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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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Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:
rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

PV=nRT

n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of  water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

P'V'=n'RT'

n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol

CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

\frac{3}{1}\times 0.6402 mol=1.9208 mol of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

\frac{1}{22.4 L}\times 26 L= 1.1607 mol

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

\frac{Experimental}{Theoretical}\times 100

\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%

60.42% is the percent yield of the reaction.

8 0
3 years ago
One method for preparing pure iron from Fe2O3 is by reaction with carbon monoxide.Fe2O3(s)+CO(g)→Fe(s)+CO2(g) How many moles of
larisa [96]

Answer:

The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.

Explanation:

Given data:

Mass of Fe₂O₃ = 122 Kg ( 122×1000 =  122000 g)

Moles of CO = ?

Solution:

Chemical equation:

Fe₂O₃ + 3CO  → 2Fe + 3CO₂

Number of moles of  Fe₂O₃:

Number of moles = mass/ molar mass

Number of moles = 122000 g /159.69 g/mol

Number of moles = 764 mol

Now we will compare the moles of Fe₂O₃ with CO.

                               Fe₂O₃            :            CO  

                                  1                  :             3

                               764                :             3×764  =2292 mol

The 2292 moles of CO are needed to react completely with 122 Kg of Fe₂O₃.

6 0
3 years ago
Choose the nonmetallic elements from the list. Check all that apply yttrium: oxygen: boron: polonium: argon: gallium: carbon:
BaLLatris [955]

Answer:

B, E, & G

Oxygen, Argon, & Carbon

7 0
3 years ago
Read 2 more answers
Each period in the periodic table corresponds to a
Rudiy27
Each correspond to a principal energy level
4 0
3 years ago
Read 2 more answers
Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
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