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Zinaida [17]
3 years ago
7

What type of bond is expected between two carbon atoms?

Chemistry
1 answer:
tatuchka [14]3 years ago
3 0
<h3>Answer:</h3><h2>Covalent Bonds</h2>

<em><u>Carbon Forms Covalent Bonds</u></em>

<em><u>Carbon Forms Covalent BondsThe most common type of bond formed by carbon is a covalent bond. In most cases, carbon shares electrons with other atoms (usual valence of 4). This is because carbon typically bonds with elements which have a similar electronegativity.Jul 28, 2019</u></em>

Explanation:

<h3><em>Hope</em><em> it</em><em> works</em><em> out</em><em>!</em></h3>

<em>ᕦᶘ ᵒ</em><em>A</em><em>d</em><em>r</em><em>i</em><em>㉨</em><em>a</em><em>n</em><em>n</em><em>a</em><em>ᵒᶅᕤ</em>

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What is the mass of a 1.68-l sample of a liquid that has a density of 0.921g/ml?
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Hey there!

Volume in mL :

1.68 L  * 1000 => 1680 mL

Density = 0.921 g/mL

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Mass = density * Volume

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How reliable is the perfect gas law in comparison with the van der waals equation?calculate the difference in pressure of 10.00g
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Answer:

The perfect gas law is reliable at the limit of pressure tending to zero. That is to say, at low pressure. The Van der Waals law instead can be used at higher pressures.

Explanation:

The perfect law equation is

P.V = n.R.T

The perfect gas law considers that the gas molecules have no interaction between each other and each molecule has null volume. This condition happens at the limit of pressure tending to zero.

The Van der Waals equation is

(P +\frac{n^{2}.a}{V^{2}})(V-n.b) = n.R.T

On the other hand, the Van der Waals law has two extra terms, one to consider the interaction of the molecules (n2a/V2) and other to consider the volume of the molecules (V-nb). These terms make a better approximation to a real gas.

Using the two equations to calculate the pressure (P) for CO2 with

<em>V = 100cm3 = 0.1L</em>

<em>T = 25ºC = 298K</em>

<em>n = 10g/(44g/mol) = 0.23mol</em>

<em>a = 3.658 atm.L2/mol2</em>

<em>b = 0.0429 L/mol</em>

With the perfect law equation:

P = \frac{n.R.T}{V} = \frac{0.23molx0.082\frac{atm.L}{K.mol}x298K }{0.1L} = 56.2atm

With the Van del Waals law:

P = \frac{n.R.T}{V-n.b} -\frac{n^{2}.a}{V^{2} }

P = \frac{0.23molx0.082\frac{atm.L}{K.mol}x298K}{0.1L - 0.23molx0.0429\frac{L}{mol} }-\frac{(0.23mol)^{2}x3.658\frac{atm.L^{2}}{mol^{2}}}{(0.1L)^{2}} = 43 atm

At this case the difference between the results of the two equations is due to the big mass of CO2 which produce a high pressure.

8 0
3 years ago
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