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Assoli18 [71]
2 years ago
14

What are the respective central-metal oxidation state, coordination number, and overall charge on the complex ion in Na2[Cr(NH3)

2 (NCS)4]?
Chemistry
1 answer:
Elden [556K]2 years ago
6 0

Answer:

Central-metal oxidation state: +2

Coordination number: 6

Overall charge: -2

Explanation:

In the ion complex:

Na₂[Cr(NH₃)₂(NCS)₄]

The central-metal is Cr, the ligands are NH₃ and NCS.

The NH₃ is a neutral ligand, while NCS is an anionic ligand with -1 in charge.

The overall charge of the complex is:

2Na⁺ +  [Cr(NH₃)₂(NCS)₄]⁻² → <em>-2</em>

As NCS gives -1 in charge and there are four NCS, The Cr must be in <em>+2</em> of oxidation state to give an overall charge of -2.

As Cr is bonded to 2 NH₃ and 4 NCS, the coordination number is 2+4 = <em>6</em>

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I hope it helps!

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How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 years ago
Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)
Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0
2 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
2 years ago
Read 2 more answers
11 Which compound has the greatest percent composition by mass of sulfur?(1) BaS (3) MgS
harina [27]
Percent composition by mass is calculated (mass of element within compound)/(mass of compound)*100. The lower the total molar mass of the compound, the greater the percent composition of sulfur. In this case, MgS would be that compound, since Mg has the lowest molar mass of the four elements bonded to S.
7 0
2 years ago
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A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solu
irinina [24]

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize 6.05\times 10^3 kg of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = 6.05\times 10^3 kg=6.05\times 10^6 g

1 kg = 10^3 g

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = \frac{95.0}{100}\times 6.05\times 10^6 g

=5,747,500 g

Moles of sulfuric acid = \frac{5,747,500 g}{98 g/mol}=58,647.96 mol

H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

\frac{1}{1}\times 58,647.96 mol=58,647.96 mol of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

106 g/mol\times 58,647.96 mol=6,216,683.76 g

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize 6.05\times 10^3 kg of sulfuric acid solution.

3 0
2 years ago
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