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Andrew [12]
3 years ago
10

According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by ________ valence electron

s.
Chemistry
1 answer:
Mazyrski [523]3 years ago
6 0
According to the octet rule, atoms tend to gain, lose, or share electrons until they are surrounded by__8__ valence electrons.
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Explain why, when the imidazole ring of histidine is protonated, the double-bonded nitrogen is the nitrogen that accepts the pro
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The important thing to note is the reason why electron react is due to the instability of the electrons. All elements wants to aim the electron configuration of the noble gases. This is the most stable form in which each of the orbitals are sufficiently filled. When it comes to bonding, the order of reactivity is: alkynes > alkenes > alkanes. Alkynes are compounds with triple bonds, alkenes with double bonds and alkanes with single bonds. The single bonds are called saturated hydrocarbons. This is because they have reached stability, so it is quite difficult to react this with reducing or oxidizing agents. Alkynes and alkenes are unsaturated hydrocarbons. They readily react with reducing and oxidizing agents so as to become saturated, as well. The underlying principle for this is that single bonds contain sigma bonds which is the head-on overlapping of electrons. These is the strongest type of covalent bond. Double and triple bonds contain pi bonds which is the side overlapping of electrons orbitals. Hence, these electrons would be easily separated making it more reactive especially during protonation.
4 0
3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

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3 years ago
A magnet moved near a coil of wire can cause a(n)
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When a magnet moves near a coil of wire it can cause an A. electric current
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3 years ago
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Hydrocarbons are separated from each other by a process called cracking distillation.
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True, cracking distillation is one process used to break hydrocarbons into smaller and more useful parts. This is achieved by using high temperatures and pressures with a catalyst. The most common catalyst used are the zeolites.
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3 years ago
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10 points!! Please give me work with it and I will mark as brainlist.
Georgia [21]

Answer: The molecular formula will be C_2H_4O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 40.0 g

Mass of O = 53.3 g

Mass of H = 6.66 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.0g}{12g/mole}=3.33moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.33}{3.33}=1

For O =\frac{3.33}{3.33}=1

For H = \frac{6.66}{3.33}=2

The ratio of C : O : H = 1: 1: 2

Hence the empirical formula is COH_2

The empirical weight of COH_2 = 1(12)+1(16)+2(1)= 30g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2

The molecular formula will be=2\times CH_2O=C_2H_4O_2

8 0
3 years ago
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