Min : sec helium and nickel are classified as _____, which are pure substances that make up all kinds of matter

1 answer:

4 0

Helium which has a chemical formula of He is considered as a noble gas while Nickel which has a chemical formula of Ni is considered as metallic. Although both have different phases but they are both classified under the general category of “Elements”.

answer:

<span>elements</span>

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Process by which plants, algae, and many types of bacteria use sunlight, water, and carbon dioxide to make food and oxygen

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Answer: Photosynthesis

Explanation:

<em>The process of photosynthesis</em> is used to help make food for plants. The inputs are sunlight, water (H2O), and carbon dioxide (CO2). Then the outputs are glucose (C6H12O6) which is food, and oxygen (O2) is released.

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2 years ago

Which sample would serve as a better buffer and therefore a better environment for aquatic life?

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Answer:

sample A

Explanation:

the first one because of the ppm value

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2 years ago

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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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How many moles of KNO3 are in 500 mL of 2. 0 M KNO3? mol KNO3.

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The moles can be defined as the mass of the substance with respect to molar mass. The moles of potassium nitrate is 1 mol.

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