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3 years ago

How do you determine if an element is metal or non-metal?

Chemistry

2 answers:

snow_tiger [21]3 years ago

5 0

Answer: If the outermost shell has 1–4 electrons, it is a metal.

If the outermost shell has 5–7 electrons, it is a non-metal.

denpristay [2]3 years ago

3 0

Metal elements have 1,2,3 or 4 electrons in the outer most shell while non metals have 5,6 or 7 electrons.
Example - Carbon
atomic number - 6
electron configuration - 2,4
Meaning it has 4 electrons in its outer most shell which makes it a METAL.

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The source of the sun's heat is A. nuclear fission. B. nuclear fusion. C. nuclear disintegration. D. nuclear separation

ExtremeBDS [4]

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Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ The source of the sun's heat is B. nuclear fusion

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

5 0

3 years ago

Read 2 more answers

What is the volume of the rock in the image in mL

Yuri [45]

Answer:

Initial volume: mL

Final volume: mL

Volume of rock: cm3

5 0

3 years ago

A sample has a pH of -0.248 (yes, it really is possible to have a negative pH). Find

dalvyx [7]

Answer:

pOH= 14.248

[H+]=1.77 M

[OH-]=5.65 x10^-15M

Explanation:

pH+pOH= 14

pOH= 14-pH

pOH=14-(-0.248)

pOH= 14.248

[H+]=10^-pH= 10^-(-0.248)=1.77 M

[OH-]=10^-pOH= 10^-14.248=5.65 x10^-15M

5 0

3 years ago

Water (H2O) is made up of molecules with a bent structure, whereas hydrogen sulfide (H2S) is also made up of molecules with a be

jeka94

Śhüt ûp and go pay attention in your class

3 0

3 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

3 years ago