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2 years ago

Lewis structure of butane

Chemistry

1 answer:

eduard2 years ago

7 0

Answer:

This is the C4H10 Lewis structure: Butane. For Butane, we have a total of 26 valence electrons. Whenever we see the ending, "ane", we know that we're going to have Carbons and Hydrogens single bonded. That makes it a little bit easier to draw the C4H10 Lewis structure. We'll put four Carbons in a row and then we'll put Hydrogens around them. Because each Carbon needs to have four single bonds--each bond having two valence electrons, that'll give it an octet--we'll have three Hydrogens on the end Carbons and two on the center, like this. There are the three on the ends, and then we'll put two Hydrogens on the central Carbons. Next we'll place a single bond between each of the atoms to show that a pair of electrons is being shared.

So we've used all 26 valence electrons for the C4H10 Lewis structure, and we can see that each Carbon has four single bonds. Since each single bond has two valence electrons, that means that each Carbon has an octet. Each Hydrogen has a single bond, so it has two valence electrons. That means that it has a full outer shell as well. So we've used all the valence electrons that we had for C4H10 and everything has an octet..','.

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- The subatomic particle possessing the largest

Ierofanga [76]

Answer:

B. electrons possess the largest charge-to-mass ratio among the subatomic particles listed in the four choices.

Explanation:

Consider the mass of each particle. Express the masses in atomic mass units:

Protons: approximately 1.007 amu each;
Neutrons: approximately 1.009 amu each;
Electrons: approximately 0.0005 amu each.

Similarly, consider the charge on each particle. Express the charges in multiples of the fundamental charge:

Protons: +1 e;
Neutrons: 0;
Electrons: -1 e.

Calculate the charge-to-mass ratio for the three species:

Protons: approximately $\rm 0.99\; e\cdot amu^{-1}$ ;
Neutrons: 0;
Electrons: approximately $\rm 1,800\;e \cdot amu^{-1}$ .

Almost all nuclei contain protons and neutrons. The only exception is the hydrogen-1 nucleus, which contains only one proton and no neutron. The mass of the nucleus is approximately the same as the sum of its components' masses. The extra neutron will only add to the mass of the nucleus (the denominator) without contributing to the charge (the numerator.) As a result, the charge-to-mass ratio of nuclei will be positive but no greater than the charge-to-mass ratio of protons.

Among the particles in the four choices, the charge-to-mass ratio is the greatest for electrons.

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3 years ago

How many molecules are in 23.47 grams of CO2?

Genrish500 [490]

Do a quick conversion: 1 grams Co2 = 0.0084841820909097 mole using the molecular weight calculator and the molar mass of Co2.

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2 years ago

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B

loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

$C_{15}H_{24}0$

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in $CO_2$ and also, all the hydrogen in BHT is present in $H_2O$ , Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of $CO_2$ = $8.990 g*(\frac{1mole}{44.01g})$

= 0.2043 moles

∴ moles of C = 0.2043 moles

moles of $H_2O$ = $2.944 g *(\frac{1mole}{18.01g} )$

= 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

= 0.327 moles

Since number of moles= $\frac{mass}{molarmass}$

number of moles of H = 0.327 moles

molar mass of H = 1.008 g/mol

∴ mass of H in the sample = 0.327 moles × 1.008 g/mol

= 0.329616g

mass of C in the sample can be calculated as = 0.2043 moles × $(\frac{12.01g}{1 mole} )$

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × $(\frac{1mole}{16.0g})$

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C H O

0.2043 0.327 0.0136

Dividing by the least number (0.0136) , we have :

$\frac{0.2043}{0.0136}$ $\frac{0.327}{0.0136}$ $\frac{0.0136}{0.0136}$

15.02 24.04 1

≅

15 24 1

Therefore, the empirical formula would be : $C_{15}H_{24}0$

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3 years ago

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Gennadij [26K]

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3 years ago

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3 years ago