All categories

3 years ago

Black hair (B) is dominant to brown hair (b). Cross a black rabbit and a brown hair. The black rabbit had a brown haired mother.

Chemistry

1 answer:

Fofino [41]3 years ago

7 0

Answer:

Genotypic ratio = 1 Bb: 1 bb

Phenotypic ratio = 1 black rabbit : 1 brown rabbit

Explanation:

This question involves a gene coding for hair color in rabbits. The allele for Black hair (B) is dominant to the allele for brown hair (b).

According to this question, a black rabbit whose mother is brown haired (bb) is crossed with a brown hair rabbit. This means that the black rabbit will have a genotype 'Bb' while the brown rabbit will have a genotype 'bb'. Each parent will produce the following gametes:

Bb - B and b

bb - b and b

Using these gametes in a punnet square (see attached image), the following proportion of offsprings will be produced:

Bb, Bb, bb and bb

This means that the;

Genotypic ratio = 1 Bb: 1 bb

Phenotypic ratio = 1 black rabbit : 1 brown rabbit

You might be interested in

What functional group is aspartic acid

Westkost [7]

The functional group of aspartic acid is -COOH.

8 0

3 years ago

Which is the IUPAC name for NO?

Vanyuwa [196]

Answer:

Nitrogen (ii) oxide

Explanation:

To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.

NOTE: The oxidation number of oxygen (O) is always – 2.

Thus the oxidation number of N in NO can be obtained as follow:

N + O = 0 (ground state)

N + (– 2) = 0

N – 2 = 0

Collect like terms

N = 0 + 2

N = +2

Thus, the oxidation number of Nitrogen (N) in NO is +2.

Therefore, the IUPAC name for NO is Nitrogen (ii) oxide

3 0

3 years ago

Read 2 more answers

To measure the speed of a car, we use miles per hour (miles/hour or mi/h or mph). To measure the rate of a reaction we use molar

arsen [322]

Answer:

Part A: 47.8 mi/h

Part B: 0.072 M/s

Part C: 0.144 M/s

Explanation:

Part A

The average speed or velocity (V) is the variation of the space divided by the variation of the time:

V = (241 - 2)/(8 -3)

V = 47.8 mi/h

Part B

As Part A, the average rate (r) of formation of I2 is the variation of the concentration divided by the variation of time:

r = (1.83 - 1.11)/(15 - 5)

r = 0.072 M/s

Part C

The rates of the substances are proportional of their number of moles (n) which are their coefficient, so:

rI2/nI2 = rHCl/nHCl

0.072/1 = rHCl/2

rHCl = 2*0.072

rHCl = 0.144 M/s

7 0

3 years ago

Which of the following is an example of a model? PV=nRT, Geographical map, I believe that aliens exist, all of the above.

aleksandrvk [35]

Answer: The correct answer is geographical map.

Explanation:

Model is defined as the three dimensional representation of a proposed structure in a smaller scale.

For the given options:

Option 1: PV = nRT

This is a proposed law. This law is known as ideal gas law. A law is defined as the rule which defined a correct procedure. For any gas behaving ideally, this law is used.

Option 2: Geographical map

Map is defined as the diagrammatic representation of an area of land or sea which also shows physical features, roads, cities etc..

A 3-D model can be prepared for geographical map.

Option 3: I believe that aliens exist

This is a hypothetical statement. It is a smart guess by the means of set of assumptions and observations. For its validation, we need to conduct some experiments.

Hence, the correct answer is geographical map.

3 0

3 years ago

Read 2 more answers

Two liquids are analyzed and found to both be 85.7% carbon and 14.3% hydrogen. At 750 mmHg and 150 C, both are gases. At these c

vodomira [7]

Answer:

Molecular formula A: C₅H₁₀

Molecular formula B: C₇H₁₄

Explanation:

It is possible to obtain empirical formula of compounds using percent composition, thus:

C: 85.7% × (1mol / 12.01g) = 7.136 moles C

H: 14.3% × (1mol / 1.01g) = 14.158 moles H

Mole ratio of H:C is:

14.158mol / 7.136mol = 2

That means in compounds A and B you have 2 hydrogens per atom of carbon and empirical formila is:

CH₂

Using PV = nRT, moles of A and B are:

Where P is pressure (750mmHg / 760 = 0.987atm), V is volume (0.8000L), R is gas constant (0.082atmL/molK), and T is temperature (150°C +273.15 = 423.15K)

Moles A and B: n = PV / RT

n = 0.987atm×0.8000L / 0.082atmL/molK×423.15K

n = 0.0228 moles of A and B.

Using the mass of A and B it is possible to find molar mass of each compound:

A = 1.60g / 0.0228mol = 70.31g/mol

B = 2.22g / 0.0228mol = 97.56g/mol

As empirical formula of both compounds is CH₂, (molar mass = 14.03g/mol). Molecular formula of compounds is:

A = 70.31g/mol / 14.03g/mol = 5 → Molecular formula: 5×CH₂ = C₅H₁₀

B = 97.56g/mol / 14.03g/mol = 7 → Molecular formula: 7×CH₂ = C₇H₁₄

8 0

3 years ago