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3 years ago

What is the direction of the polarity of the indicated bond in h3c−oh marked by δ+ and δ−?

Chemistry

1 answer:

Sedbober [7]3 years ago

6 0

Answer:- The direction of the polarity of the indicated bond is from carbon to oxygen.

Explanations:- There are two types of covalent compounds, polar and non polar. If the bond is between two same atoms for example, H-H, Cl-Cl etc then the bond is non polar. If the bond is between two different atoms then the bond would be polar. The direction of the polarity is from loss electron negative atom to more electron negative atom.

Oxygen is more electron negative than carbon. So, being more electron negative, the bonding electrons are more towards oxygen and it cases partial negative charge on oxygen and partial positive charge on carbon. The direction of the polarity is from less electron negative carbon to more electron negative oxygen.

It is shown in the diagram below:

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Which reaction type is really part of another type of reaction but treated separately? Explain

Novosadov [1.4K]

Answer:

Oxidation - reduction reaction which is called redox reaction.

Explanation:

The type of reaction that is a part of another type but treated separately is called Oxidation - reduction reaction.

Thus is because they usually involve 2 separate half reactions which are oxidation reactions that involve loss of electrons and then reduction equations which involve gain of electrons.

They are treated separately for example when sodium reacts with chlorine to form sodium chloride.

2Na + Cl2 = 2NaCl

Now, sodium undergoes oxidation by loosing elctrons and it's half reaction is;

2Na → 2Na^(+) + 2e^(-)

Meanwhile chlorine undergoes reduction by gaining electrons and its half reaction is;

Cl2 + 2e^(-) → 2Cl^(-)

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2 years ago

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Answer:

Explanation:

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3 years ago

What function do a brain cell and liver cell have in common?

Inessa05 [86]

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3 years ago

Read 2 more answers

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

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3 years ago

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erastova [34]

Answer:

The chemical equation needs to be balanced so that it follows the law of conservation of mass. A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

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