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Leviafan [203]
3 years ago
15

The ease with which a raw material can be molded , flattened , or bent is known as its ?

Chemistry
2 answers:
Leya [2.2K]3 years ago
6 0
A. Malleability

Ductility is being able to be drawn into wires
Elasticity is being able to resist stress
Resilience is being able to spring back into shape (kind of like elasticity)
Angelina_Jolie [31]3 years ago
4 0

Answer:

Malleability

Explanation:

Malleability is the capability of a substance, especially a metal, to be reshaped or molded into a different shape. For people in the field of chemistry, the malleability of a metal provides an important means of describing the specific features of a metal and relating it to the arrangement of the atoms within the metal.

Malleability in metals happens due to the metallic bonds that keep the atoms in place. Metallic bonds, is made up of  a 'sea' of electrons that easily move from atom to another, allow the metal atoms to slide past each other if a force is applied. The force can come from a blow from a hammer, the impact from a fall, high pressure from being squeezed, or from a collision.

The degree of malleability differs widely among metals as well as mixtures of different metals, also known as alloys. Multiple factors can affect the malleability of a metal or alloy, but two fundamentally important factors are the strength of the metallic bond and the temperature.

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Which pair must represent atoms of the same element?
andrew-mc [135]
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4 0
3 years ago
A chemist makes of nickel(II) chloride working solution by adding distilled water to of a stock solution of nickel(II) chloride
Lady_Fox [76]

Answer:

0.0900 mol/L

Explanation:

<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>

Step 1: Given data

  • Initial concentration (C₁): 0.135 mol/L
  • Initial volume (V₁): 220. mL
  • Final concentration (C₂): ?
  • Final volume (V₂): 330. mL

Step 2: Calculate the concentration of the final solution

We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L

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