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IgorLugansk [536]
2 years ago
10

Một hỗn hợp X gồm CH3OH; C2H5OH; phenol có khối lượng 28,9 gam phản ứng vừa hết với 100ml dung dịch NaOH 2M. Phần trăm theo khối

lượng phenol là ? Biết C = 12; O = 16; H = 1; Na = 23. *
Chemistry
1 answer:
slavikrds [6]2 years ago
3 0

Answer:

nNaOH = 0,2 mol

Trong hỗn hợp các chất đề bài cho chỉ có phenol tác dụng với NaOH

C6H5OH + NaOH → C6H5ONa + H2O

   0,2  ←     0,2 (mol)

⟹ mphenol = 0,2.94 = 18,8 gam

⟹  

%mC6H5OH=18,8/28,9 x100%= 65,05%

Explanation:

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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
Which atom would be expected to have a half-filled 6s subshell
eduard

Answer: I think the answer is Cesium (Cs)

Explanation:

A half-filled 6s subshell would be 6s^1

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Max Planck discovered the quantum of action

Explanation:

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If you have 40 grams of potassium nitrate in 100 grams of water at 20 C:
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Answer:

Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:

[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.

So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.

To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.

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How can the requirements of Na (sodium) solve the requirements of Cl (chloride) instability achievement?
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