the constructor

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distances to the object and image, respectively

lens with f₁ = 55mm = 0.55cm

         \frac{1}{q} = \frac{1}{f} - \frac{1}{p}

         \frac{1}{q_1} = \frac{1}{0.55} - \frac{1}{10}

          \frac{1}{q_1} = 1.718

          q₁ = 0.582 m

lens with f₂ = 200mm = 2m

           \frac{1}{q_2} =   \frac{1}{2} - \frac{1}{10}

            \frac{1}{q_2} = 0.4

            q₂ = 2.5 m

the magnification of a lens is given by

            m = \frac{h'}{h} = -  \frac{q}{p}

             h ’= - \frac{q}{p} \ h

let's calculate for each lens

f = 55mm

             h '= - 0.582 / 10 1.7

             h ’= 0.0989 m

             h ’= -9.89 cm

f = 200 mm

             h '= - 2.5 / 10 1.7

             h ’= -0.425 m

             h ’= 42.5 cm

The negative sign indicates that the image is real and inverted

4 0
3 years ago
What are parasites? Give some example​
Inga [223]

Answer:

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4 0
3 years ago
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Dmitriy789 [7]

Answer:

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8 0
3 years ago
The moon Ariel orbits Uranus at a distance of 1.91 x 108 m once every 2.52 days. Use that data to calculate the mass of Uranus.
PIT_PIT [208]

Mu = 8.66 × 10^25 kg

Explanation:

centripetal force = gravitational force

m \frac{ {v}^{2} }{r}  = (grav.const) \frac{m \times mu}{ {r}^{2} }

where

m = mass of moon Ariel

mu = mass of Uranus

r = radius of Ariel's orbit

v = Ariel's velocity around Uranus

To find the velocity, we need to find the circumference of the no orbit and then divide it by the period (2.52 days):

circumference = 2πr = 2π×(1.91 × 10^8 m)

= 1.2 × 10^9 m

period = 2.52 days × (24 h/1 day)×(3600 s/1 hr)

= 2.18 × 10^5 s

v = (1.2 × 10^9 m)/(2.18 × 10^5 s)

= 5.5 × 10^3 m/s

(5.5 × 10^3 m/s)^2/(1.91 × 10^8 m) = (6.67 × 10^-11 m^3/kg-s^2)Mu/(1.91 × 10^8 m)^2

Solving Mu,

Mu = 8.66 × 10^25 kg

4 0
3 years ago
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