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zalisa [80]
2 years ago
6

An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy. True False

Chemistry
1 answer:
andrew11 [14]2 years ago
8 0

The statement "An atom with high ionization energy will form a positive ion more easily than an atom with low ionization energy" is false.

In this context , we will define ionization energy as the minimum energy required to remove a valence electron from a neutral atom in it's gaseous state.  In a sense the ionization energy is a measure the amount of 'difficulty' of making something an ion. A high ionization energy means that it takes a lot of energy to remove a valence electron from that atom. A low ionization energy means that it is easy to remove a valence electron from the atom. It is known that group 1 elements generally have a low ionization energy.  On the other hand, it is harder for noble gasses and group 7 atoms to loose electrons because they have higher ionization energy.

To form a positive ion, you have to remove an electron. When an electron is removed from an atom, there ion formed has more positive charges than negative charges in it, making it net positive. We have established that atoms with low ionization energy loose elections much more easily. We have also established that atoms with high ionization energy do not loose electrons easily. From this we can gather that the statement is false. An atom with high ionization energy will not form a positive ion more easily that an atom with low ionization energy.

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I went for a walk the other day. I went four blocks east, then seven blocks south, then one block west and finally
nalin [4]

Answer:

a) distance is 4+7+1+8=20 blocks

b) displacement is 10 blocks

Explanation:

find displacement: x and y

x axis displacement = 4-1 = 3 blocks

y axis displacement = -7+8= 1 block

displacement = the square root of 3^2 + 1^2

= 9+1 = 10 blocks.

You can find the angle of displacement with respect to the initial position using trig identities, if you wish.

4 0
2 years ago
What did Kwang Jeon discover when researching amoeba infected by bacteria?
defon
<span>Kwang Jeon observed that Amoeba had been attacked by a bacterial infection, and lots of the Amoeba had died. However, some survived and continued to reproduce. After investigating the remaining Amoeba and their offspring, he noticed they were very healthy. He thought maybe they were able to fight off the bacteria, but instead, he found they were still infected with the bacteria but were not dying. The bacteria were no longer making the Amoeba sick. Then, he killed off the bacteria using antibiotics and was surprised to see that the Amoeba also died. It seemed the Amoeba and bacteria had formed a relationship in which they both needed each other to survive. After researching, Jeon found that the bacteria made a protein that the Amoeba needed to survive. </span>
6 0
2 years ago
After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid
nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\

Then, we add the half reactions:

2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0
2 years ago
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