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2 years ago

How to classify an unknown element as a metal, metalloid, or non-metal

1 answer:

7 0

You can classify a metal: when a subject can surely conduct electricity.

examples: if a problem says the unknown object is malleable, or says "on the left side of the periodic table", is a conductor, is shiny, etc.

Metalloids: when a subject is alongside the staircase, uses word "semi-conductor", it depends on if it's malleable or ductile, and depends on whether it's shiny or not, etc.

Nonmetal: is located on the right side of the periodic table, can ONLY be an insulator, is not malleable, is not shiny, etc.

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A sodium atom has 11 protons, 10 electrons, and 12 neutrons. What is its charge

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2 years ago

Which is stronger- the attractive forces between water molecules and chromium and chloride ions, or the combined ionic bond stre

Nutka1998 [239]

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When chromium chloride, CrCl2, is dissolved in water, the temperature of the water decreases. ... The attractive forces between water molecules and chromium and chloride ions is stronger, because the reaction is endothermic means the energy released in formation is less than the energy required in breaking bond.

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2 years ago

Write down your observations of what the pieces of cabbage in the images look like

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1 year ago

What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm

lesya [120]

<u>Answer:</u> The mass of sucrose required is 69.08 g

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 8.80 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL (Density of water = 1 g/mL)

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

$8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g$

Hence, the mass of sucrose required is 69.08 g

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2 years ago