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goldfiish [28.3K]
3 years ago
12

A substance that conducts electricity, is malleable, ductile, and has luster would be classified as a A) compound B) element C)

metal D) nonmetal
Chemistry
2 answers:
IgorLugansk [536]3 years ago
6 0

Answer: Option (C) is the correct answer.

Explanation:

The properties of a metal are as follows.

  • Metals are malleable that is they can be drawn into thin sheets.
  • They have luster.
  • They are good conductor of electricity.
  • They are ductile that is metals can be drawn into wires.

Thus, we can conclude that a substance that conducts electricity, is malleable, ductile, and has luster would be classified as a metal.

Sati [7]3 years ago
5 0
C. Metal

You can think of aluminum cans foils
They are both malleable (to morph easily)
They both are ductile (to flatten out)
They both have luster (shiny)
And they both conduct electricity.
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What is the molarity of 0.26 mol of H2SO4 dissolved in 0.3 L of solution? *
irina1246 [14]

Answer:

A

Explanation:

molarity=moles of solute/liter of solution

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molarity=0.87molar

7 0
3 years ago
What is the mass of 1.2 x 1023 atoms of arsenic?
Gre4nikov [31]

Answer:

14.93 g

Explanation:

First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:

  • 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As

Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:

  • 0.199 mol * 74.92 g/mol = 14.93 g

Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.

6 0
3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
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