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gizmo_the_mogwai [7]
3 years ago
7

List all possible genotypes for a purple flower:

Chemistry
2 answers:
irina [24]3 years ago
5 0
Im not sure if this is right but it may even be RB for red and blue and the person above is correct!
DiKsa [7]3 years ago
4 0
The genotype of the purple-flowered plant is PP
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PLZ HELP ME. This is my second time posting my question because the first time a person put a random answer. If you do not know
kupik [55]

Answer:

PLZ HELP ME. This is my second time posting my question because the first time a person put a random answer. If you do not know the question please do not answer it and leave it to someone else. Thank you, and my question is on the attached image below.

5 0
3 years ago
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An ideal liquid mixture of benzene and toluene (50% by volume each) is used in a plant at 27°C and 1 atm. Determine (a) the mixt
cestrela7 [59]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

7 0
3 years ago
Draw the structure of bromous acid, hbro2. optimize formal charges.
ArbitrLikvidat [17]

STRUCTURE OF BROMOUS ACID: H–O–Br=O

<span>In this structure, all the elements have a formal charge of zero. The formal charge of each element is calculated below: </span><span>

H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
<span>O: 6 – 1/2(4) – 4 = 0</span></span>

8 0
3 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
slava [35]

Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation  

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA  = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

8 0
3 years ago
Consider the following reaction and situations 1 through 10. In the spaces provided, clearly indicate the best response to each
olchik [2.2K]

Answer:

1. C. no change

2. A. increase

3. E. shift to the right

4. A. increase

5. E. shift to the right

6. A. increase

7. F. cannot be determined

8. B increase

9. D. shift to the left

10 F. cannot be determined

Explanation:

<em>According to Le Chaterlier principle, when a reaction is in equilibrium and one of the constraints that affect reactions is applied, the equilibrium will shift so as annul the effects of the constraints.</em>

From the equation: C(s) + H2O(g) ⇌ CO(g) + H2(g),

H is greater than 0, meaning that the system is endothermic, that is energy is absorbed.

1. If the pressure of the system is increased, there would be no change to the system because there are equal number of moles of products and reactants.

2. If H2 concentration is decreased, the equilibrium will shift to the right and more products will be formed. Hence, the concentration of CO will increase.

3. If H2 concentration is decreased, the equilibrium will shift to the right to annul the effects of the decrease in the concentration of a product.

4. If the concentration of H2 is increased, the equilibrium will shift to the left to annul the effects of increased concentration of a product. Hence, more H2O would be formed.

5. If H2 (a product) is removed, and C (a reactant) is added, more of the products will be formed in order to annul the effects of the actions. Hence, equilibrium will shift to the right.

6. If the amount of C (a reactant) is increased, the equilibrium will shift to the right. Hence, more H2 will be formed.

7. The reaction is endothermic, hence an increase in temperature will ordinarily shift the equilibrium to the right. However, the addition of H2 (a product) is supposed to shift the equilibrium to the left. Hence, the effects of simultaneous addition of the two actions become indeterminate.

8. Since the reaction is endothermic, increase in the temperature of the system will shift the equilibrium to the right. Hence, more CO will be formed.

9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

5 0
3 years ago
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