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Luba_88 [7]
2 years ago
7

In 1909, Ernest Rutherford performed an experiment to explore the atomic structure. In his experiment, he projected high-speed α

‎ particles onto a thin gold foil. He found that all α‎ particles did not follow the same path. Most of the particles passed through the foil without any scattering, implying that most of the space in an atom is empty. Some particles were scattered at a large angle, and very few of them scattered back in the direction from which they had come. Based on these observations, Rutherford proposed an atomic model, which is known as Rutherford’s atomic model.
O True
O False
Chemistry
1 answer:
xz_007 [3.2K]2 years ago
6 0

Answer:

True

Explanation:

In 1909 Ernest Rutherford proposed to Geiger and Marsden to conduct an experiment in which they would have to launch alpha particles from a radioactive source against a gold foil a few atoms thick.

The aim of the experiment was to corroborate Rutherford's idea that the particles would pass through the metal foil with little deviation.

Rutherford's interpretation of the experimental results gave rise to a new atomic model, and concluded that the mass of the atom was concentrated in a small region of positive charges that impeded the passage of alpha particles. He suggested a new model in which the atom had a nucleus or center in which the mass and the positive charge are concentrated, and that in the extra nuclear zone are the negatively charged electrons.

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On the periodic table, in which period is copper?
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3 0
2 years ago
if a plot weight (in g) vs. volume (in ml) for a metal gave the equation y= 13.41x and r^2=0.9981 what is the density of the met
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8 0
3 years ago
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The compound adrenaline contains 56.79% c, 6.56% h, 28.37% o, and 8.28% n by mass. what is the empirical formula for adrenaline?
Phoenix [80]
To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:<span>
         mass        moles
C       56.79        4.73
H       6.56          6.50
O       28.37        1.77
N      8.28           0.59

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

</span>         moles                 ratio
C       4.73     / 0.59       8
H       6.50     / 0.59      11
O       1.77     / 0.59       3
N        0.59    / 0.59       1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
8 0
3 years ago
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