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2 years ago

Explain how atoms are held together in covalent bonds. Give an example of a covalent compound.

Chemistry

1 answer:

asambeis [7]2 years ago

5 0

Answer:

Explanation:

Atoms are held together by covalent bonds when they share electrons between themselves.

Covalent bonds are bonds that are formed between non-metals usually with a low electronegative difference between them. In this bond type, two non-metals donate electrons which are shared between the combining atoms and this makes them both like the corresponding noble gases. The shared electrons is what forms the covalent bonds.

An example of covalent bond is HCl, H₂S, SO₂, CO₂, O₂ etc

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How many moles of water are in a beaker with 50 mL?

tatiyna

Answer:

Number of moles = 2.8 mol

Explanation:

Given data:

Number of moles of water = ?

Volume of water = 50 mL

Density of water = 1.00 g/cm³

Solution:

1 cm³ = 1 mL

Density = mass/ volume

1.00 g/mL = mass/ 50 mL

Mass = 1.00 g/mL× 50 mL

Mass = 50 g

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 50 g / 18 g/mol

Number of moles = 2.8 mol

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2 years ago

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madam [21]

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2 years ago

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Water exhibits high levels of cohesion due to the polarity of the molecule (positively and negatively charged sides). given this

nalin [4]

Covalent bonds and it is between the hydrogens (hydrogen bonds)

7 0

2 years ago

Compare and contrast conduction, convection, and<br> radiation as part of heat energy

Zinaida [17]

Answer:

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1 year ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

3 years ago