Complete the graphic organizer. Basic Parts of an Atom

]If a sample of pure iodine contains 4.69 x 1022 atoms of iodine, how many moles of iodine are in the sample? A=0.08 B=0.13 C=7.

Lorico [155]

It is B, and also for a moment I didn't understand that 4.69 x 10^22. I almost did this whole problem wrong.

6 0

3 years ago

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Magnesium and iron are metallic elements. How does a mole of magnesium compare with a mole of iron? A mole of iron has more atom

grandymaker [24]

Answer:

They both have the same number of atoms

Explanation:

The number that indicates the amount of particles in a compound is the Avogadro's number (NA).

It does not matter the mass of compound we have, If we have 1 mol we will be sure that we are talking about 6.02×10²³ particles

6.02×10²³ represents the amount of atoms in twelve grams of 12-pure carbon and it is considered a reference to measure the amount of all kinds of substances present in a given system.

8 0

3 years ago

A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8

aivan3 [116]

Answer:

16.06 L was the initial volume of the balloon.

Explanation:

Initial moles of freon in ballon = $n_1=9.79 mol$

Initial volume of freon gas in ballon = $V_1=?$

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = $n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol$

Final volume of freon gas in ballon = $V_2=21.8 L$

Using Avogadro's law:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ( at constant pressure and temperature)

$V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L$

16.06 L was the initial volume of the balloon.

4 0

3 years ago

PLEASEEEE HELP MEEEEE

fiasKO [112]

Answer:

3.82 x 10²¹ molecules As₂O₃

Explanation:

To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).

1 kilogram = 2.2 lb

Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)

Molar Mass (As₂O₃): 197.978 g/mol

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb

1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole

6 0

2 years ago

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Calculate the electrical energy per gram of anode material for the following reaction at 298 K:

solmaris [256]

43.8 kJ

<h3>Explanation</h3>

There are two electrodes in a voltaic cell. Which one is the anode?

The lithium atom used to have no oxygen atoms when it was on the reactant side. It gains two oxygen atoms after the reaction. It has gained more oxygen atoms than the manganese atom. Gaining oxygen is oxidation. As a result, lithium is being oxidized.

Oxidation takes place at the anode of a cell. Therefore, the anode of this cell is made of lithium.

Lithium has an atomic mass of 6.94. Each gram of Li would contain 1/6.94 = 0.144 moles of Li atoms. Each Li atom loses one electron in this cell. Therefore, the number of electron transferred, n, equals 0.144 moles for each gram of the anode.

Let $w_\text{max}$ represents the electrical energy produced.

$w_\text{max} = n \cdot F \cdot E_\text{cell}$ , where

n is the number of moles electrons transferred,
F is the Faraday's constant,
E $_\text{cell}$ is the cell potential,

n = 0.144 mol, as shown above, and

F = 96.486 kJ / ( $\text{V} \cdot \text{mol} \; \text{e}^{-}$ ).

Therefore,

$w_\text{max} = n \cdot F \cdot E\\\phantom{w_\text{max}} = 0.144 \times 96.486 \times 3.15 \\\phantom{w_\text{max}} = 43.8 \; \text{kJ}$ .

3 0

3 years ago