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Ad libitum [116K]
3 years ago
13

The definition of oxidation is loss of electrons. The definition is formation of additional bonds between carbon and oxygen (or

in the number of carbon-hydrogen bonds.) These are similar because when forms a bond with electronegative __________, it essentially some of its attraction for its electrons.
Chemistry
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

Oxidation can be defined in many ways:

In terms of electron, <em>Oxidation</em> can be defined as loss of electrons by an atom or ion or a molecule during a reaction.

In terms of oxygen transfer, <em>oxidation</em> maybe defined as a substance that gains oxygen or loses hydrogen.

It can also be defined as a process in which a carbon atom gains bond to more electronegative atom such as oxygen.

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When light cannot travel trough a material, the light has been
Yuki888 [10]
The answer would be absorbed
8 0
3 years ago
A forensic chemist is given a white powder for analysis. She dissolves 0.50 g of the substance in 8.0 g of benzene. The solution
Sholpan [36]

Answer:

The given compound cannot be cocaine.

Explanation:

The chemist can comment on the nature of compound being cocaine or not from the depression in freezing point.

Depression in freezing point of is related to molality as:

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant = 4.90°C/m

depression in freezing point = normal freezing point - freezing point of solution

depression in freezing point = 5.5-3.9 = 1.6°C

1.6°C = 4.90 X molality

molality=\frac{1.6}{4.90} = 0.327 m

we know that:

molality=\frac{moles}{mass of solvent(kg)}=\frac{moles}{0.008kg}

therefore

moles = 0.327X0.008 = 0.00261 mol

moles=\frac{mass}{molarmass}

molarmass=\frac{mass}{moles}=\frac{0.50}{0.00261}= 191.57g/mol

The molar mass of cocaine is 303.353

So the given compound cannot be cocaine.

3 0
3 years ago
Which group would Ns2np2 go into
FrozenT [24]

Answer:

The group 15 elements: the pnicogens

Explanation:

The group 15 elements, nitrogen, phosphorus, arsenic, antimony and bismuth, all have the general valence shell electronic configuration ns2np3. They can all exist in the +3 or +5 oxidation state, with the +3 state increasing in stability as we move vertically down the group.

3 0
3 years ago
Read 2 more answers
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
2 years ago
Investigate: Note the empty jars on the shelf that can be filled by using the slider. Set the amount to 2.000 moles of carbon (m
balandron [24]

Answer:

The molar mass of carbon

Explanation:

Before the mass (in grams) of two moles of carbon can be determined, <u>the molar mass of the element would be needed.</u>

<em>This is because the number of mole of an element is the ratio of its mass and the molar mass</em>. That is,

number of mole = mass/molar mass

Hence, the mass of elements can be obtained by making it the subject of the formular;

mass = number of mole x molar mass

<em>Therefore, the molar mass of carbon would be needed before the mass of 2 moles of the element can be determined.</em>

3 0
2 years ago
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