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TiliK225 [7]
2 years ago
9

Atoms or helium nickel would represent _____

Chemistry
1 answer:
den301095 [7]2 years ago
6 0

Answer:

they represent their own atoms

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In 1911, Ernest Rutherford tested the atomic model existing at the time by shooting a beam of alpha particles (42He, helium nucl
STatiana [176]

Answer:

At the time of Rutherford's experiment, the accepted model for the atom was the Thomson plum-pudding model of the atom, in which the atom consists of a "sphere" of positive charge distributed all over the sphere, with tiny negative particles (the electrons) inside this sphere.

In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

- The positive charge of the atom was all concentred in a tiny nucleus

- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

5 0
3 years ago
What mass of oxygen (O2) forms in a reaction that forms 15.90 g C6H12O6? (Molar mass of O2 = 32.00 g/mol; molar mass of C6H12O6
Dafna11 [192]

The answer is: the mass of oxygen is 16.95 grams.

The overall balanced photosynthesis reaction:  

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.  

m(C₆H₁₂O₆) = 15.90 g; mass of glucose.

n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).

n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.

n(C₆H₁₂O₆) = 0.088 mol; amount of glucose.

From chemical reaction: n(C₆H₁₂O₆) : n(O₂) = 1 : 6.

n(O₂) = 6 · 0.088 mol.

n(O₂) = 0.53 mol; amount of oxygen.

m(O₂) = 0.53 mol · 32.00 g/mol.

m(O₂) = 16.95 g; mass of oxygen.

5 0
3 years ago
Read 2 more answers
For the following reaction, calculate how many moles of NO2 forms when 0.356 moles of the reactant completely reacts.
Nesterboy [21]

Answer:

0.712 mol of NO₂ are formed .

Explanation:

For the reaction , given in the question ,

2 N₂O₅ ( g )  →  4 NO₂ ( g ) + O₂ ( g )

From the above balanced reaction ,

2 mol of N₂O₅  reacts to give 4 mol of NO₂

Applying unitary method ,

1 mol of N₂O₅  reacts to give 4 / 2 mol of NO₂

From the question , 0.356 mol of N₂O₅ are reacted ,

<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>

Since ,

1 mol of N₂O₅  reacts to give 4 / 2 mol of NO₂

0.356 mol of N₂O₅  reacts to give 4 / 2 * 0.356 mol of NO₂

Calculating ,

0.712 mol of NO₂ are formed .

7 0
3 years ago
Sugar dissolving in warm water is chemical or phyiscal change
BlackZzzverrR [31]
Hey there! Great question;) Answer:Physical change Explanation: When sugar mixes with water, at the end, the chemical formulas are the same. Nothing has changed! I hope this helps;)
5 0
3 years ago
43. A stock glucose standard has a concentration of 1,000 mg/dL. A 1/5 dilution of this standard is made. What would be the fina
Nat2105 [25]

The final concentration of the diluted standard is 0.2 mg/dL.

<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>

Using the dilution formula:

  • C1V1 = C2V2

where

  • C1 is initial concentration
  • V1 initial volume
  • C2 is final concentration
  • V2 is final volume.

Assuming a final volume of 100 mL, and since a 1/5 dilution is made:

C1 = 1.00 mg/dL

V1 = 20

C2 = ?

V2 = 100 mL

C2 = C1V1/V2

C2 = 20 × 1/100

C2 = 0.2 mg/dL

Therefore, the final concentration of the diluted standard is 0.2 mg/dL.

Learn more about dilution at: brainly.com/question/24881505

6 0
2 years ago
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