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dmitriy555 [2]
3 years ago
14

What do the elements of any given group in the periodic table have in common with each other?

Chemistry
1 answer:
Alika [10]3 years ago
4 0
They have the same number of electrons
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Science help for 6th grade advanced?
tino4ka555 [31]
16287.50 I think? I just googled it though so I’m not sure if it’s correct.
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2 years ago
When electrons are lost, a____ion is formed.
kati45 [8]

Answer:

Positive; negative

Explanation:

When electrons are lost, a positive ion is formed. A positive ion is called a cation.

When electrons are gained, a negative ion is formed. A negative ion is called an anion.

3 0
3 years ago
Read 2 more answers
What is the formula weight (amu) of the molecule H2O? Use atomic masses of H and O as 1.008 amu and 16.00 amu respectively. Repo
vaieri [72.5K]

Answer:

Formula weight of H₂O molecule is  18.02 amu.

Explanation:

Given data:

Formula weight of H₂O = ?

Atomic mass of H = 1.008 amu

Atomic mass of O = 16.00 amu

Solution:

Formula weight:

"It is the sum of all the atomic weight of atoms present in given formula"

Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu

Formula weight of H₂O = 18.02 amu

Thus, formula weight of H₂O molecule is  18.02 amu.

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3 years ago
I WILL GIVE BRAINLIEST
Sergeeva-Olga [200]

Answer:

waste gas he should use algae

4 0
3 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0
3 years ago
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