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Arlecino [84]
2 years ago
13

What method is used for chaff from grain​

Chemistry
1 answer:
Marta_Voda [28]2 years ago
4 0

Answer:

Winnowing

Explanation:

Wind blows the lighter(in terms of mass) chaff from the whole grains,which are heavier(in terms of mass)

You might be interested in
CHEMISTRY HELP PLEASE *answer all questions please*
Lerok [7]

Answer:

<u><em>Question 1: </em></u>

A) 0.289 moles.

B) 1.74 x 10²³ atoms.

<u><em>Question 2:</em></u>

A) 0.30 moles.  

B) it contains 0.3 moles of both Na and Cl.

C) it contains 6.023 x 10²³ atoms of both Na and Cl.

<u><em>Question 3:</em></u>

A) The number of moles of sucrose (C₁₂H₂₂O₁₁) ≅ 0.0228 moles.

B) The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2763 mole of C atoms.

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = 0.5016 mole of H atoms.

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2508 mole of O atoms.

C) The number of C atoms = 1.65 x 10²³ atoms.

The number of H atoms = 3.02 x 10²³ atoms.

The number of O atoms = 1.51 x 10²³ atoms.

Explanation:

<u><em>Question 1:</em></u>

A) The number of moles of Au in 57.01 g sample:

n = mass / molar mass,

mass = 57.01 g and molar mass = 196.966 g/mol e.

The number of moles of Au in the sample = (57.01 g) / (196.966 g/mole) = 0.289 moles.

B) The number of atoms of Au in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of Au → 6.023 x 10²³ atoms

0.289 mole of Au → ???? atoms

<em>using cross multiplication: </em>

The number of atoms of Au in the sample = (6.023 x 10²³ x 0.289 mole) / (1.0 mole) = 1.74 x 10²³ atoms.


<u><em>Question 2:</em></u>

A) The number of moles of 17.45 g of NaCl:

n = mass / molar mass,

mass = 17.45 g and molar mass = 58.44 g/mole.

The number of moles of NaCl = (17.45 g) / (58.44 g/mole) = 0.298 mole ≅ 0.30 moles.

B) The number of moles of each element in NaCl  

NaCl → Na + Cl

Each mole of NaCl contains one mole of Na and one mole of Cl.

<em><u>using cross multiplication: </u></em>

1.0 mole NaCl → 1.0 mole Na

0.3 mole NaCl → ??? mole Na

The number of moles of Na atoms in NaCl = (1.0 mole Na x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Na atoms.

by the same way; the number of moles of Cl atoms = (1.0 mole Cl x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Cl atoms.

C) The number of atoms of each element in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of NaCl → 6.023 x 10²³ molecules

0.3 mole of NaCl → ???? molecules

<em><u>using cross multiplication:</u></em>

The number of molecules in 0.3 mole of NaCl = (6.023 x 10²³ x 0.3 mole) / (1.0 mole) = 1.8069 x 10²³ molecules.

Every molecule of NaCl contains one atom of Na and one atom of Cl.

So, it contains 6.023 x 10²³ atoms of both Na and Cl.


<u><em>Question 3:</em></u>

A) The number of moles of 7.801 g of sucrose (C₁₂H₂₂O₁₁):

n = mass / molar mass,

mass = 7.801 g and molar mass = 342.3 g/mole.

The number of moles of sucrose (C₁₂H₂₂O₁₁) = (7.801 g) / (342.3 g/mol) = 0.022789 mol ≅ 0.0228 moles.

B) The number of moles of each element in sucrose (C₁₂H₂₂O₁₁):

C₁₂H₂₂O₁₁ → 12C + 22H + 11O

Each mole of sucrose contains 12 moles of C, 22 moles of H, and 11 moles of O.

  • <em><u>using cross multiplication: </u></em>

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 12.0 moles C

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles C

The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = (12.0 moles C x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2763 mole of C atoms.

  • By the same way; the number of moles of H atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 22.0 moles H

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles H

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = (22.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.5016 mole of H atoms.

  • Also; the number of moles of O atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 11.0 moles O

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles O

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = (11.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2508 mole of O atoms.

C) The number of atoms of each element in the sucrose (C₁₂H₂₂O₁₁) sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 6.023 x 10²³ molecules

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ???? molecules

<em><u>using cross multiplication: </u></em>

The number of molecules in 0.0228 mole of sucrose (C₁₂H₂₂O₁₁) = (6.023 x 10²³ x 0.0228 mole) / (1.0 mole) = 1.273 x 10²² molecules.

Each molecule of sucrose contains 12 atoms of C, 22 atoms of H, and 11 atoms of O.

So, the number of each atom that the sucrose (C₁₂H₂₂O₁₁) sample contains are:

The number of C atoms = (12 x 1.273 x 10²² molecules) = 1.65 x 10²³ atoms.

The number of H atoms = (22 x 1.273 x 10²² molecules) = 3.02 x 10²³ atoms.

The number of O atoms = (11 x 1.273 x 10²² molecules) = 1.51 x 10²³ atoms.

6 0
3 years ago
What is the concentration of a solution made by diluting 25ml of 6.0M HCI to a final volume of 750
Varvara68 [4.7K]

Answer:

Explanation:

The

7 0
3 years ago
46. True or false. Chemical reactions do not involve changes in the chemical bonds that join
Naddika [18.5K]

Answer:

Chemical reactions do not involve changes in the chemical bonds that join

atoms in compounds :

<u>False</u>

Explanation:

Chemical reaction are the reaction in which old bonds break and new  bonds are formed . The formation of new bond result in formation of new compounds . This happen because new bond are result of linking different atoms by the bond.

For example : Water formation from Oxygen and Hydrogen is a chemical process :

2H_{2}+O_{2}\rightarrow 2H_{2}O

Original(old) bonds are :

H-H bond in H2 and O-O bonds in O2

In H2 = Hydrogen is joined to Hydrogen

IN O2 = Oxygen is joined to oxygen

New Bonds =

O-H bonds in water (H2O)

Oxygen is joined to hydrogen = New Bond formation

Hence,

<u>Chemical reactions do involve changes in the chemical bonds that join </u>

<u>atoms in compounds</u>

<u />

3 0
3 years ago
Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0
3 years ago
How many grams of iron oxide, Fe2O3 will be produced if 165 g of O2 gas is supplied? (follow the same steps as mol to mol, only
forsale [732]

Answer:

m_{Fe_2O_3}=549gFe_2O_3

Explanation:

Hello there!

In this case, according to the given chemical reaction for this problem about stoichiometry:

4Fe+3O_2\rightarrow 2Fe_2O_3

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3}  \\\\m_{Fe_2O_3}=549gFe_2O_3

Regards!

8 0
3 years ago
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