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2 years ago

What direction are the groups (families) in the periodic table?

Chemistry

2 answers:

Soloha48 [4]2 years ago

6 0

Answer:

left

Explanation:

Troyanec [42]2 years ago

6 0

The answer to the question is left

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How many moles are in 68.16 grams of NH3 ?

seropon [69]

Answer:

4.002 mol is in 68.16 grams of NH3

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2 years ago

SOMEONE HELO PLZ!!!!!

umka21 [38]

Answer:

2 8 8 2 is a correct answer

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2 years ago

WILL MARK BRAINLY! Which chemical element has the shortest name?

Aleonysh [2.5K]

Answer:

boron I guess

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2 years ago

Let this be an easy 5 points for your brainly account all is ask for in return is a "thank you"

Svet_ta [14]

Answer:

The number of neutrons present in one atom of isotope of Silicon of mass 28 amu is<u> 14 neutrons</u>

Explanation:

Symbol of Si isotope

$_{14}^{28}\textrm{Si}$

<u>Number of Neutron = Mass number - Atomic Number</u>

Mass number = Total number of protons and neutrons present in the nucleus of the atom.For Si = 28 amu

Atomic Number = Total number of Protons present in the nucleus.

Si = 14

Number of neutron = 24 - 14

= 14

4 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago