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dem82 [27]
4 years ago
8

The effective nuclear charge experienced by the outermost electron of Na is different than the effective nuclear charge experien

ced by the outermost electron of Ne. This difference best accounts for which of the following
Chemistry
1 answer:
Svetach [21]4 years ago
4 0

Answer:

B. Na has a lower first ionization energy than Ne.

Explanation:

Ionization energy is the energy needed to move an electron from a molecule or gaseous molecule.The factors that affect the ionization energy are:

  1. The size of the positively charged nuclear-the more positively charged the molecule is the more energy needed to remove an electron.
  2. Size of the atom- an increase in the molecule/atom size, the less ionization energy needed to remove an electron, because with an increase in size the attraction between the positive nucleus and the electron decreases.
  3. Shielding effects of the inner shell electron-increase in shielding results in the decrease of the ionization energy.

Ne atom is bigger than the Na atom.

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Give the chemical symbols for the following elements: (a) potassium, (b) tin, (c) chromium, (d) boron,(e) barium, (f) plutonium,
Ilia_Sergeevich [38]

Answer:

A) Alkili Metal

B) Basic Metal

C)  Transition Metal

D) Semimetal

E) Alkiline Earth

F) Actinide

G) Nonmetal

H) Noble Gas

I) Transitional Metal  

Hope This Helps

8 0
4 years ago
3. The graph below shows the temperature changes occurring in two hot packs that are made from different materials. The same amo
nika2105 [10]

Answer:

LDPE would be the best choice for a hand warmer designed to release heat quickly.

Explanation:

Judging from the graph, the LDPE plastic releases heat quicker than the Latex. (and I also did the gizmo-)

4 0
3 years ago
A gas is confined in a 0.3m diameter cylinder by a piston, on which rests a weight. The mass of the piston is 85 kg. The local a
iris [78.8K]

Explanation:

Force applied on the gas will be as follows.

                   F_{gas} = F_{atm} + (m + M) g

As,   F = pressure × area. Hence, calculate the forces as follows.

                  F_{gas} = pressure × area

                         = 1.4 \times 10^{5} Pa \times \pi \times (\frac{0.3}{2})^{2}

                          = 1.979 \times 10^{4} N

                  F_{atm} = pressure × area

                            = 1.0133 \times 10^{5} \times \pi \times (\frac{0.3}{2})^{2}

                            = 1.432 \times 10^{4} N

      F_{gas} - F_{atm} = 5.47 \times 10^{3} N

Substituting the calculated values into the above formula as follows.

                F_{gas} = F_{atm} + (m + M) g

              F_{gas} - F_{atm} = (m + M) g

              5.47 \times 10^{3} N = (m + 85) \times 9.8    

              5.47 \times 10^{3} N = 9.8m + 833

                               m = 472.76 kg

Thus, we can conclude that the mass is 472.76 kg.  

5 0
4 years ago
How many grams do 6.534e+24 molecules of phosphoric acid weigh?
fredd [130]

Answer:

1,063 grams H₃PO₄

Explanation:

To find the mass of phosphoric acid (H₃PO₄), you should (1) convert molecules to moles (via Avogadro's number) and then (2) convert moles to grams (via molar mass from periodic table).

Molar Mass (H₃PO₄): 3(1.008 g/mol) + 30.974 g/mol + 4(15.998 g/mol)

Molar Mas (H₃PO₄): 97.99 g/mol

6.534 x 10²⁴ molecules H₃PO₄                       1 mole                         97.99 g
---------------------------------------------  x  -------------------------------------  x  --------------
                                                            6.022 x 10²³ molecules          1 mole

= 1,063 grams H₃PO₄

3 0
2 years ago
g Suppose you are titrating vinegar, which is an acetic acid solution of unknown concentration, with a sodium hydroxide solution
Elza [17]

Answer: The molar concentration of acetic acid in the vinegar is 0.539 M.

Explanation:

The formula used is:

M_1V_1=M_2V_2

where,

M_1 and V_1 are the concentration and volume of base.

M_2 and V_2 are the concentration and volume of an acid.

Given:

Molar concentration of NaOH = 0.1798 M

Volume of NaOH = 30.01 mL

Volume of acetic acid = 10.0 mL

Now putting all the given values in the above formula, we get:

M_1V_1=M_2V_2\\\\0.1798M\times 30.01mL=M_2\times 10.0mL\\\\M_2=0.539M

Thu, the molar concentration of acetic acid in the vinegar is 0.539 M.

4 0
3 years ago
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