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2 years ago

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Which is a correct set of values of m for one of the subshells of n = 2? –1, 0, 1 –1, –2, 0, 1, 2 –1, –2, –3, 0, 1, 2, 3 –1, –2,

–3, –4, 0, 1, 2, 3, 4

1 answer:

Alexandra [31]2 years ago

7 0

For n=2 the values of I are from 0 to (n-1).

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Please help me balance:<br>_NaNO3 + _PbO_ ➡️ Pb(NO3)2 + _Na2O

KengaRu [80]

Answer:

The balanced reaction is given by,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

Explanation:

The reaction is as given.

Lets count the number of each elements in the reaction.

<em>In reactant side, number of sodium atoms are 1 , lead are 1, nitrogen are 1 and oxygen are 4.</em>

<em>in product side, number of sodium atoms are 2 , lead are 1 , nitrogen are 2 and oxygen are 7.</em>

<em>So we need to balance sodium and oxygen atoms in the reaction.</em>

<em>There is deficient of sodium and oxygen atoms on reactant side</em>.

Thus, multiply (NaNO3) by 2.

<em>Thus, sodium atoms become 2 , nitrogen 2 and oxygen 6. Total 7 oxygen atoms.</em>

Thus, the balanced reaction is,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

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2 years ago

What is the formula for CO3 2- and Al3+ ?

Goryan [66]

Answer is: formula is Al₂(CO₃)₃.

Aluminium carbonate (Al₂(CO₃)₃) has neutral charge. Because aluminium cation has positive charge 3+ and carbonate anion has negative charge 2-, for right chemical formula, we need two aluminium cations and three carbonate anion:

charge of the molecule = 2 · (3+) + 3 · (-2).

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3 years ago

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Why is hydrogen a non metal, usually placed with group I element in the periodic table, even though it does not show a metallic

podryga [215]

because it has ns1 electron configuration like the alkali metals

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2 years ago

What can group numbers on the periodic table help you determine?

Ilia_Sergeevich [38]

Http://www.chem4kids.com/files/elem_pertable.html this should help.

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2 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

2 years ago