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3 years ago

1. What are the possible genotypes of offspring when crossing BB, Bb parents?

Chemistry

2 answers:

ELEN [110]3 years ago

8 0

_______oBBBBB8o______oBBBBBBB

_____o8BBBBBBBBBBB__BBBBBBBBB8________o88o,

___o8BBBBBB**8BBBB__BBBBBBBBBB_____oBBBBBBBo,

__oBBBBBBB*___***___BBBBBBBBBB_____BBBBBBBBBBo,

_8BBBBBBBBBBooooo___*BBBBBBB8______*BB*_8BBBBBBo,

_8BBBBBBBBBBBBBBBB8ooBBBBBBB8___________8BBBBBBB8,

__*BBBBBBBBBBBBBBBBBBBBBBBBBB8_o88BB88BBBBBBBBBBBB,

____*BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB8,

______**8BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB*,

___________*BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB8*,

____________*BBBBBBBBBBBBBBBBBBBBBBBB8888**,

_____________BBBBBBBBBBBBBBBBBBBBBBB*,

_____________*BBBBBBBBBBBBBBBBBBBBB*,

______________*BBBBBBBBBBBBBBBBBB8,

_______________*BBBBBBBBBBBBBBBB*,

________________8BBBBBBBBBBBBBBB8,

_________________8BBBBBBBBBBBBBBBo,

__________________BBBBBBBBBBBBBBB8,

__________________BBBBBBBBBBBBBBBB,

__________________8BBBBBBBBBBBBBBB8,

__________________*BBBBBBBBBBBBBBBB,

__________________8BBBBBBBBBBBBBBBB8,

_________________oBBBBBBBBBBBBBBBBBB,

________________oBBBBBBBBBBBBBBBBBBB,

________________BBBBBBBBBBBBBBBBBBBB,

_______________8BBBBBBBBBBBBBBBBBBB8,

______________oBBBBBBBBB88BBBBBBBBB8,

______________8BBBBBBBBB*8BBBBBBBBB*,

______________BBBBBBBBB*_BBBBBBBBB8,

______________BBBBBBBB8_oBBBBBBBBB*,

______________8BBBBBBB__oBBBBBBBB*,

______________BBBBBBB*__8BBBBBBB*,

_____________8BBBBBB*___BBBBBBB*,

____________8BBBBBB8___oBBBBBB8,

___________8BBBBBB8____8BBBBBB*,

__________oBBBBBB8____BBBBBBB8,

__________BBBBBBB8___BBBBBBBB*,

_________oBBBBBBB8___BBBBBBBB,

_________8BBBBBB8____BBBBBBB*,

_________BBBBBB*_____8BBBBB*,

________oBBBB8_______BBBBB*,

________oBBB8________BBBB*,

______8BBBB*_______*BBBBBBBB8o,

______BBBBB*____________*88BBBo

marissa [1.9K]3 years ago

4 0

Answer:

1, C: BB, Bb, Bb, BB

2. C: Hybrid

Explanation:

1. If u do a punnet square for BB and Bb you will get: BB, Bb, Bb, Bb

B| B|

B| BB. BB

b| Bb Bb

2. You do a punnet square for BB and bb and you'll get: Bb, Bb, Bb, Bb, which means all the children are hybrids of Dominant alleles ans recessive alleles

<u>B </u> <u>B</u>

b| Bb Bb

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Explanation:

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3 years ago

Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

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3 years ago

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2 years ago

A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t

kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

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