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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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Mg(s) + HCl (l) --> H2 (g) +MgCl2 (s) A student conducted an experiment with 2g of magnesium and 3 mL of 6M hydrochloric acid

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3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

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Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

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3 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

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Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

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