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alexandr402 [8]

3 years ago

14

When one form of energy is converted to another form in any physical or chemical change, energy input alwasy equals energy outpu

t?

1 answer:

aliya0001 [1]3 years ago

5 0

In the context of chemistry, yes. Energy input is always equal to the energy output.

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How many electrons in the same atom can share the quantum number n = 2?

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4 years ago

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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 vanadium $a_v=0.0785 mol\times N_A$

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

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

6 0

4 years ago