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2 years ago

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

Chemistry

2 answers:

ELEN [110]2 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]2 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

B  B

b| Bb Bb

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effect. RbClO(s) → Rb+(aq) + ClO−(aq) HClO(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + ClO−(aq) The degree of dissociatio

Lemur [1.5K]

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:

$RbClO(s)\rightarrow Rb^{+}(aq.)+ClO^{-}(aq.)$

$HClO(aq)+H_2O\rightleftharpoons H_3O^+(aq.)+ClO^{-}(aq.)$

According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.

Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.

The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.

8 0

3 years ago

. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH

Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

Given data :

Volume of water ( V ) = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI + H₂O ⇄ NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = $\sqrt{Ka.C}$ and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺ in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰

= 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

= 0.359 mol / L * 0.25 L = 0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

= 0.08979 * 53.5 g/mol

= 4.80 g ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

8 0

2 years ago

NEED ANSWER FAST!

lorasvet [3.4K]

Answer:

Explanation:

Molarity = 0.010M

Volume = 2.5L

Applying mole-concept,

0.010mole = 1L

X mole = 2.5L

X = (0.010 × 2.5) / 1

X = 0.025moles

0.025moles is present in 2.5L of NaOH solution.

Molar mass of NaOH = (23 + 16 + 1) = 40g/mol

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Mass = 0.025 × 40

Mass = 1g

1g is present in 2.5L of NaOH solution

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3 years ago

Why do some elements (ex: Neon) have a large number of spectral lines, compared to Hydrogen?

Doss [256]

The production of absorption and emission lines . The first theory of the atom to provide an explanation of hydrogen's observed spectral lines those two energy levels, the electron can exist only in certain sharply defined . The number of protons in the nucleus of an atom determines the element that

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2 years ago

DENIUS [597]

1. Homogeneous substance
2. Homogeneous mixture cannot be separated into their respective constituents similar to elements that cannot be separated into simpler substance (not sure)

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3 years ago