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vesna_86 [32]
3 years ago
14

Axial flowers are a dominant trait in pea plants, and terminal flowers are a recessive trait. Which genotype represents a plant

that has homozygous dominant genes for axial flowers?
aa
Aa
AA
AAaa
Chemistry
2 answers:
kolbaska11 [484]3 years ago
3 0
As the plant is Homozygous dominant, it must contains, two capital "A"s mentioned above.

In short, Your Answer would be Option C

Hope this helps!
Lapatulllka [165]3 years ago
3 0

Answer:

AA

Explanation:

In genetics a dominant trait is represented with a capital letter. And the recessive trait always with lowercase. When we are talking about a specimen being homozygous, it means that it has both equal alleles (AA or aa). And if you talk about it being heterozygous, the two alleles are different (Aa).

In this case the dominant allele would be "A".

And since it is homozygous, it corresponds to both alleles being equal (AA).

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PLEASE HELP ME ASAP! CHEMISTRY TUTOR<br><br> SEE ATTACHED
Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

8 0
3 years ago
02. Using Tables 1 to 3, what pattern do you observe in terms of the phase and number of carbon atoms of the alkanes, alkenes, a
BabaBlast [244]

Hey there mate ;), Im Benjemin and lets solve your question.

★ (Alkanes) : forms single bonds between carbon atoms.

The first four elements are gases and others are liquid in state.

★(Alkenes) : forms double bonds between carbon atoms.

The first three alkenes are gases and rest are liquid.

★ (Alkynes) : forms triple bonds between carbon atoms.

First three are gases and the last one is liquid.

According to boiling point :

The larger structure of the hydrocarbons, the higher the boiling points they have.

In the 3 tables, we can see that the boiling point increases.

5 0
3 years ago
Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

8 0
3 years ago
In the following reaction, what element is gaining mass?
vagabundo [1.1K]

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

8 0
3 years ago
Help me plz, I need help on this.
kirza4 [7]
Omg i lost everything ugh
To do it again

1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol

2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol

3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol

4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol

5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g

6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g

7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g

8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g

I cant believe i had to do this all over

4 0
3 years ago
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