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3 years ago

You have 4.72x10^25 atoms of a substance with MW of 136.94 g/mol How many moles do you have

Chemistry

1 answer:

slega [8]3 years ago

4 0

Number of moles = 78.41

<h3>Further explanation </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

4.72 x 10²⁵ atoms of a substance, then moles :

$\tt n=\dfrac{N}{No}\\\\n=\dfrac{4.72\times 10^{25}}{6.02\times 10^{23}}\\\\n=78.41$

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Marat540 [252]

Observation consists of receiving knowledge of the outside world through our senses,or recording information using scientific tools and instruments.Any data recorded during an experiment can be called and observation.

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3 years ago

Using table 9.4, calculate an approximate enthalpy (in kj) for the reaction of 1.02 g gaseous methanol (ch3oh) in excess molecul

Tanya [424]

<u>Given:</u>

Mass of methanol = 1.02 g

<u>To determine:</u>

Enthalpy for the reaction of 1.02 g of methanol with excess O2

<u>Explanation:</u>

Balanced equation-

2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

The reaction enthalpy is given as:

ΔHrxn = ∑nH°f(products) - ∑nH°f(reactants)

where n = number of moles

H°f = standard enthalpy of formation.

ΔHrxn = [2H°f(CO2(g)) + 4H°f(H2O(g))] - [2H°f(CH3OH(g)) + 3H°f(O2(g))]

= [2(-393.5) + 4(-241.8)]-[2(-201.5) + 3(0)] = -1351.2 kJ

Now, 1 mole of CH3OH = 32 g

The calculated ΔHrxn corresponds to 2 moles of CH3OH. i.e.

The enthalpy change for 64 g of Ch3OH = -1351.2 kJ

Therefore, for 1.02g gaseous methanol we have:

ΔH = 1.02 * -1351.2/64 = -21.5 kJ

Ans: The enthalpy for the given reaction is -21.5 kJ

6 0

3 years ago

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following me

Anika [276]

Answer:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Explanation:

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.

Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

6 0

3 years ago

when mercury ii oxide is heated it decomposes into mercury and oxygen how many molecules of oxygen gas are produced if 12.5g of

natali 33 [55]

The chemical reaction would be written as:

2HgO = 2Hg + O2

We use this reaction and the amount of the reactant to calculate for the moles of oxygen produced. THen, we use avogadro's number to convert it to molecules. We do as follows:

12.5 g HgO (1 mol / 216.59 g) (1 mol O2 / 2 mol HgO) ( 6.022x10^23 molecules O2 / 1 mol O2 ) = 1.74x10^22 molecules O2 produced

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3 years ago

A liquid has a density of 1g/mL. If you have 50mL of the liquid, what would its mass be?

Svetlanka [38]

Answer:

<h2>The answer is 50 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of liquid = 50 mL

density = 1g/mL

The mass of the liquid is

mass = 50 × 1

We have the final answer as

50 g

Hope this helps you

8 0

3 years ago