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3 years ago

Which metal atom below cannot form a cation of several different charges?

1 answer:

4 0

Missing question: Cr, Mn, Fe, Ba, Cu.
Answer is: Ba (barium).
Barium is metal from second group of periodic table of elements. Barium looses two electrons and has oxidation number +2.
Chromium has different oxidation numbers, usually +3 and +6, iron has oxidation numbers +2 and +3, copper +1 and +2 in their compounds.

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Help me someone plz

Anastasy [175]

Been awhile since I done these but i believe it looks like this

8 0

3 years ago

Decide which element probably forms a compound with oxygen that has a chemical formula most and least similar to the chemical fo

mafiozo [28]

Answer:

Most similar ----- Lithium

Least similar ---- Nitrogen

Explanation:

Cesium is an element on the periodic with the atomic number 133. It lies in group 1 (i.e., the alkali metals) and period 6 on the periodic table. The oxidation state of group 1 metals is +1. Cesium forms an oxide with oxygen as $Cs_2O$ .

The most similar compound to this chemical compound is Lithium because Lithium happens to be in the same group one metal with Cesium and forms the compound $Li_2O$ with the oxygen

The least similar compound nitrogen due to fact that it is an oxide that is covalent in nature and lies between-group 3 -17 to form an $NO_2$ with oxygen.

8 0

2 years ago

65.39 Atomic # = Atomic Mass = # of Protons = # of Neutrons = # of Electrons =

ratelena [41]

Answer:

The answer to your question is below

Explanation:

Atomic mass = 65.39 g

Searching this number in the periodic table we find that the element is Zinc.

Then:

# of Protons = 30

# of neutrons = atomic mass - # of protons

= 65.39 - 30

= 35.39

# of electrons = # of protons = 30

8 0

3 years ago

How many moles of C2H2 are needed to react completely with 84 mol O2

Butoxors [25]

Balanced chemical equation:

2 C2H2 + 5 O2 = 4 CO2 + 2 H2O

2 moles C2H2 ---------------- 5 moles O2
moles C2H2 ------------------ 84 moles O2

moles C2H2 = 84 * 2 / 5

molesC2H2 = 168 / 5 => 33.6 moles of C2H2

4 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

5 0

2 years ago