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fomenos
3 years ago
6

Where are elements that form molecules of two of the same atoms commonly found on the periodic table?

Chemistry
2 answers:
timofeeve [1]3 years ago
8 0
The seven exceptions to that are the seven elements that are in gaseous form as a diatomic molecule, that is, two atoms of the same element attached to each other. The list of these elements is best memorized. They are: hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. Was this helpful??
Sveta_85 [38]3 years ago
6 0
Elements that form diatomic molecules, or molecules of two atoms each, are commonly found on the right side of the periodic table.
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I’m confused how to do this
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Answer:

B. P (Phosphorus)

Explanation:

The element is be phosphorus.

This is true because phosphorus has a larger atomic radius than carbon. If you move than the group, elements gain additional electron shell. That additional electron shell keeps electrons far from the nucleus of the atom. This actually increases the atomic radius. Also, phosphorus is more electronegative than aluminum because phosphorus itself is a non-metal and non-metals are generally electronegative while aluminum is a metal and electropositive.  

Phosphorus has a lower ionization energy than argon because the ionization energy across the period (i.e from left to right) increases and across period 3, you will find phosphorus first before argon in the periodic table.

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3 years ago
Which of the following compounds is soluble in water?
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Lead Nitrate is highly soluble in water. 37.65g/100 mL at 20*C
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Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
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Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
3 years ago
A gas system has an initial number of moles of 0.693 moles with the volume unknown. When the number of moles changes to 0.928 mo
lara [203]

Answer:

The initial volume in mL is 5959.2 mL

Explanation:

As the number of moles of a gas increases, the volume also increases.  Hence, number of moles and volumes are directly proportional i.e

n ∝ V

Where n is the number of moles and V is the volume

Then, n = cV

c is the proportionality constant

∴n/V = c

Hence n₁/V₁ = n₂/V₂

Where n₁ is the initial number of moles

V₁ is the initial volume

n₂ is the final number of moles

and V₂ is the final volume.

From the question,

n₁ = 0.693 moles

V₁ = ?

n₂ = 0.928 moles

V₂ = 7.98 L

Putting the values into the equation

n₁/V₁ = n₂/V₂

0.693 / V₁ = 0.928 / 7.98

Cross multiply

∴ 0.928V₁ = 0.693 × 7.98  

0.928V₁ = 5.53014

V₁ = 5.53014/0.928

V₁ = 5.9592 L

To convert to mL, multiply by 1000

∴ V₁ = 5.9592 × 1000 mL

V₁ = 5959.2 mL

Hence, the initial volume in mL is 5959.2 mL

5 0
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