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4 years ago

What is the complex Ions for CuO

1 answer:

8 0

The ion or molecule that forms the bond with the transition metal is called a ligand, e.g. :Cl-, :CN-, H2O:, :NH3. More than one ligand can bind to a transition metal ion to form a complex ion. The copper ion, Cu2+ (aq), in aqueous solution, forms a blue complex ion, [Cu(OH2)6]2+, with 6 water molecules.

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What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose?

LiRa [457]

Answer:

$T_f=-2.58\°C$

Explanation:

Hello,

In this case, we can compute the the freezing point depression by using the following formula:

$T_f-T_0=-i*m*Kf$

Whereas the freezing point of pure water is 0 °C van't Hoff factor for glucose is 1, the molality is computed as shown below and the freezing point constant of water is 1.86 °C/m:

$m=\frac{25.0g\ glucose*\frac{1mol\ glucose}{180g\ glucose} }{100g*\frac{1kg}{1000g} }\\ \\m=1.39m$

Thus, the freezing point of the solution is:

$T_f=T_0-i*m*Kf\\\\T_f=0\°C-1*1.39m*1.86\frac{\°C}{m}\\ \\T_f=-2.58\°C$

Regards.

4 0

4 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

Citrus2011 [14]

Answer: The mass of lead iodide produced is 9.22 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

6 0

3 years ago

If an electron is gained the atom has what kind of charge

Mrrafil [7]

The answer maybe negative charge

6 0

3 years ago

Emprical formula for c6h6

Vesna [10]

The emperical formula for benzene (C6H6) is “CH”.

7 0

3 years ago

If 5.85 moles of CaCO3 are used in an experiment, how many moles of carbon dioxide are created? (Be sure to include units in you

azamat

Answer:

5.85 moles of carbon dioxide are created.

Explanation:

The balanced reaction is:

CaCO₃ → CaO + CO₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CaCO₃: 1 mole
CaO: 1 mole
CO₂: 1 mole

Then you can apply the following rule of three: if by stoichiometry 1 mole of CaCO₃ produces 1 mole of CO₂, 5.85 moles of CaCO₃ will produce how many moles of CO₂?

$moles of CO_{2}=\frac{5.85 moles of CaCO_{3} *1mole of CO_{2} }{1 mole of CaCO_{3}}$

moles of CO₂= 5.85

5.85 moles of carbon dioxide are created.

7 0

3 years ago