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3 years ago

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Is anyone good with science?

1 answer:

valkas [14]3 years ago

4 0

Answer:

Yes

Explanation:

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Benzene is an organic chemical compound with the molecular formula C6H6. In the benzene molecule, carbon atoms form a ring with

Sergeu [11.5K]

Answer:

$sp^2$

Explanation:

Hello,

The six-carbon benzene ring contains two types of bonds: C-C and C-H bonds, that are $sp^2$ -hybridized σ bonds, and the six π bonds that form the aromatic ring. The σ bonds form from one $s$ orbital and two $p$ orbitals from each carbon, which then bond the carbon to the two carbons on either side and the carbon's single hydrogen. The remaining $p$ orbital from each carbon atom sticks out above and below the plane of the ring; these $p$ orbitals overlap sideways, rather than lengthwise, to form the aromatic π bond system.

Best regards.

8 0

2 years ago

What is the formula for carbon tetrachloride?<br> a. CCl<br> b. CCl2<br> c. CCl3<br> d. CCl4

Fittoniya [83]

The <span>formula for carbon tetrachloride is CCl4. The answer is letetr D. The rest of the choices do not answer the question above.</span>

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3 years ago

If the solubility of a N30 gas is 2.26g/l at 1.26atm of pressure, what is the solubility of

natali 33 [55]

The answer for your question is 0.27g/L

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2 years ago

The Transition Elements can be divided into two types. List each type.

jenyasd209 [6]

Main group elements
transition metals
lanthanides
actindes.

7 0

3 years ago

10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0

3 years ago