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katovenus [111]
3 years ago
13

How do the fusion reaction in the sun compare to the fusion occurring in large atars and supernovas

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
6 0
The fusion reaction in the sun is a combination of hydrogen atoms fusing to create helium. The fusion reaction in larger stars involve much heavier elements like oxygen and iron. In supernovas, often elements like gold are produced
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HELP PLEASE THE OTHER 'ANSWER' ISNT EVEN AN ANSWER!
hodyreva [135]

Answer:

most likely that (2) the replicated experiment was performed incorrectly.

Why, u ask? u dare question me:

1- The initial experiment invalidness cannot be proven.

2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>

3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.

4- Already knowing the data and errors would increase the precision of the replicated experiment.

5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.

happy learning!

4 0
3 years ago
Which of the following is the correct chemical formula for cs and br? csbr cs2br csbr2
nalin [4]
The correct chemical formulae is CsBr
4 0
3 years ago
Read 2 more answers
Which pair has identical electron configurations?
Thepotemich [5.8K]

Identical electron configurations : K⁺ and Cl⁻

<h3>Further explanation  </h3>

In an atom, there are levels of energy in the shell and sub-shell  

This energy level is expressed in the form of electron configurations.  

Charging electrons in the sub-shell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

  • S²⁻ and CI

S²⁻ : [Ne] 3s²3p⁶

Cl : [Ne] 3s²3p⁵

  • K⁺ and CI⁻

K⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶

Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶

  • S and Ar

S :[Ne] 3s²3p⁴

Ar : [Ne] 3s²3p⁶

  • Cl⁻ and K

Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶

K : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹

6 0
3 years ago
At the melting point of a substance, temperature ______ as heat is being added and the substance is changing from a solid to a l
madam [21]
The temperature Increases
4 0
2 years ago
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Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi
Marizza181 [45]
  • Answer:  3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced.  Explanation:  Ostwald process  is a multi-step for manufacturing Nitric Acid.  First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow.  Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l)  Step 2: NO(g) + O2(g) ⟶ NO2(g)  Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g)  For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms.  So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides.   Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)  Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g)  Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g)   Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg  and Molecular Weigth = 63,01g/mol  NH3  Molecular Weight= 17,031 g/mol  Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights.  7,839Kg  of HNO3  x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3   The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3
7 0
3 years ago
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