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e-lub [12.9K]
3 years ago
13

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.00mm away. What is

the initial acceleration of the proton after it is released?
Chemistry
1 answer:
myrzilka [38]3 years ago
3 0
<span>F = m.a
   = mass x acceleration

So,
a = F / m
Where,
F is coulomb repulsion.
k q q / r^2
with,
k = 9 x 10^9 in SI units (q in C and r in m)

Hence,
</span><span>a = 2.40 × 10^11 m/s^2 </span>
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nordsb [41]

Answer:

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

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To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01  . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01  . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

6 0
4 years ago
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Which of the following word best describe the diagram
adoni [48]

does it have multiple choice

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4 years ago
_Fe2O3 + 2CO —&gt; _Fe + _CO2
soldi70 [24.7K]
<h3>Answer:</h3>

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<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

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