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Ratling [72]
2 years ago
14

How many moles are in 19.6 g of Sodium (Na)? And the conversion factor?

Chemistry
1 answer:
Snezhnost [94]2 years ago
6 0

Answer:

n=0.852 moles

Explanation:

Given mass is, m = 19.6 g

The molar mass of sodium is, M = 22.99 u

We need to find the no of moles in 19.6 of Sodium. We know that, no of moles is equal to given mass divided by molar mass.

n=\dfrac{m}{M}\\\\n=\dfrac{19.6}{22.99}\\\\n=0.852

So, there are 0.852 moles in 19.6 g of Sodium.

You might be interested in
1) How many aluminum atoms are there in 3.50 grams of Al2O3?
drek231 [11]

Answer:

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms =  moles aluminium * Number of Avogadro

Aluminium atoms =  0.0686 * 6.022 * 10^23

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

3 0
2 years ago
Why is it important to have regular supervision of the weight and measurements in the market
Sergio [31]

Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

4 0
2 years ago
What is the name for molecular compound N4O8?
zmey [24]
Tetranitrogen octoxide
8 0
2 years ago
Read 2 more answers
Determine the number of moles and mass requested for each reaction in Exercise 4.42.
suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(c) 8 mol NaNO₃ and 680g NaNO₃

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

  1. Step 1: Write the balanced equation
  2. Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
  3. Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.

Step 3:

10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}

10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.

Step 3:

1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}

1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.

Step 3:

128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}

128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.

Step 3:

20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2

[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2[/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.

Step 3:

1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}

5 0
2 years ago
Please Help Me!!
Cerrena [4.2K]

Answer:

1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.

2. Rainwater or Damp/moist air

3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws

Explanation:

1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.

In order for galvanic corrosion to occur, three elements are required.

i. Two metals with different corrosion potentials (anode and cathode)

ii. Direct metal-to-metal electrical contact

iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.

For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.

2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.

3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.

All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.

3 0
2 years ago
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