What property of copper makes it ideal to create thin sheets of copper?

1 answer:

8 0

Copper is a soft, ductile, and malleable metal that bends and stretches easily without breaking. Copper is a versatile metal for construction and manufacturing because of this.

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Alguien sabe el nombre del compuesto etilisopropilacetileno?

Paladinen [302]

Noooooooooooooooooooooo

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3 years ago

12. In the modern periodic table, which of the following describes the elements with similar

V125BC [204]

B. same group I believe

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If 0.2 g of nitrobenzene are added to 10.9 g of naphthalene, calculate the molality of the solution. (given: molar mass of nitro

In-s [12.5K]

Molality is defined as 1 mole of a solute in 1 kg of solvent.

Molality=

$\frac{Number of moles of solute}{Mass of solvent in kg}$

Number of moles of solute, n=

$\frac{Given mass of the substance}{Molar mass of the substance}$

Given mass of the nitrobenzene=0.2 g

Molar mass of the substance= 123.06 g mol⁻¹

Number of moles of nitrobenzene,

$n= \frac{0.2 }{123.06}$

Number of moles of nitrobenzene, n= 0.0016 mol

Mass of 10.9 g of naphthalene in kg=0.0109

$Molality= \frac{0.0016}{0.0109 }$

Molality= 0.146 m

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3 years ago

A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask ab

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .