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geniusboy [140]
3 years ago
12

What is the lewis structure and formal charge of CH2O

Chemistry
1 answer:
aksik [14]3 years ago
3 0

Diagram A shows the Lewis structure (LS) of CH_2O. The formal charge on each atom is zero.

To get the formal charge (FC) on the atoms, cut each bond in half, as in <em>Diagram B</em>.  Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE  

<em>On O: </em>

VE = 6

BE = 2 lone pairs 2 + 2 bonding electrons = 4 + 2 = 6

FC = 6 – 6 = 0.

<em>On H: </em>

VE = 1

BE = 1 bonding electron

FC = 1 – 1 = 0

<em>On C: </em>

VE = 4

BE = 1 in each single bond + 2 in the double bond = 2 + 2 = 4

FC = 4 - 4 = 0

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<h3>Answer:</h3>

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

<h3>Explanation:</h3>

The balanced chemical equation for the combustion of the hydrocarbon in question is;

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

  • A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
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A monatomic ideal gas that is initially at 1.50 * 105 Pa and has a volume of 0.0800 m3 is compressed adiabatically to a volume o
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Answer:

300000Pa or 3×10^5 Pa

Explanation:

Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.

Using Boyle's law

P1V1 = P2V2

P1 is the initial pressure = 1.5×10^5Pa

V1 is the initial volume = 0.08m3

P2 is the final pressure (required)

V2 is the final volume = 0.04 m3

From the formula, P2 = P1V1/V2

P2 = 1.5×10^5 × 0.08 ÷ 0.04

= 300000Pa or 3×10^5 Pa.

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Answer:

Option D. 230 J

Explanation:

We'll begin by calculating the temperature change of the iron. This can be obtained as follow:

Initial temperature (T₁) = 50 °C

Final temperature (T₂) = 75 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 75 – 50

ΔT = 25 °C

Thus, the temperature change of the iron is 25 °C.

Finally, we shall determine the amount of heat energy used. This can be obtained as follow:

Mass (M) = 20 g

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Specific heat capacity (C) = 0.46 J/gºC

Heat (Q) =?

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Q = 20 × 0.46 × 25

Q = 230 J

Thus, the amount of heat used was 230 J

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