The formulas N2, H20, and CO2 all represent molecules, which are defined as what?

Good Morning I need help with this science work, please help it would mean a lot.

Kay [80]

That’s easy it would be B.

8 0

3 years ago

Give the formulas of the compounds in each set (a) lead(l|) oxide and lead(lV) oxide; (b) lithium nitride, lithium nitrite, and

Ivanshal [37]

Answer:

For a: The formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b: The formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c: The formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d: The formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

Explanation:

All the given compounds are ionic compounds. This means that between the atoms complete sharing of electrons takes place.

For a:

Lead is the 82th element of periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^2$ .

To form $Pb^{2+}$ ion, this element will loose 2 electrons and to form $Pb^{4+}$ ion, this element will loose 4 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lead (II) oxide is PbO and for lead (IV) oxide is $PbO_2$

Thus, the formula of lead (II) oxide and lead (IV) oxide is $PbO\text{ and }PbO_2$ respectively.

For b:

Lithium is the 3rd element of periodic table having electronic configuration of $[He]2s^1$ .

To form $Li^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

Nitrite ion is a polyatomic ion having chemical formula of $NO_2^{-}$

Nitrate ion is a polyatomic ion having chemical formula of $NO_3^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for lithium nitride is $Li_3N$ , for lithium nitrite is $LiNO_2$ and for lithium nitrate is $LiNO_3$

Thus, the formula of lithium nitride, lithium nitrite and lithium nitrate is $Li_3N,LiNO_2\text{ and }LiNO_3$ respectively.

For c:

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Hydrogen is the 1st element of periodic table having electronic configuration of $1s^1$ .

To form $H^{-}$ ion, this element will gain 1 electron and is named as hydride ion.

Hydroxide ion is a polyatomic ion having chemical formula of $OH^{-}$

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium hydride is $SrH_2$ and for strontium hydroxide is $Sr(OH)_2$

Thus, the formula of Strontium hydride and strontium hydroxide is $SrH_2\text{ and }Sr(OH)_2$ respectively.

For d:

Magnesium is the 12th element of periodic table having electronic configuration of $[Ne]3s^2$ .

To form $Mg^{2+}$ ion, this element will loose 2 electrons.

Manganese is the 25th element of the periodic table having electronic configuration of $[Ar]3d^54s^2$

To form $Mn^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for magnesium oxide is MgO and for manganese (II) oxide is MnO.

Thus, the formula of magnesium oxide and manganese (II) oxide is $MgO\text{ and }MnO$ respectively.

7 0

3 years ago

What could be the fourth quantum number for one of the electrons in the 4p energy sublevel of bromine?

Anton [14]

Ms = -1/2 Negative one-half

7 0

3 years ago

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Please help me on this question I'm confusied on it

Korolek [52]

Answer:

So you need to look for the average speed of each car, which would allow you to then enter the numbers.

Explanation:

Hope this helps! Have a great say! :)

3 0

3 years ago

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To investigate the reactivity of metals a student uses four metals. Each time they added 1 g of the metal to 25 cm³ of sulfuric

Mazyrski [523]

Answer:

Volume of the sulfuric acid (25cm³), same mass of each metal (1g)

Explanation:

In an experiment, the CONTROL VARIABLE also known as constant is the variable that is kept unchanged for all groups in an experiment. This is done in order not to influence the outcome of the experiment.

In this case, students are trying to investigate the reactivity of four different metals. They added 1 g of each metal to 25cm³ of sulfuric acid and recorded the temperature change. Based on the explanation of control variable above, the VOLUME OF SULFURIC ACID (25cm³) and the MASS OF EACH METAL (1g) are the CONTROL VARIABLES because they are the same or unchanged in this experiment.

5 0

3 years ago