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wariber [46]
1 year ago
8

Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of

its aircraft. A commercial aircraft skin material having a density of 2.42 g/cm3 is desired. Compute the concentration of Li (in wt%) that is required. The densities of aluminum and lithium are 2.71 and 0.534 g/cm3, respectively.
Chemistry
1 answer:
gtnhenbr [62]1 year ago
6 0

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y <em>(1)</em>

<em>Where X is molar fraction of Al and Y is molar fraction of Li.</em>

X + Y = 1 <em>(2)</em>

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= <em>3,47%</em>

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goldenfox [79]

Answer:

m C2H5OH = 191.52 g

Explanation:

  • proof ≡ (2)(%v/v C2H5OH)

∴ %v/v C2H5OH = ( v C2H5OH / v sln)×100

⇒ 48 proof = (2) (%v/v C2H5OH)

⇒ 48/2 = %v/v C2H5OH

⇒ 24 = %v/v C2H5OH

⇒ 24/100 = v C2H5OH / v sln = 0.24

∴ v sln (gin) = 1.00 L

⇒ v C2H5OH = ( 0.24 )( 1.00 L )

⇒ v C2H5OH = 0.24 L = 240 mL

⇒ m C2H5OH = (240 mL)(0.798 g/mL)

⇒ m C2H5OH = 191.52 g

7 0
2 years ago
Calculate the nuclear binding energy for 5525mn in megaelectronvolts per nucleon (mev/nucleon). express your answer numerically
VARVARA [1.3K]

Answer:

8.533 Mev/nucleon

Explanation:

The given elements is:- ^{55}_{25}Mn

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, the number of protons = 25

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

55 =  25 + Number of neutrons

Number of neutrons = 30

Mass of neutron = 1.008665 amu

Mass of proton = 1.007825 amu

Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron

Thus,  

Calculated mass = (25*1.007277 + 30*1.008665) amu = 55.441875 amu

Actual mass = 54.9380451 amu

Mass defect = Δm = |54.9380451 - 55.441875| amu = 0.5038299 amu

The conversion of amu to MeV is shown below as:-

1 amu = 931.5 MeV

So, Energy = 0.5038299*931.5 MeV= 469.317552 MeV

Total number of nucleons in the atoms = 55

<u>So, Energy = 469.317552/55 MeV/nucleon = 8.533 Mev/nucleon</u>

5 0
2 years ago
The temperature of 6.24 l of a gas is increased from 25.0°c to 55.0°c at constant pressure. the new volume of the gas is
bixtya [17]
At constant pressure, temperature is directly proportional to the volume and vise versa. The formula will be

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature and V2 and T2 are the final volume and temperature. The temperature is in Kelvin and to convert Celsius to Kelvin add 273.

so, 6.24L/298 = V2/328
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6 0
2 years ago
Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. The initial rea
Dima020 [189]

Answer : The correct option is, (C) ^{58}\textrm{Fe}

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma

Process 2 : Beta emission.

_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e

Process 3 : Neutron capture.

_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma

From this we conclude that, the initial reactant in the production of cobalt-60 is _{26}^{58}\textrm{Fe}

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7 0
2 years ago
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To determine what elements are represented by the electron configuration given above, we need to know the sum of the exponents of each term or subshell involved in the configuration as this represent the atomic number of the element.

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<span>1s2 2s2 2p6:                                      2 + 2 + 6 = 10                         neon
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1s2 2s2 2p6 3s2 3p6 4s1: </span>2 + 2 + 6 + 2 + 6+1 = <span>19                    potassium
1s2 2s2 2p6 3s2 3p6 4s2 3d8:              20 + 8 =  28                    nickel
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3 0
2 years ago
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