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aleksandrvk [35]
2 years ago
12

Chlorine can bond with fluorine to form CIF. Chlorine can also bond with lithium to form LiCI Which compound will have a greater

partial charge?
Chemistry
1 answer:
Sergio [31]2 years ago
5 0

Answer: ClF will have a greater partial charge.

Explanation:

A polar covalent bond is defined as the bond which is formed when there is a low difference of electronegativities between the atoms, thus resulting in charge difference. Example: ClF

Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms and thus there is no charge difference. Example: F_2

Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal. The electronegative difference between the elements is high. The charges on cation and anion neutralise each other. Example: LiCl

Thus as ClF will have greater partial charge.

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How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction? (HCl molar mass is 36.46 g/mol).
Scilla [17]

Answer:

Moles of BCl₃ needed = 0.089 mol

Explanation:

Given data:

Moles of BCl₃ needed = ?

Mass of HCl produced = 10.0 g

Solution:

Chemical equation:

BCl₃ + 3H₂O  →     3HCl + B(OH)₃

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 36.46 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of HCl with BCl₃.

               HCl             :           BCl₃

                 3               :             1

             0.27             :            1/3×0.27 = 0.089 mol

6 0
2 years ago
The incomplete table below shows selected characteristics of gas laws.
Elenna [48]

Explanation :

In the given case different law related to gas is given. The attached figure shows the required solution.

Boyle's law states that the pressure is inversely proportional to the volume of the gas i.e.

P\propto \dfrac{1}{V}

PV=k

k is a constant.

Charle's law states that the volume of directly proportional to the temperature of the gas.

V\propto T

V=kT

Combined gas law is the combination of the pressure, volume and the temperature of the gas i.e.

\dfrac{PV}{T}=k

Hence, this is the required solution.

6 0
2 years ago
Read 2 more answers
What is the molality of a solution in which 3.0 moles of NaCl is dissolved in 1.5 Kg of water?
Nesterboy [21]
Density H2O = 1g/cm³
1,5 kg H2O = 1500g = 1500cm³             (1dm³ = 1000cm³)

3moles of NaCl-----in---------1500cm³ H2O
x moles of NaCl ----in--------1000cm³ H2O
x = 2moles of NaCl

answer: 2 mol/dm³
5 0
2 years ago
The volume of a gas is 36.0 ml at 10.0°c and 4.50 atm. at what temperature (°c) will the gas have a pressure of 3.50 atm and a v
galben [10]
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
5 0
2 years ago
If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le
ElenaW [278]

The volume of sphere can be calculated using the following formula:


V=\frac{4}{3}\pi r^{3}


Here, r is radius of the sphere which is 3.5 cm. Putting the value,


V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}


This is equal to the volume of lead, density of lead is 11.34 g/cm^{3} thus, mass of lead can be calculated as follows:


m=d×V


Putting the values,


m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g


Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.


Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:


m=0.686×0.866=0.5940 g


Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

m=\frac{92.5}{100}\times 0.5940=0.54945 g

1 g of lead obtain from\frac{1}{0.54945}=1.82 grams of ore.


Thus, 2041.2 g of lead obtain from:


2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg


Therefore, mass of ore required to make lead sphere is 3.715 kg.


6 0
2 years ago
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