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SashulF [63]
1 year ago
7

Question 8 of 32

Chemistry
1 answer:
Anna007 [38]1 year ago
3 0
Sodium, atomic radius increases as we go down and right in the periodic table, which means that cesium is the element with the biggest atomic radius as it is located as the farthest bottom right of the periodic table, anyways, the atomic number of aluminum is 13 which means that the distribution of electrons here is K:2, L:8, M:3,, number of levels = the atomic period which the atom is located in, last number in the last electronic level (or the valency) = the atomic group which the atomic is located in, so in this case aluminum is located in group 3 period 3 at the periodic table, we have to find the atom which is located the nearest to the bottom right using the same method, sodium is located in group 1 period 3, boron is located in group 3 period 2, silicon is located in group 4 period 3, nitrogen is located in group 5 period 2,,, so the correct answer is sodium as it’s located at the left of aluminum and closer to the left of the periodic table, albeit being in the same period
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Help asap on a time crunch
marusya05 [52]

Answer:

<u>only sodium chloride compounds</u> MOST LIKELY products of the reaction between sodium metal and chlorine gas.

Correct answer - C

Explanation:

A solid sodium metal reacts with gaseous form of chlorine gas to form  a solid sodium chloride.

This formation of sodium chloride arises due to transfer of one valence electron from sodium to chlorine atom and the both atoms get stable octet electronic configuration.

As a result, an ionic compound solid sodium chloride is formed.

The chemical reaction of formation sodium chloride is as follows.

2Na(s)+Cl_{2}(g) \rightarrow 2NaCl(s)

Therefore, <u>only sodium chloride compounds</u> MOST LIKELY products of the reaction between sodium metal and chlorine gas.

7 0
3 years ago
Two isotopes of hydrogen fuse to form a neutron plus the larger element,A) beryllium.B) carbon.C) deuterium.D) helium.
Nezavi [6.7K]

Answer:  D) helium.

Explanation:

Nuclear fission is a process which involves the conversion of a heavier nuclei into two or more small and stable nuclei along with the release of energy.

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{56}^{143}Ba+_{36}^{90}Kr+3_0^1\textrm{n}

Nuclear fusion is a process which involves the conversion of two small nuclei to form a heavy nuclei along with release of energy.

Example: _1^2\textrm{H}+_1^3\textrm{H}\rightarrow _2^4\textrm{He}+_0^1\textrm{n}+\text{energy}

Thus when deuterium and tritium , the two isotopes of hydrogen are fused, a heavier nuclei helium is being formed from two smaller nuclei releasing a neutron.

3 0
3 years ago
Determine the number of atoms of O in 92.3 moles of Cr3(PO4)2
Irina18 [472]

739 is the number of atoms of O in 92.3 moles of Cr3(PO4)2.

Explanation:

Molecular formula given is  = Cr3(PO4)2

number of moles of the compound is 92.3 moles

number of 0 atoms in 92.3 moles =?

From the chemical formula 1 mole of the compound has 8 atoms of oxygen

So, it can be written as

1 mole Cr3(PO4)2 has 8 atoms of oxygen

92.3 moles of Cr3(PO4)2 has x atoms of oxygen

\frac{8}{1} = \frac{x}{92.3}

x = 8 x 92.3

x = 738.4 atoms

There will be 739 oxygen atoms in the 92.3 moles of Cr3(PO4)2.

4 0
3 years ago
Liquid nitrogen is used for: electrosurgery cryosurgery laser surgery
Otrada [13]
Cryosurgery. You automatically know this because 'cryo' means ice or something cold, thus you can assume that it is that
4 0
3 years ago
The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

Now put all the given values in this expression, we get:

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

7 0
2 years ago
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