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pochemuha
3 years ago
6

Which of the following is a chemical property of iron? It

Chemistry
1 answer:
astraxan [27]3 years ago
4 0

Answer:

is capable of combining with oxygen to form iron oxide

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White phosphorous has the chemical formula p4(s). a p4 molecule has 20 valence electrons. draw a lewis formula for a white phosp
aev [14]
The Lewis structure of P₄ is shown in 3-D form. The two bottom corner P atoms are facing right in front of us, one P atom behind the two, and one P above it. Each line represents 2 electrons. When you add the lone electrons, you get a total of 20 valence electrons.

Formal charge of each P:  5 - (2 +1/2*6) = 0

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3 years ago
Forces that act between two molecules are referred to as intermolecular forces true or false
julsineya [31]
Answer:
            TRUE: <span>Forces that act between two molecules are referred to as Intermolecular Forces.

Explanation:
                   Those forces which are present within the molecule among atoms are called as Intramolecular Forces, while, The forces which are present between two molecules are called as Intermolecular Forces. Intermolecular Forces are as follow,

                1) Hydrogen Bond Interactions

                2) Dipole-Dipole Interactions

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3 years ago
Potassium + Chlorine --&gt; Potassium Chloride
ArbitrLikvidat [17]

Answer:

The balanced equation is 2K(s) + Cl2(g)→2KCl(s)

3 0
3 years ago
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An orbital is best defined as
kaheart [24]
The answer is the most probable location of electrons in an atom
8 0
2 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

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

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
3 years ago
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