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morpeh [17]
3 years ago
10

A recessive gene is described as one that

Chemistry
1 answer:
djyliett [7]3 years ago
5 0

Answer:

<h2>C) may be hidden </h2>

Explanation:

A gene is a unit of heredity which is transferred from parents to offspring and it control the traits.

Alternative form of gene are known as alleles.

Gene may be dominant or recessive

Dominant gene expresses in both homozygous as well as in heterozygous condition, while recessive gene expresses in homozygous condition only, when it present with dominant gene, its effect is hidden by dominant gene. Dominant gene is represented by capital letter while recessive gene is represented by small letter.

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Density is the ratio of a sample's mass to its volume. A bar of lead has a mass of 115.2 g. When it is submerged in 25.0 mL of w
Margarita [4]

Answer:

$10.97 \ g/cm^3$

Explanation:

Given :

Mass of a bar of lead = 115.2 g

Initial water level $\text{in the graduated cylinder}$ = 25 mL

Final water level $\text{in the graduated cylinder}$ = 35.5 mL

Difference in the water level = 35.5 - 25

                                               = 10.5 mL

                                               = 10.5 \ cm^3

We know that when a body is submerged in water, it displaces its own volume of water.

Therefore, the volume of the lead bar = volume of the water displaced = 10.5 mL  = 10.5 \ cm^3

We know that mathematically, density is the ratio of mass of body to its volume.

Density of the lead bar is given by :

$\rho =\frac{\text{mass}}{\text{volume}}$

$\rho =\frac{\text{115.2 g}}{\text{10.5 cm}^3}$

  = $10.97 \ g/cm^3$

3 0
3 years ago
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
Silver has two naturally isotopes and has an atomic mass of 107.868 amu. One isotope is Ag-109 isotope (108.905 amu) and has a n
Triss [41]

Answer:  106.905

Explanation:  If there are only 2 isotopes, and 1 of them is 48.16%, the second must, by default, be (100 - 48.16%) = 51.84%  The final, averaged, atomic mass is 107.868.  This is made up of each isotope's atomic mass times the percentage of that isotope in the total sample.  The weighted value of the known isotope (109) plus that of the unknown must come to the observed value of 107.868 amu.  (107.868 - 52.45 = 55.42).  Divide that by the % for that isotope (55.42/0.5184) = 106.90 amu for the second isotope.

<u>Atomic Mass</u>  <u>% of Sample</u> <u>Weighted Value</u>

    108.905         48.16%              52.45

          X               51.84%              <u>55.42</u>

                                                     107.87

      X = (55.42/0.5184) = 106.90 amu

5 0
3 years ago
In what beaker does the sugar dissolve the fastest? ( particle size, temp. of water, condition(stir/no stir), etc).
Vsevolod [243]
Temperature. In case you have ever baked, you can relate to this. For example, when you make caramel, you simply put the caramel on a boiling hot pan and sit back and watch it. It always goes fastest this way.
7 0
3 years ago
phosphorus-32 ahs a half-life of 14.0 days. Starting with 8.00 g of 32 P, how many grams will remain after 56.0 days
sertanlavr [38]
0.500 grams would be left after 56.0 days.

After 14 days=4.00 g
After 28 days=2.00 g
After 42 days=1.00 g
After 56 days=0.500 g

This is after four half lives.
3 0
3 years ago
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