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4 years ago

Which example represents sexual reproduction?

Chemistry

2 answers:

sergey [27]4 years ago

8 0

Answer:

the answer is (A) the lioness giving birth

Explanation:

sexual reproduction is when two parents of different orientation come together to have a baby. e.g. a male and female having a baby

Asexual is when you only need one parent so like with plants and stuff.

Levart [38]4 years ago

3 0

Answer:

Explanation:

lioness giving birthe is not the process

amoeba is asexual

its either c or d but most likely d

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You carefully weigh out 14.00 g of CaCO3 powder and add it to 56.70 g of HCl solution. You notice bubbles as a reaction takes pl

zepelin [54]

For the answer to the question above, let us assume that all Co2 is given off and this are the mass that is lost:

mass at start = 14.00 = 56.70 = 70.70g
<span>mass at end = 64.96g </span>
<span>mass lost = 5.74g </span>
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</span><span>I hope my answer helped you. Have a nice day!</span>

7 0

4 years ago

Read 2 more answers

Why does heating increase the speed at which a solute dissolves in water?

saw5 [17]

Answer: C

Explanation:

It gives kinetic energy to the molecules and it breaks the bonds faster because they jiggle more

8 0

3 years ago

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula

Murrr4er [49]

Answer: The empirical formula will be $SF_6$

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 S= 3.21 g

Mass of F = 11.4 g

Step 1 : convert given masses into moles.

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{3.21g}{32g/mole}=0.1moles$

Moles of F = $\frac{\text{ given mass of F}{\text{ molar mass of F}}= \frac{11.04g}{19g/mole}=0.6moles$

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

For S= $\frac{0.1}{0.1}=1$

For F = $\frac{0.6}{0.1}=6$

The ratio of S: F= 1: 6

Hence the empirical formula is $SF_6$

7 0

3 years ago

What is the equilibrium expression for:

julia-pushkina [17]

Answer:

Explanation:

in kc we only consider gases and aquas not solids and liquids

6 0

2 years ago

Draw structures for all constitutional isomers with the molecular

ycow [4]

Answer:

(a)

Cyclobutane.
Methylcyclopropane.

(b)

1-Butene.
2-Butene (two spatial isomers.)
2-Methyl-1-propene.

Refer to the diagram attached for the structures.

Explanation:

<h3>Number of rings and double bonds in each molecule</h3>

Calculate the degree of unsaturation to find the number of rings and multiple bonds (including double bonds) in those isomer molecules. The presence of each ring and each double bond would add one to the degree of unsaturation of that molecule.

Let $C$ , $N$ , $X$ , and $H$ denote the number of carbon, nitrogen, halogen, and hydrogen atoms in each isomer molecule, respectively.

The degree of unsaturation of one such molecule would be:

$\displaystyle \frac{2\, C + 2 + N - X - H}{2}$ .

For $\rm C_4 H_8$ :

$C = 4$ (four carbon atoms in each molecule.)
$N = 0$ (no nitrogen atom.)
$X = 0$ (no halogen atom.)
$H = 8$ (eight hydrogen atoms in each molecule.)

Therefore, the degree of unsaturation of a molecule with the chemical formula $\rm C_4 H_8$ would be:

$\begin{aligned}&\text{degree of unsaturation} \\ &= \frac{2\, C + 2 + N - X - H}{2} \\ &= \frac{2\times 4 + 2 + 0 - 8}{2} = 1\end{aligned}$ .

In other words, the degree of unsaturation is $1$ for all isomers with this particular chemical formula. Therefore, each isomer would contain either a ring or a double bond, but not both.

If there is no double bond in one such molecule, there must be exactly one ring per molecule.

The minimum number of carbon atoms in a ring is three. With four carbon atoms in each molecule, there would be either a three-member ring (with one methyl group attached) or a four-member ring.

Four-membered ring with single bonds only: cyclobutane.
Three-membered ring with single bonds only and one methyl group attached: methylcyclopropane.

If there is one double bond in one such isomer molecule, there would not be a ring. Because of the degree of unsaturation, there would be no more than one double bond in each of these molecules.

With four carbon atoms, there are two possible backbones to consider: backbones with three carbon atoms each, and backbones with four carbon atoms each.

The backbone of those molecules might contain three carbon atoms. There would be one double bond in the backbone and one methyl group attached to the carbon atom at the center of the backbone. The corresponding isomer molecule would be 2-methyl-1-propene.

Alternatively, the backbone of those molecules might contain four carbon atoms. There are two possible locations for the double bond:

Between the first and the second carbon atoms: 1-Butene.
Between the second and third carbon atoms: 2-Butene.

Notice, that the name 2-Butene refers to two distinct spatial isomers. Unlike carbon-carbon single bonds, groups on the two sides of a carbon-carbon double bond are unable to rotate relative to one another. Therefore, the first and fourth carbon atoms would either be:

on the same side of the $120^\circ$ double bond: (2Z)-2-Butene, or
on opposite sides of that double bond: (2E)-2-Butene.

6 0

3 years ago