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Dafna11 [192]
3 years ago
5

If the average atomic mass of hydrogen in nature is 1.0079, what does that tell you about the percent composition of H-1 and H-2

in nature
Chemistry
1 answer:
vovangra [49]3 years ago
4 0

Answer:

That the isotope H-1 is the most abundant in nature.

Explanation:

Hello!

In this case, since the average atomic mass of an element is computed considering the mass of each isotope and the percent abundance each, for hydrogen we would set up something like this:

m_H=m_{H_1}*\%abund_{H_1}+m_{H_2}*\%abund_{H_2}

Moreover, since the isotope notation H-1 and H-2 means that the atomic mass of H-1 is 1 amu, that of H-2 is 2 amu and the average one is 1.0079 amu, we can infer that the most of the hydrogen in nature is H-1 as the most of it composes the average hydrogen atom.

Best regards!

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As an electron gains energy, it becomes excited and jumps to a lower energy level?
Maru [420]

Yes if you add an energy to an electron the electron will become excited, and it will jump to its highest level then go back down releasing  energy


3 0
3 years ago
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If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac
MakcuM [25]

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

5 0
3 years ago
How many moles of carbonate are there in sodium carbonate​
vaieri [72.5K]

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>
  • The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

  • no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

  • Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

  • = 46 + 12 + 48 = 106g/mol

  • mass of CO3 = 12 + 48 = 60g

  • no. of moles of CO3 = 60/106

  • no. of moles of CO3 = 0.566mol

  • Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

5 0
2 years ago
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35. Explain what the Triple Point is and why the Triple
Lady_Fox [76]

Answer:

In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases of that substance coexist in thermodynamic equilibrium. It is that temperatureand pressure at which the sublimation curve, fusion curve and vaporisation curve meet.

7 0
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The question is in the picture below
Rus_ich [418]

Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0
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