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2 years ago

Brainliest if CORRECT 11

Chemistry

2 answers:

Marianna [84]2 years ago

7 0

The B is the answer to the question pretty sure

Klio2033 [76]2 years ago

6 0

Answer:

Model B

Explanation:

Model A shows that light has been reflected, whereas Model B shows the light has been refracted. Light will be refracted as the light enters a more dense medium (glass), from a less dense medium (air).

This website has a really useful, thorough explanation for why this occurs:

(Check the explanation under the heading 'Explaining refraction - Higher')

https://www.bbc.co.uk/bitesize/guides/zw42ng8/revision/2

Hope this helps!

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What is the name of the isotope produced when Californium-251 emits an alpha particle?

Charra [1.4K]

<h3>Answer:</h3>

Curium-247 <em>i.e.</em> ²⁴⁷₉₆Cm

<h3>Explanation:</h3>

Alpha decay is given by following general equation,

ᵃₓA → ⁴₂He + ᵃ⁻⁴ₓ₋₂B

Where;

A = Parent Isotope

B = Daughter Isotope

ᵃ = Mass Number

ₓ = Atomic Number

Californium-251 is the parent isotope in our case and it has 98 protons (atomic number) and is given as,

²⁵¹₉₈Cf

The alpha decay reaction of Californium-251 will be as,

²⁵¹₉₈Cf → ⁴₂He + ²⁴⁷₉₆B

The symbol for B with atomic number 96 was found to be the atom of Curium (Cm) by inspecting periodic table. Hence, the final equation is as follow,

²⁵¹₉₈Cf → ⁴₂He + ²⁴⁷₉₆Cm

3 0

3 years ago

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.

6 0

2 years ago

Read 2 more answers

Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b.

Ilia_Sergeevich [38]

Answer:

a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe

b. Glu-Ala-Phe + Gly-Ala-Tyr

Explanation:

In this case, we have to remember which peptidic bonds can break each protease:

-) <u>Trypsin</u>

It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.

-) <u>Chymotrypsin</u>

It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.

With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the <u>"Lis"</u> and <u>"Arg"</u> (See figure 1).

In "peptide b", the peptidic bond that would be broken is the one in the <u>"Phe"</u> (See figure 2). The second amino acid that can be broken is <u>tyrosine</u>, but this amino acid is placed in the <u>C terminal spot</u>, therefore will not be involved in the <u>hydrolysis</u>.