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frutty [35]
3 years ago
15

Determine the chemical formula for the compound, tetracarbonylplatinum(iv) chloride. [pt(co)4cl4] 2- [ptcl4](co)4 [pt(co)4]cl4 [

pt(co)4cl4] 4- [pt(co)4]cl2
Chemistry
1 answer:
shusha [124]3 years ago
4 0
Analyze the Name of complex Compound.

                              T<span>etracarbonylplatinum(iv) chloride

So, there are,
                               4 Carbonyl groups = 4 CO = (CO)</span>₄
                               1 Platinum Metal    = 1 Pt  = Pt
                               Unknown Chloride atoms = ?
In complexes positive part is always named first, so the sphere containing Pt and carbonyl ligands is written first,

                                     [Pt (CO)₄]

The charge on sphere is +4 because CO ligand is neutral, and Pt has a Oxidation state of four as written in name (IV),
So,
                                             [Pt (CO)₄]⁴⁺

Now, in order to neutralize +4 charge we should add 4 Chloride ions, So,

                                            [Pt (CO)₄] Cl₄
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o-na [289]

O the amount of heat required to increase the temperature of 1 gram by 1°C

Explanation:

The specific heat of a substance is the amount of heat required to increase the temperature of 1 gram of a substance by 1°C. It is an intensive property that is specific to every substance.

The unit is given as J/g⁻¹°C⁻¹ or J/g⁻¹K⁻¹

This related to the quantity of heat using the expression below:

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7 0
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4. Suppose 8.00 g of CH4 is allowed to burn in the presence of 16.00 g of oxygen. CH4(g)+2O2(g)--&gt;CO2(g)+2H2O(g)
umka21 [38]
<h3>Answer:</h3>

No masses of CH₄ and O₂ remained after the reaction, while 22.005 g of CO₂ and 18.02 g of H₂O remained

<h3>Explanation:</h3>

The combustion of methane is given by the reaction;

CH₄(g)+2O₂(g) → CO₂(g)+2H₂O(g)

We are given, 8 g of CH₄ and 16.00 g of O₂

Required to determine the mass of CH₄, O₂, CO₂ and H₂O that remained after the complete reaction.

<h3>Step 1: Moles of CH₄ and O₂ in the mass given </h3>

Moles = mass ÷ molar mass

Molar mass of CH₄ = 16.04 g/mol

Moles of CH₄ = 8.00 g ÷ 16.04 g/mol

                      = 0.498 moles

                      = 0.5 moles

Molar mass of O₂ = 16.0 g/mol

Moles of O₂ = 16.00 g ÷ 16.00 g/mol

                    = 1 mole

From the reaction, 1 mole of CH₄ reacts with 2 moles of O₂

CH₄ is the limiting reactant since it is way less than the amount of O₂

Therefore, 0.5 moles of CH₄ will react with 1 mole of oxygen.

This means there will be no amount of O₂ and CH₄ that remains.

<h3>Step 2: Moles of CO₂ and H₂O that were produced.</h3>

From the reaction 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O.

Therefore,

In our case, 0.5 moles of CH₄ will react with 1 mole of O₂ to produce 0.5 moles of CO₂ and 1 mole of H₂O.

<h3>Step 3: Mass of CO₂ and H₂O produced </h3>

Mass = Moles × Molar mass

Molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.5 mol × 44.01 g/mol

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Molar mass of H₂O = 18.02 g/mol

Moles of H₂O = 1 mole × 18.02 g/mol

                       = 18.02 g

Therefore, no masses of CH₄ and O₂ remained after the reaction, 22.005 g of CO₂ and 18.02 g of H₂O remained

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Explanation:

Neutralization is a chemical reaction in which an acid and a base reacts to form salt and water.

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:

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