Cows need oxygen given off by plants. Plants need carbon dioxide from the cow. This cycle is an example of:

The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV

astra-53 [7]

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

7 0

4 years ago

when a compound containing c h and o is completely combusted in air what reactant besides the hydrocarbon is involved in the rea

Arada [10]

When a compound containing C, H and O is completely combusted in air what reactant besides the hydrocarbon is involved in the reaction is Oxygen.

<h3>What is hydrocarbon?</h3>

A hydrocarbon is an organic molecule composed completely of hydrogen and carbon in organic chemistry. Group 14 hydrides include hydrocarbons. Hydrocarbons are often colourless and hydrophobic, with scents that are weak or exemplified by gasoline and lighter fluid. They exist in a wide range of molecular forms and phases, including gases (like methane and propane), liquids (like hexane and benzene), low melting solids (like paraffin wax and naphthalene), and polymers (such as polyethylene and polystyrene). Hydrocarbon refers to naturally occurring petroleum, natural gas, and coal, as well as their hydrocarbon derivatives and refined forms, in the fossil fuel industry. The primary source of energy on the planet is the combustion of hydrocarbons.

To learn more about hydrocarbons visit:

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7 0

2 years ago

A solution prepared by dissolving 0.100 mole of propionic acid in enough water to make 1.00 L of solution is observed to have a

torisob [31]

Answer:

C) $k_a=1.3\times 10^{-5}$

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,

pH = - log [H⁺]

The expression of the pH of the calculation of weak acid is:-

$pH=-log(\sqrt{k_a\times C})$

Where, C is the concentration = 0.5 M

Given, pH = 2.94

Moles = 0.100 moles

Volume = 1.00 L

So, $Molarity=\frac{Moles}{Volume}=\frac{0.100}{1.00}\ M=0.100\ M$

C = 0.100 M

$2.94=-log(\sqrt{k_a\times 0.100})$

$\log _{10}\left(\sqrt{k_a0.1}\right)=-2.94$

$\sqrt{0.1}\sqrt{k_a}=\frac{1}{10^{2.94}}$

$k_a=1.3\times 10^{-5}$

4 0

3 years ago

Which of the following are true statements about equilibrium systems?For the following reaction at equilibrium:2 H2(g) + O2(g) ?

VMariaS [17]

These are five questions about equilibrium systems each with its complete answer.

Question 1. For the following reaction at equilibrium:

2 H₂(g) + O₂(g) ⇄ 2 H2O(g), the equilibrium will shift to the left if the volume is doubled?

Answer: TRUE

Explanation:

When a force disturbs a chemical equillibrium, the system will shift toward the direction that reduces the effect. This is Le Chatelier's principle.

As per Bolye's law, at constant temperature, the volume and the pressure of a fixed amount of gas are inversely related.

Also, the pressure of the system is directly related to the number of particles (atoms or molecules). Hence, more molecules, more pressure; less molecules, less pressure.

Now, you can reason in this way: if the volume of the given system is doubled, then the pressure is lowered, and the system will try to alleviate this disturbance by shifting the reaction to the side that produces more molecules, to restore the pressure. Because on the left side three molecules can be produced from the reaction of two molecules of H₂O on the rihgt, the system will shift to the left. And this proves the truth of the statement.

Question 2. For the following reaction at equilibrium:

H₂(g) + F₂(g) ⇄ 2HF(g), removing H₂ will decrease the amount of F₂ present once equilibrium is reestablished.

Answer: FALSE.

Explanation:

Note that, since the temperature and other conditions have not changed, the equilibrium constant, Ke, has not changed. And, for the given equilibrium, Ke is given by the following equation.

Ke = [ H₂] [F₂] / [HF]²

Hence, to keep Ke unchanged, when removing H₂, the amount of F₂ present once equilibrium is reestablished will have to increase.

This is the opposite of the stated on the question, so the statement is false.

Question 3. Increasing the temperature of an exothermic reaction shifts the equilibrium position to the right.

Answer: FALSE.

Explanation:

You can write an exothermic equlibrium placing heat as a product on the right side of the equation; in this way:

A + B ⇄ C + D + heat

There, treating the heat as another product, you can reason that increasing the temperature, which is equivalent to supplying heat, will shift the equilibrium to the left side to consume heat, instead to the proposed by the statement. So, this is a false statement.

Question 4. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), adding more CaCO₃ will shift the equilibrium to the right.

Answer: TRUE.

Explanation:

CaCO₃(g) is the only reactant of the forward reaction.

Adding more CaCO₃ may be seen as a disturbance against which the system will act by consuming it and producing more CaO and CO₂.

So, the forward reation will be favored and you conclude that adding more CaCO₃ will shift the equilibrium to the right.

Question 5. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), increasing the total pressure by adding Ar(g) will have no effect on the equilibrium position.

Answer: TRUE.

Explanation:

In accordance to Le Chatelier's principle, increasing the pressure should be addresed by the equilibrium by shifting to the side where such pressure increase could be released.

That is possible when the number of molecules of gases on both sides are different: the equilibrium will shift to the side where more molecules less molecules are produced.

But, when the stoichiometry of the reaction shows the same number of molecules on both sides, which is the case in the given equilibrium, increasiing (or decreasing) the pressure will have no effect on the equilibrium position. Then, the answer is true.

8 0

3 years ago

A neutralization reaction between an acid and a metal hydroxide produces

Vikki [24]

Answer:

A neutralization reaction between an acid and a metal hydroxide produces salt and water.

Explanation:

In a neutralization reaction, an acid and a base are combined according to the following example (hydrochloric acid and sodium hydroxide):

HCl + NaOH ---> H20 + NaCl

They are generated as products: water and a salt, in this case sodium chloride.

3 0

4 years ago