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Ad libitum [116K]
3 years ago
15

Rutherford and his colleagues fired a beam of small, positively charged alpha particles at thin gold foil. Some of these alpha p

articles were deflected at angles as they passed through the foil.
What conclusion did Rutherford and his colleagues draw from their results?
A) The atom must contain some neutral charge spread throughout the atom. He called these particles neutrons.
B) The atom must contain some centrally located negative charge. He called this core of negative charge the nucleus.
C) The atom must contain some centrally located positive charge. He called this core of positive charge the nucleus.
D) Thomson's plum pudding model of the atom was correct. The atom contains positive charge spread throughout its structure.
Chemistry
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

C

Explanation:

The questions states that some particles were deflected meaning both the nucleus and the particles were the same charge because they repelled. Using common knowledge, we know that the nucleus is positively charged. Using this information, you can immediately eliminate choices A and B. Choice D says that the positive charge was spread throughout its structure. This is false because protons and neutrons are only found in the nucleus. Electrons are the subatomic particles found outside the nucleus, thus you can get rid of choice D, only leaving you with choice C.

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Answer:

Solar energy.

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2 years ago
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Identify the products formed in this Brønsted-Lowry reaction.<br><br> HPO2−4+NO−2↽−−⇀acid+base
klemol [59]

Answer: The products formed in this Bronsted-Lowry reaction are HNO_{2} and PO^{2-}_{4}.

Explanation:

According to Bronsted-Lowry, acids are the species which donate hydrogen ions to another specie in a chemical reaction.

Bases are the species which accept a hydrogen ion upon chemical reaction.

For example, HPO^{2-}_{4} + NO^{-}_{2} \rightleftharpoons HNO_{2} + PO^{2-}_{4}

Here, the products formed in this Bronsted-Lowry reaction are HNO_{2} and PO^{2-}_{4}.

Thus, we can conclude that the products formed in this Bronsted-Lowry reaction are HNO_{2} and PO^{2-}_{4}.

5 0
3 years ago
The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p
earnstyle [38]

Answer:

The frequency is  f =  2,01 * 10^{15} \  Hz

Explanation:

From the question we are told that

   The energy required to ionize boron is E_b  =  801 KJ/mol

Generally the ionization energy of boron pre atom is mathematically represented as

     E_a  =  \frac{E_b}{N_A}

Here  N_A is the Avogadro's constant with value N_A  =  6.022*10^{23}

So

      E_a  =  \frac{801}{6.022*10^{23}}

=>     E_a  =  1.330 *10^{-18} \  J/atom

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

       E =  hf  =  E_a

=>     hf  =  E_a

Here h is the Planks constant with value h = 6.626 *10^{-34}

So

       f =  \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}

=>      f =  2,01 * 10^{15} \  Hz

8 0
2 years ago
meteorologist preparing a talk about global warming compiled a list of weekly low temperatures (in degrees Fahrenheit) he observ
Setler [38]

Answer:

Regarding Measures of center, the Mean will be smaller, and Median will not be affected. Thus, the median will not change, since the temperature was already low initially and only decreased. However, the median takes the average of all values into account, and if you substitute a value for a smaller value, the average decreases.

In relation to Measures of spread, the range and standard deviation will be larger, and the IQR will not change. Therefore, the range is the highest minus the lowest number; if the lowest number decreases sharply, the range increases. On the other hand, the IQR only counts the first and third quartile values. Therefore, since the value was already the lowest, if it is decreased, it will not change the IQR. However, the standard deviation will increase significantly, as the decreasing value of the lowest data point will increase the overall spread.

Explanation:

6 0
2 years ago
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without
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When ammonium chloride NH4Cl is added to water and stirred, it dissolves spontaneously (this is the basis for ΔG) for and the resulting solution feels cold (endothermic, the basis for ΔH). Without doing any calculations, we can easily deduce the signs of ΔG, ΔH, and ΔS for this process based on the observations.

ΔG < 0 (it is spontaneous)
ΔH < 0 (because the process is endothermic - it absorbs energy)
ΔS > 0 (entropy increases because of the dissolution of NH4Cl in water
7 0
2 years ago
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