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mote1985 [20]
3 years ago
10

What mass of copper is equivalent to 0.7moles?​

Chemistry
1 answer:
Shkiper50 [21]3 years ago
7 0
I hope thus equations help


1. How many atoms are in 6.5 moles of zinc?
1 mole = atomic mass (g)
6.5 moles 6.02 x 1023 atoms = 3.9 x 1024 atoms 1 mole
2. How many moles of argon are in a sample containing 2.4 x 1024 atoms of argon?
2.4 x 1024 atoms of argon 1 mole
6.02 x 1023 atoms
3. How many moles are in 2.5g of lithium?
2.5 grams Li 1 mole 6.9 g
4. Findthemassof4.8molesofiron.
4.8 moles 55.8 g 1 mole
= 4.0 mol
= 0.36 mol
= 267.84 g = 270g

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)
MASS (g)
1 mole = molar mass (look it up on the PT!)
MOLE
1 mole = 6.02 x 1023 atoms
PARTICLES (ATOM)
Two Step Problems:
1. What is the mass of 2.25 x 1025 atoms of lead?
2.25 x 1025 atoms of lead 1 mole 207.2g = 6.02 x 1023 atoms 1 mole
2. How many atoms are in 10.0g of gold?
10 g gold 1 mole 6.02 x 1023 atoms = 197.0g 1 mole
PRACTICE PROBLEMS:
1. How many moles are equal to 625g of copper?
625g of copper 1 mol = 9.77 mol Cu 64 g Cu
7744.19g = 7740g
3.06 x 1022 atoms
2. How many moles of barium are in a sample containing 4.25 x 1026 atoms of barium?
4.25 x 1026 atoms of barium 1 mol = 706 mol 6.02 x 1023 atoms
3. Convert 2.35 moles of carbon to atoms.
2.35 moles 6.02 x 1023 atoms = 1.41 x 1024 atoms 1 mole
4. How many atoms are in 4.0g of potassium?
4.0 g 1 mole 6.02 x 1023 atoms = 6.2 x 1022 atoms 39.1 g 1 mole

1 mole = 6.02 x 1023 atoms 1 mole = atomic mass (g)
5. Convert 9500g of iron to number of atoms in the sample.
9500 g Fe 1 mole 6.02 x 1023 atoms = 55.8g 1 mole
6. What is the mass of 0.250 moles of aluminum?
0.250 moles 27.0g = 6.75 g Al 1 mole
7. How many grams is equal to 3.48 x 1022 atoms of tin?
3.48 x 1022 atoms 1 mole 118.7g 6.02 x 1023 atoms 1 mole
8. What is the mass of 4.48 x 1021 atoms of magnesium?
4.48 x 1021 atoms 1 mole 24.3g 6.02 x 1023 atoms 1 mole
9. Howmanymolesis2.50kgoflead?
2.50kg 1000 g 1 mol 1 kg 207.2 g
1.0 x 1026 atoms
= 6.86 g Sn
= 0.181 g Mg or 1.8 x 10-1
= 12.1 mol
10.Find the mass, in cg, of 3.25 x 1021 atoms of lithium.
3.25 x 1021 atoms 1 mol 6.9g 100cg 6.02 x 1023 atoms 1 mol 1 g
= 3.7 cg
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swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

                      B                                   N                         H

mass          40.28 g                         52.20 g                 7.53 g

number of moles  

                 40.28 g / 11 g/mol      52.20 g / 14 g/mol    7.53 g / 1 g/mol

                = 3.662 mol                  = 3.729 mol             = 7.53 mol

divide the number of moles by the least number of moles, that is 3.662

                3.662 / 3.662              3.729 / 3.662              7.53 / 3.662

               = 1.000                           = 1.018                        = 2.056

the ratio of the elements after rounding off to the nearest whole number is

B : N : H = 1 : 1 : 2

therefore empirical formula for the compound is B₁N₁H₂          

that can be written as BNH₂    

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Radon (Rn) is the heaviest, and only radioactive, member of Group 8A(18) (noble gases). It is a product of the disintegration of
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Given data                Atomic mass of Ra= 226g/mol

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no. of atoms in 0.044moles

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= 0.044x 6.022 x10^23                  = 0.264968 x 10^22

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n(Rn)=5,25*10−9                              pV=nR*T

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V=5.25∗10^−9 ∗ 8.314 ∗ 273.15  /  101325

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Explanation:

How do I describe an ion?

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