Ask question

Ask question

All categories

3 years ago

6

Which statements describe independent assortment? Check all that apply. Independent assortment occurs during meiosis II. Indepen

dent assortment happens during fertilization. Independent assortment results in different combinations of genes in sex cells. Independent assortment makes offspring identical to their parents. Independent assortment occurs when chromosomes randomly line up in the middle of the cell.

2 answers:

kap26 [50]3 years ago

8 0

Answer:

well...

Explanation:

Acoording to my brain i think that the answer would be C

ser-zykov [4K]3 years ago

7 0

Answer:

The CORRECT answer is:

C. Independent assortment results in different combinations of genes in sex cells.

E. Independent assortment occurs when chromosomes randomly line up in the middle of the cell.

You might be interested in

A raft is made of 14 logs lashed together. Each log is 42 cm in diameter and a length of 6.4 m. 42% of the log volume is above t

Pavel [41]

Answer:

Explanation:

Given

No of logs $n=14$

diameter of log $d=42\ cm$

Length of log $L=6.4\ m$

42 % of log volume(V) is above water when no one is on raft

so 58 % of log volume(V) is submerged in the water

Weight of 14 log

$W=14\times \rho _{log}\times V\times g$

Buoyancy force on 14 logs $F_b=14\times \rho _{water}\times 0.58V\times g$

as system is in equilibrium so

$W=F_b$

$14\times \rho _{log}\times V\times g=14\times \rho _{water}\times 0.58V\times g$

$\rho _{log}=0.58\rho _{water}$

$\frac{\rho _{log}}{\rho _{water}}=0.58$

Specific gravity of log $=0.58$

7 0

3 years ago

A 1.00 L flask is filled with 1.30 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres

vichka [17]

Answer :

Part A : The partial pressure of argon is, 0.795 atm

Part B : The partial pressure of ethane is, 0.505 atm

Explanation :

<u>Part A :</u>

First we have to calculate the moles of argon.

Molar mass of argon = 39.95 g/mole

$\text{Moles of argon}=\frac{\text{Mass of argon}}{\text{Molar mass of argon}}=\frac{1.30g}{39.95g/mol}=0.0325mole$

Now we have to calculate the partial pressure of argon.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of argon = ?

V = Volume of argon = 1.00 L

n = number of moles of argon = 0.0325 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of argon = $25^oC=273+25=298K$

Putting values in above equation, we get:

$P_{Ar}=\frac{nRT}{V}$

$P_{Ar}=\frac{(0.0325mol)\times (0.0821L.atm/mol.K)\times (298K)}{1.00L}=0.795atm$

The partial pressure of argon is, 0.795 atm

<u>Part B :</u>

Now we have to calculate the partial pressure of ethane.

As we know that,

Total pressure = Partial pressure of argon + Partial pressure of ethane

1.300 atm = 0.795 atm + Partial pressure of ethane

Partial pressure of ethane = 1.300 - 0.795

Partial pressure of ethane = 0.505 atm

The partial pressure of ethane is, 0.505 atm

3 0

3 years ago

Which of the following concepts underlies all the others: electron affinity, ionization energy, effective nuclear charge, atomic

AlekseyPX

The concept that underlies the other would be :
effective nuclear charge zeff
It control Ip and electron affinity by determining attractive force between atom and the electron
it underlies atomic size because it draws the valence electron closer

hope this helps

6 0

3 years ago

Read 2 more answers

This is a strong, short-lived magnet created by a looped electrical current.

Fed [463]

Answer:elctromagnetic :D

Explanation:

4 0

3 years ago

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think

dangina [55]

Answer:

4.44 rpm

Explanation:

$\omega$ = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = $\frac{3138000}{2}\ m$

R = Radius of arm = 6 m

The acceleration due to gravity is given by

$g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2$

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

$a_c=\omega^2R$

$a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s$

Converting to rpm

$1\ rad/s=\frac{60}{2\pi}\ rpm$

$0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm$

The angular speed of the arm is 4.44 rpm

8 0

3 years ago