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podryga [215]
3 years ago
9

10th grade Physical Science.... Help plss

Physics
1 answer:
Alekssandra [29.7K]3 years ago
5 0

1) A negatively charged ion is chloride

2) Moving from left to right, valence electrons increase by one.

3) The period number gives information about how many energy levels it has

4) Fluorine has a charge of 1–

5) Potassium and iodine form an ionic bond

The periodic table is an arrangement of elements into groups and periods based on their periodic properties.

In the periodic table, elements are arranged in groups and periods. There are 18 groups and 8 periods.

Chlorine is in group 17, there have seven outermost electrons hence the chlorine atom needs only one more electron in order to attain a stable octet. This is done by accepting one electron to form the negatively charged chloride ion.

As we move from one period to another, one extra electron is added to the outermost shell of elements. Hence, the valence electrons increases by one.

The period to which an element belongs shows you the number of shells or energy levels in the atom of that element.

Fluorine is in group 17. One electron is needed to achieve a stable octet. When an atom accepts one electron, its charge is 1–.

Bonding based on ionic charges occurs between metals and nonmetals. Potassium is a metal of group 1 and iodine is a non metal of group 17 hence they can bond together based on their ionic charges.

Learn more:brainly.com/question/23277186

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Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

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Explanation:

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If Brian would mow the lawn by himself in 1010 minutes more than it would take Andy, this means B=1010A...eqn 2.

Substituting eqn 2 into eqn 1

Equation 1 becomes

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