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marishachu [46]
4 years ago
11

Any body help me? please Q 17

Chemistry
1 answer:
iVinArrow [24]4 years ago
8 0

Question 17

 The subscript  2  is the  atomic number   for helium  and subscript  4  is the mass   number for Helium


 <u><em>explanation</em></u>

  when  writing   down  an isotope  of   an  element the mass  number of that isotope  is written  in upper   side  of the chemical  symbol,  while atomic  number  is written  on the  lower  side.

Since 4  is on upper side it is the mass  number of He, while 2  is  the atomic  number since  it is  on lower side.


Question 18

The type of radiation  emitted  when polonium -212  form lead 208  is  an alpha  particle


<u><em>explanation</em></u>

Alpha  decay  is a decay in which  an  atomic  nuclei  emit an alpha  particles and be transformed  into different atomic  nuclei  with    a mass  number that is  reduced by 4

polonium  212  is reduced by 4  to  form lead 208  therefore it undergo  alpha  particle  decay.

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Answer:

False

Explanation:

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These Roman numerals are used to depict the oxidation state of the central atom in the molecule and not to show how many of that ion appear in the formula.

For instance, in carbon IV oxide, the Roman numeral IV shows that the central atom in the compound-carbon is in the +4 oxidation state.

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse
amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given  by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

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