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3 years ago

What is the mass of 2.23 x 1023 atoms of sulfur?

Chemistry

1 answer:

Minchanka [31]3 years ago

5 0

Answer:

11.9 g S

General Formulas and Concepts:

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

2.23 × 10²³ atoms S

<u>Step 2: Define conversions</u>

Avogadro's Number

Molar Mass of S - 32.07 g/mol

<u>Step 3: Dimensional Analysis</u>

<u /> $2.23 \cdot 10^{23} \ atoms \ S(\frac{1 \ mol \ S}{6.022 \cdot 10^{23} \ atoms \ S} )(\frac{32.07 \ g \ S}{1 \ mol \ S} )$ = 11.8758 g S

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules.</em>

11.8758 g S ≈ 11.9 g S

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How are radio waves transmitted

kotegsom [21]

Attractive antennas that pull the waves tword it.

7 0

3 years ago

3) How many moles of solute are dissolved in 1.5 L of a 0.80 M solution? * 10 points

Mrrafil [7]

Answer:

1.2

Explanation:

Molarity = moles/volume

to find the moles, you must multiply both sides by volume

so, take 1.5 L * 0.80 M

this equals 1.2 :)

7 0

2 years ago

The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answe

zimovet [89]

Answer:

Mostly Para

Explanation:

First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).

Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.

Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.

Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

4 0

3 years ago

How many particles are in 1.40 x 10^7 mol of aluminum (Al)?

SpyIntel [72]

Answer:

<h3>Hlo there !! </h3>

<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.</u>

<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.So 1.04*107 mol of Al contains 1.40*107 * 6.022*1023 = 8.43*1030 structural units (in case of Al – atoms).</u>

<h3><u>8.43*1030 particles Al.</u></h3>

Explanation:

<h3>Hope this helps !!</h3>

5 0

2 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago