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AnnyKZ [126]
3 years ago
6

PLZZZ HELP

Chemistry
1 answer:
azamat3 years ago
7 0
The cooling of water vapor enough to return to a liquid state is condensation.
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The common titanium alloy known as T-64 has a composition of 90 weight% titanium 6 wt% aluminum and 4 wt% vanadium. Calculate th
Anna007 [38]

Explanation:

Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.

Mass of titanium = 90 g

Moles of titanium = \frac{90 g}{47.87 g/mol}=1.8800 mol

Total number of atoms of titanium ,a_t=1.8800 mol\times N_A

Mass of aluminum = 6 g

Moles of aluminium = \frac{6 g}{26.98 g/mol}=0.2223 mol

Total number of atoms of aluminium,a_a=0.2223 mol\times N_A

Mass of vanadium  = 4 g

Moles of vanadium= \frac{4 g}{50.94 g/mol}=0.0785 mol

Total number of atoms of vanadiuma_v=0.0785 mol\times N_A

Total number of atoms in an alloy = a_t+a_a+a_v

Atomic percentage:

Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100

Atomic percentage of titanium:

:\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%

Atomic percentage of Aluminium:

:\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%

Atomic percentage of vanadium

:\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%

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ivolga24 [154]
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lara31 [8.8K]

Answer:

Explanation:

think the answer is exact criteria that can be used to assess a particular assignment but i'm not 100% sure

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