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Rufina [12.5K]
3 years ago
12

The radius of a niobium atom is 131 pm. how many niobium atoms would have to be laid side by side to span a distance of 2.40 mm

Physics
1 answer:
pochemuha3 years ago
7 0
First, we must know the following conversions:

1 pm = 10⁻¹² m
1 mm = 10⁻³ m

Therefore,
1 pm = 10⁻⁹ mm

The radius of the atom is 131 pm, which means that the span of one atom, the diameter, is 262 pm.

Number of atoms needed = distance to be covered / span of one atom
Number of atoms needed = 2.4 / 262 x 10⁻⁹
Number of atoms needed = 9.16 x 10⁷

9.16 x 10⁷ atoms of niobium will be needed
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Answer:

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Given:

Temperature, T = 3.13 K

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Putting the values in the above formula:

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3 years ago
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0.557 s

Explanation:

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3 years ago
What are the very small particles that make up matter​
mrs_skeptik [129]
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A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t
user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) =  m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

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if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg *  0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

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7 0
3 years ago
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If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

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To know more about velocity, refer: brainly.com/question/12413963

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