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777dan777 [17]
3 years ago
8

What is the mass (in grams) of 169 mole of CF4

Chemistry
1 answer:
Sergeu [11.5K]3 years ago
3 0
The answer will depend on the number of significant figures you want in your answer, so I can't give you a solid answer, but the process is still the same: you convert from grams to moles by using the molar mass.

You can calculate the molar mass of carbon tetrafluoride from the periodic table, by adding the molar mass of 1 atom of C and 4 atoms of F. Round the total molar mass to the appropriate number and multiply 169 moles by the molar mass you just calculated.

Hope this helps!
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7 0
3 years ago
A 45ml of a 4M solution of CaBr2 contains how many grams of CaBr2?
natta225 [31]

The required mass of calcium bromide is 35.98 grams.

<h3>What is molarity?</h3>

Molarity is any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V, where

  • M = molarity = 4M
  • V = volume = 45mL = 0.045L

Moles will be calculated by using the above equation as:

n = (4)(0.045) = 0.18 mole

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g

Hence required mass of CaBr₂ is 35.98 grams.

To know more about molarity, visit the below link:
brainly.com/question/22283918

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6 0
2 years ago
Answer asap with at least 3 or more sentences!
mixas84 [53]

Answer:

no.

Explanation:

The reason this has

never happened is due to the source of magnetic fields:  moving electric

charges.  When electric charges (e.g. electrons) move in circles, they

produce a magnetic field.  In a piece of iron, it is very easy to line up

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magnet.

For each of these circles, one side is the north pole and one side is the

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6 0
3 years ago
Read 2 more answers
what happens if you fail a quarter in high school, does that credit not count anymore ? and do i have to retake the class if i p
vampirchik [111]
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6 0
3 years 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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