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Setler79 [48]
3 years ago
13

What does inductive effect mean?​

Chemistry
1 answer:
Serggg [28]3 years ago
7 0

Answer:

Tbh I looked it up but here is what I found. I hope it helps

Explanation:

In chemistry, the inductive effect is an effect regarding the transmission of unequal sharing of the bonding electron through a chain of atoms in a molecule, leading to a permanent dipole in a bond.

You might be interested in
Which statements about reducing sugars are true? D‑Glucose (an aldose) is a reducing sugar. The oxidation of a reducing sugar fo
qaws [65]

Answer:

The true statements are given below.

Explanation:

1 D glucose is a reducing sugar

2 The oxidation of reducing sugar forms a carboxylic acid sugar.

D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.

 The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).

4 0
3 years ago
If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced
ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • P₄: 1 mole
  • H₂: 6 moles
  • PH₃:4 moles

Being the molar mass of the compounds:

  • P₄: 124 g/mole
  • H₂: 2 g/mole
  • PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

  • P₄: 1 mole* 124 g/mole= 124 g
  • H₂: 6 mole* 2 g/mole= 12 g
  • PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

  • P= 2 atm
  • V= 10 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 298 K

Replacing:

2 atm*10 L= n*0.082 \frac{atm*L}{mol*K} *298 K

and solving you get:

n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

  • If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3}  }{124grams of P_{4}}

moles of PH₃=0.2742

  • If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2}  }{124grams of P_{4}}

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0
3 years ago
rusting of steel can be prevented by coating the steel with a layer of zinc. Explain, in terms of electron transfer, why steal d
Luba_88 [7]

Answer:

Wouldn't rust because zinc will lose electrons more readily than iron and will therefore oxidize first.

Explanation:

This process whereby rusting of steel is prevented by coating the steel with a layer of zinc is known as galvanization.

Now, in this process, the steel object will be coated in a thin layer of zinc. This coating will prevent oxygen and water from reaching the underneath metal since the zinc will also act as a sacrificial metal.

Now, Zinc is used because it has a lower reduction potential than iron and thus it will get easily more oxidized than iron. Which means the zinc will lose electrons more readily than iron.

Also, since zinc has a lower reduction potential, it is therefore the more active metal. Thus, even if the zinc coating is scratched and the steel is exposed to moist air, the zinc will still get to oxidize before the iron.

7 0
2 years ago
In a water molecule (H2O), the oxygen atom has two unshared pairs of electrons and two bonding pairs of electrons. How do the un
vladimir1956 [14]

Answer:

They reduce the bond angle to be slightly lower than the tetrahedral bond angle, approximately 104.45 degrees.

Explanation:

The unshared pair of electrons or lone pair electrons in order to have the minimum repulsion possible with each other pushes the other bonding pairs closer together making the bond angle smaller or bent.

The bond angle is slightly lower than the tetrahedral bond angle of 108 degrees, leaving the water molecule with a bent molecular geometry.

4 0
3 years ago
how many moles of water are in 1.23 x 10^18 water molecules? I need the answer in scientific notation ?
Ann [662]

Answer:

Moles=2.04×10^(-6)

Explanation:

No. Of moles=no. Of particles/ Avogadro's no

(Where no. Of particles may be atoms molecules or compounds)

Moles=1.23×10^18/6.022×10^23

Moles=0.204×10^(-5)

Moles=2.04×10^(-6)

8 0
3 years ago
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