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ivann1987 [24]
3 years ago
5

Which statement is true regarding sexual reproduction as a method of producing offspring?

Chemistry
2 answers:
DerKrebs [107]3 years ago
8 0

The answer is D, I hope this helps!

7nadin3 [17]3 years ago
5 0

Answer:

B. is the answer.

Explanation: don't quote me on it ,but most plants use asexual reproduction.

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The molecular formula of butane is C4H10. It is obtained from petroleum and is used commonly in LPG (Liquefied Petroleum Gas) cy
guajiro [1.7K]

Answer:

See attached picture.

Explanation:

Hello.

In this case, since butane has two common occurring structures, n-butane and isobutane, there is a way in which the tert-butyl radical can be formed upon the removal of a hydrogen from the isobutane form of butane as shown on the attached picture, wherein you can see that the radical is named by "tert" since the central carbon is bonded to three carbon atoms, that is why we classify it as tertiary. Moreover, it is a radical due to the presence of the bolded dot next to the tertiary carbon suggesting that it is very likely to bond with an other atom.

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3 years ago
Which ionic compound is used as a building material?
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Read 2 more answers
For each of the following nuclei, determine the binding energy per nucleon (in MeV). (For all masses, keep six places beyond the
garik1379 [7]

Answer: (a) BE = 1.112 MeV

               (b) BE = 7.074 MeV

               (c) BE = 7.767 MeV

               (d) BE = 8.112 MeV

Explanation: Binding energy per nucleon is the average energy necessary to remove a proton or a neutron from the nucleus of an atom. It is mathematically defined as:

BE = \frac{\Delta m.c^{2}}{A}

Where

Δm is a difference in mass known as <u><em>mass</em></u><em> </em><u><em>defect</em></u>

A is atomic mass of an atom.

Mass Defect is determined by:

\Delta m =Zm_{p}+(A-Z)m_{n} - m_{nuc}

where:

Z is atomic number

m_{p} is mass of proton

m_{n} is mass of neutron

m_{nuc} is mass of the nucleus

Mass of proton is 1.007825u.

Mass of neutron is 1.008665u.

The unit u is equal to 931.5MeV/c².

(a) 2H(deuterion): Given: Z = 1; A = 2; m_{nuc} = 2.014102u

\Delta m =1(1.007825)+1(1.008665) -2.014102

\Delta m =0.002388u

BE = \frac{0.002388.c^{2}}{2}.931.5\frac{MeV}{c^{2}}

BE = 1.112MeV

(b) 4He (Helium): Given: Z = 2; A = 4; m_{nuc} = 4.002603

\Delta m =2(1.007825)+2(1.008665) -4.002603

\Delta m =0.030377u

BE = \frac{0.030377.c^{2}}{4}.931.5\frac{MeV}{c^{2}}

BE = 7.074MeV

(c) 18O (Oxygen): Given: Z = 8; A = 18; m_{nuc} = 17.999160

\Delta m =8(1.007825)+10(1.008665) -17.999160

\Delta m =0.15009u

BE = \frac{0.15009.c^{2}}{18}.931.5\frac{MeV}{c^{2}}

BE = 7.767MeV

(d) 23Na (Sodium): Given: Z = 11; A = 23; m_{nuc} = 22.989767

\Delta m =11(1.007825)+12(1.008665) -22.989767

\Delta m =0.200288u

BE = \frac{0.200288.c^{2}}{23}.931.5\frac{MeV}{c^{2}}

BE = 8.112MeV

3 0
3 years ago
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