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4 years ago

How do scientist determine by using frogs to study their environment

Chemistry

2 answers:

NemiM [27]4 years ago

7 0

To show there environment

Lilit [14]4 years ago

6 0

They use frogs because it easy to get and shows what is going on by what they eat what happened to them and other things.

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What is the correct formula for Triphosphorous hexachloride?

zlopas [31]

Answer:

Explanation:

Im pretty sure hope this helps

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2 years ago

Identify the statement that correctly describes light and how it travels?

expeople1 [14]

What statement?
Light moves faster through a vacuum because no particles are existent to absorb it.

8 0

4 years ago

What parts of a divers body are most affected by pressure changes?

Yanka [14]

I would say the head and heart because if you swim deep in the ocean you will feel pressure in your heart and head because the muscles in your body are closing in on them

hope i helped

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3 years ago

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

The natural distribution of the isotopes of a hypothetical element is 60.795% at a mass of 281.99481 u, 22.122% at a mass of 283

fenix001 [56]

Answer: The average atomic mass of the element is 283.291 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 281.99481 amu

Percentage abundance of isotope 1 = 60.795 %

Fractional abundance of isotope 1 = 0.60795

For isotope 2:

Mass of isotope 2 = 283.99570 amu

Percentage abundance of isotope 2 = 22.122 %

Fractional abundance of isotope 2 = 0.22122

For isotope 3:

Mass of isotope 3 = 286.99423 amu

Percentage abundance of isotope 3 = [100 - (60.795 + 22.122)] = 17.083 %

Fractional abundance of isotope 1 = 0.17083

Putting values in equation 1, we get:

$\text{Average atomic mass of element}=[(281.99481\times 0.60795)+(283.99570\times 0.22122)+(286.99423\times 0.17083)]\\\\\text{Average atomic mass of element}=283.291amu$

Hence, the average atomic mass of the element is 283.291 amu

5 0

4 years ago