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Ad libitum [116K]
3 years ago
11

How many atoms are in 35.1 g of sodium chloride (NaCl) ?

Chemistry
1 answer:
hoa [83]3 years ago
8 0

Answer:

I believe that there are 70.2

Explanation:

There are 2 atoms in sodium chloride so I x 35.1 by 2 and got 70.2

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What group of professionals specialize in designing, testing, and building the products of technology?
swat32

Answer:

2

Explanation:

4 0
3 years ago
By what mechanism does cyclohexanol react when treated in sulfuric acid and what compound results?A) E 1; methoxycyclohexane B)
tensa zangetsu [6.8K]

Answer:

E 1: cyclohexene

Explanation:

This reaction is an example of the dehydration of cyclic alcohols. The reaction proceeds in the following steps;

1) The first step of the process is the protonation of the cyclohexanol by the acid. This now yields H2O^+ attached to the cyclohexane ring.

2) the water molecule, which a good leaving group now leaves yielding a carbocation. This now leaves a cyclohexane carbocation which is highly reactive.

3) A water molecule now abstracts a proton from the carbon adjacent to the carbocation leading to the formation of cyclohexene and the regeneration of the acid catalyst. This is an E1 mechanism because it proceeds via a carbocation intermediate and not a concerted transition state, hence the answer.

8 0
3 years ago
Determine the molecular
jeyben [28]

Answer:

MgO- magnesium oxide

Cu(NO3)2- copper(11)nitrate

Li2CO3- lithium carbonate

7 0
3 years ago
How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

Learn more about precipitation here:

brainly.com/question/13859041

#SPJ4

6 0
1 year ago
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
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