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3 years ago

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Hydrogen bonding is most often seen _____. see concept 2.3 (page 38) view available hint(s) hydrogen bonding is most often seen

_____. see concept 2.3 (page 38) when multiple carbon atoms are present when the bonds within the molecule are nonpolar covalent bonds in molecules whose three-dimensional shape is tetrahedral if the molecule consists of three or fewer atoms when hydrogen is covalently bonded to an electronegative atom

1 answer:

Evgesh-ka [11]3 years ago

3 0

<h3><u>Answer;</u></h3>

When hydrogen is covalently bonded to an electronegative atom

<h3><u>Explanation;</u></h3>

Hydrogen bonding is a special type of dipole-dipole attraction between molecules. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom.
Highly electronegative atoms attract shared electrons more strongly than hydrogen does, resulting in a slight positive charge on the hydrogen atom. The slightly positive hydrogen atom is then attracted to another electronegative atom, forming a hydrogen bond.

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How many moles of O2 are needed to burn 2.56 moles of CH3OH?

lukranit [14]

Answer:

$n_{O_2}=3.84molO_2$

Explanation:

Hello!

In this case, since the combustion reaction of methanol is:

$CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O$

In such a way, since there is 1:3/2 mole ratio between methanol and oxygen, we can compute the moles of oxygen that are needed to burn 2.56 moles of methanol as shown below:

$n_{O_2}=2.56molCH_3OH*\frac{\frac{3}{2}molO_2}{1molCH_3OH} \\\\n_{O_2}=3.84molO_2$

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2 years ago

How many moles are in 48.0g of H2O2?

Tju [1.3M]

Answer:

1.41 moles H2O2(with sig figs)

Explanation:

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= 34. 02 g/mol

48.0g H2O2* 1 mol H2O2/ 34.02 g H2O2= 1.41 mol H2O2

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2 years ago

How many moles of chlorine are used up in a reaction that produces 0.35kg of BCl3

maw [93]

4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.

(a) The <em>balanced chemical equation </em>is

2B + 3Cl2 → 2BCl3

(b) Convert kilograms of BCl3 to moles of BCl3

MM: B = 10.81; Cl = 35.45; BCl3 = 117.16

Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3

(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.

Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2

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2 years ago

You have a gas stored in a rigid container (fixed volume). Explain the relationship between a decrease in temperature, collision

alexandr1967 [171]

Answer:

A decrease in temperature would decrease kinetic energy, therefore decreasing collisions possible.

Explanation:

A gas at a fixed volume is going to have collisions automatically. If you decrease the temperature (same thing as decreasing kinetic energy) you are cooling down the molecules in the container which gives them less energy and "relaxes" them. This decrease in energy causes them to move around much slower and causing less collisions, at a much slower rate. In a perfect world, these collisions do not slow down the molecule but we know that they do, just a very very small unmeasurable amount.

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3 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago