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e-lub [12.9K]
4 years ago
12

List the sub-atomic particles in the order in which they were discovered and tell who discovered each one.

Chemistry
1 answer:
harkovskaia [24]4 years ago
7 0
<span>Photons were the first sub-atomic particles detected, but not quite discovered as they could not be explained. Photons were first detected by Johann Wilhelm Ritter, Victor Schumann, and Winhelm Rontgen. The next, and first sub-atomic particle discovered, was the electron. The electron was discovered by J. J. Thompson in the late 1800s. The next two sub-atomic particle discoveries were the alpha particle and photon, discovered by Ernest Rutherford and Paul Villard respectively. Rutherford also discovered the proton and in 1932, James Chadwick discovered the neutron.</span>
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How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
As a gas changes to a solid, energy<br> A. is not used.<br> B. is released.<br> C. is absorbed.
pogonyaev

Answer:

B

Explanation:

Energy is released, when a gas changes to solid

7 0
3 years ago
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Assoli18 [71]

We know that there are 100 cm in 1 m, so we can use this to convert to meters:

(\frac{2.41x10^{2}cm}{1})*(\frac{1m}{100cm} )=2.41m

Therefore we know that 2.41x10^{2} cm is equal to 2.41 m.

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Simora [160]

Answer:

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• Its orbit covers 90 percent of Earth's surface.

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Explanation:

Correct on edge

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Answer:

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3 0
2 years ago
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