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jeyben [28]
3 years ago
15

What is the mass defect of lithium? Assume the following:

Chemistry
1 answer:
Bumek [7]3 years ago
7 0

Answer:

Δm = 3.0684

Explanation:

Data Given:

Atomic number of lithium = 3

Atomic mass of lithium = 7.0144 amu

Mass of 1 proton = 1.0073 amu

Mass of 1 neutron = 1.0087 amu

Solution:

Mass Defect:

Mass defect is the difference of mass number of an atom and its atomic number.

Formula used

Δm = [Z (mass of proton + mass of nutron)  + ( A − Z ) mass of nutron] − m of atom

where:

Δm = mass defect (amu)

Z = atomic number

A = mass number

Put values in formula

Δm = [3 ( 1.0073 amu + 1.0087 amu)  + ( 7 − 3 ) 1.0087 ] − 7.0144

Δm = [3 (2.016)  + (4) 1.0087 ] − 7.0144

Δm = [(6.048)  + (4.0348) ] − 7.0144

Δm = 10.0828 − 7.0144

Δm = 3.0684

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Kisachek [45]

Answer:

P_{solution} = 85.3Torr

Explanation:

Raoult's law is a tool that allows to determine vapour pressure of solutions. The formula is:

P_{solution} = X_{solvent}P_{solvent} <em>(1)</em>

Where

P is Pressure of solution and solvent and X is mole fraction.

Moles of solute and solvent are:

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11.5g×(1mol /154.21g) = <em>0.0746mol</em>

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Replacing in (1):

P_{solution} = 0.846*100.84Torr

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I hope it helps!

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What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

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