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1 year ago

Acrostic poem for selective breeding

Chemistry

1 answer:

Roman55 [17]1 year ago

7 0

Answer:

Selective Breeding

Weeds are bad,

But why is this so?

Just like grass,

They too grow.

Weeds are disliked,

Like the drink that's spiked.

But why is this so?

Who's to say they shouldn't grow?

The weed is unwanted,

Like a murderer hunted.

But what crime did it commit?

Why should it not, be free to sit?

Thickly they may grow,

But grass even more so.

Who's to say they should not grow?

Why should their seed not be sow?

Weeds are bad,

But why is this so?

Just like grass,

They too grow.

If all things living, were for cared,

The land we own, for all shared.

The weed let live,

And life not carefully selected by- siv.

Selective breeding is unfair,

No better than plain warfare...

Explanation:

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The equilibrium constant for A + 2B → 3C is 2.1 * 10^-6

Mariana [72]

Answer:

b- 4.4 * 10^-12.

Explanation:

Hello.

In this case, as the reaction:

A + 2B → 3C

Has an equilibrium expression of:

$K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}$

If we analyze the reaction:

2A + 4B → 6C

Which is twice the initial one, the equilibrium expression is:

$K_2=\frac{[C]^6}{[A]^2[B]^4}$

It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:

$K_2=K_1^2$

Thus, the equilibrium constant of the second reaction turns out:

$K_2=(2.1 * 10^{-6})^2\\\\K_2=4.4x10^{-12}$

Therefore, the answer is b- 4.4 * 10^-12.

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3 years ago

Describe the number of signals and their splitting in the 1H NMR spectrum of (CH3)2CHOCH3.

yanalaym [24]

Answer:

d. 3 signals: a singlet, a doublet, and a septet

Explanation:

In this case, we can start with the structure of $(CH_3)_2CHOCH_3$ . When we draw the molecule we will obtain 2-methoxypropane (see figure 1).

In 2-methoxypropane we will have three signals. The signal for the $CH_3$ groups in the left, the $CH$ and the $CH_3$ in the right. Lets analyse each one:

-) $CH_3$ in the right

In this carbon, we dont have any hydrogen as neighbors. Therfore we will have singlet signal in this carbon.

-) $CH$

In this case, we have 6 hydrogen neighbors ( the two methyl groups in the left). So, if we follow the n + 1 rule (where n is the amount of hydrogen neighbors):

$multiplicity~=~n+1~=~6~+~1~=~7$

For this carbon we will have a septet.

-) $CH_3$ in the left

In this case we have only 1 hydrogen neighbor (the hydrogen in $CH$ ). So, if we use the n+1 rule we will have:

$multiplicity~=~n+1~=~1~+~1~=~2$

We will have a doublet

With all this in mind the answer would be:

d. 3 signals: a singlet, a doublet, and a septet 

See figure 2 to further explanations

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3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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3 years ago

How many moles are present in a 5.8 g sample of copper?

Oksanka [162]

Explanation:

Molar mass of Cu = 63.5g/mol

Moles of Cu

= 5.8g / (63.5g/mol) = 0.091mol (B)

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2 years ago

Select all correct answers.

Jlenok [28]

Answer: It showed that all atoms contain electrons.

Explanation:

J.J. Thomson's experiments inside a cathode ray tube in the presence of an electric field showed that all atoms contain tiny negatively charged subatomic particles "electrons".

Also, Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup."

Furthermore, Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny positively-charged nucleus.

Then, Rutherford proposed the nuclear model of the atom based on these results.

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3 years ago